7 Second-Order Differential Equations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/7-second-order-differential-equations

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Introduction

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/7-introduction

Chapter Outline

• 7.1 Second-Order Linear Equations
• 7.2 Nonhomogeneous Linear Equations
• 7.3 Applications
• 7.4 Series Solutions of Differential Equations

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7.1 Second-Order Linear Equations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/7-1-second-order-linear-equations

7.1 Second-Order Linear Equations

• 7.1.1 Recognize homogeneous and nonhomogeneous linear differential equations.
• 7.1.2 Determine the characteristic equation of a homogeneous linear equation.
• 7.1.3 Use the roots of the characteristic equation to find the solution to a homogeneous linear equation.
• 7.1.4 Solve initial-value and boundary-value problems involving linear differential equations.

When working with differential equations, usually the goal is to find a solution. In other words, we want to find a function (or functions) that satisfies the differential equation. The technique we use to find these solutions varies, depending on the form of the differential equation with which we are working. Second-order differential equations have several important characteristics that can help us determine which solution method to use. In this section, we examine some of these characteristics and the associated terminology.

Homogeneous Linear Equations

Consider the second-order differential equation

$$xy^{''} + 2x^{2}y^{\prime} + 5x^{3}y = 0.$$

Notice that *y* and its derivatives appear in a relatively simple form. They are multiplied by functions of *x*, but are not raised to any powers themselves, nor are they multiplied together. As discussed in Introduction to Differential Equations, first-order equations with similar characteristics are said to be linear. The same is true of second-order equations. Also note that all the terms in this differential equation involve either *y* or one of its derivatives. There are no terms involving only functions of *x*. Equations like this, in which every term contains *y* or one of its derivatives, are called homogeneous.

Not all differential equations are homogeneous. Consider the differential equation

$$xy^{''} + 2x^{2}y^{\prime} + 5x^{3}y = x^{2}.$$

The $x^{2}$ term on the right side of the equal sign does not contain *y* or any of its derivatives. Therefore, this differential equation is nonhomogeneous.

A second-order differential equation is linear if it can be written in the form

$$a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x),$$ (7.1)

where $a_{2}(x),$ $a_{1}(x),$ $a_{0}(x),$ and $r(x)$ are real-valued functions and $a_{2}(x)$ is not identically zero. If $r(x) \equiv 0$—in other words, if $r(x) = 0$ for every value of *x*—the equation is said to be a homogeneous linear equation. If $r(x) \neq 0$ for some value of $x,$ the equation is said to be a nonhomogeneous linear equation.

Visit this website to study more about second-order linear differential equations.

In linear differential equations, $y$ and its derivatives can be raised only to the first power and they may not be multiplied by one another. Terms involving $y^{2}$ or $\sqrt{y^{\prime}}$ make the equation nonlinear. Functions of $y$ and its derivatives, such as $\text{sin}\mspace{2mu} y$ or $e^{y^{\prime}},$ are similarly prohibited in linear differential equations.

Note that equations may not always be given in standard form (the form shown in the definition). It can be helpful to rewrite them in that form to decide whether they are linear, or whether a linear equation is homogeneous.

Classifying Second-Order Equations

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.

1. $y^{''} + 3x^{4}y^{\prime} + x^{2}y^{2} = x^{3}$

2. $(\text{sin}\mspace{2mu} x)y^{''} + (\text{cos}\mspace{2mu} x)y^{\prime} + 3y = 0$

3. $4t^{2}x^{''} + 3txx^{\prime} + 4x = 0$

4. $5y^{''} + y = 4x^{5}$

5. $(\text{cos}\mspace{2mu} x)y^{''} - \text{sin}\mspace{2mu} y^{\prime} + (\text{sin}\mspace{2mu} x)y - \text{cos}\mspace{2mu} x = 0$

6. $8ty^{''} - 6t^{2}y^{\prime} + 4ty - 3t^{2} = 0$

7. $\text{sin}(x^{2})y^{''} - (\text{cos}\mspace{2mu} x)y^{\prime} + x^{2}y = y^{\prime} - 3$

8. $y^{''} + 5xy^{\prime} - 3y = \text{cos}\mspace{2mu} y$

Solution

1. This equation is nonlinear because of the $y^{2}$ term.

2. This equation is linear. There is no term involving a power or function of $y,$ and the coefficients are all functions of $x\text{.}$ The equation is already written in standard form, and $r(x)$ is identically zero, so the equation is homogeneous.

3. This equation is nonlinear. Note that, in this case, *x* is the dependent variable and *t* is the independent variable. The second term involves the product of $x$ and $x^{\prime},$ so the equation is nonlinear.

4. This equation is linear. Since $r(x) = 4x^{5},$ the equation is nonhomogeneous.

5. This equation is nonlinear, because of the $\text{sin}\mspace{2mu} y^{\prime}$ term.

6. This equation is linear. Rewriting it in standard form gives

$$8t^{2}y^{''} - 6t^{2}y^{\prime} + 4ty = 3t^{2}.$$

With the equation in standard form, we can see that $r(t) = 3t^{2},$ so the equation is nonhomogeneous.

7. This equation looks like it’s linear, but we should rewrite it in standard form to be sure. We get

$$\text{sin}(x^{2})y^{''} - (\text{cos}\mspace{2mu} x + 1)y^{\prime} + x^{2}y = -3.$$

This equation is, indeed, linear. With $r(x) = -3,$ it is nonhomogeneous.

8. This equation is nonlinear because of the $\text{cos}\mspace{2mu} y$ term.

Visit this website that discusses second-order differential equations.

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.

1. ${(y^{''})}^{2} - y^{\prime} + 8x^{3}y = 0$

2. $(\text{sin}\mspace{2mu} t)y^{''} + \text{cos}\mspace{2mu} t - 3ty^{\prime} = 0$

Later in this section, we will see some techniques for solving specific types of differential equations. Before we get to that, however, let’s get a feel for how solutions to linear differential equations behave. In many cases, solving differential equations depends on making educated guesses about what the solution might look like. Knowing how various types of solutions behave will be helpful.

Verifying a Solution

Consider the linear, homogeneous differential equation

$$x^{2}y^{''} - xy^{\prime} - 3y = 0.$$

Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of $x$ associated with higher-order derivatives of $y\text{.}$ Show that $y = x^{3}$ is a solution to this differential equation.

Solution

Let $y = x^{3}.$ Then $y^{\prime} = 3x^{2}$ and $y^{''} = 6x\text{.}$ Substituting into the differential equation, we see that

$$\begin{array}{cl}

{x^{2}y^{''} - xy\prime - 3y} & {= x^{2}\left( {6x} \right) - x\left( {3x^{2}} \right) - 3\left( x^{3} \right)} \\

& {= 6x^{3} - 3x^{3} - 3x^{3}} \\

& {= 0.}

\end{array}$$

Show that $y = 2x^{2}$ is a solution to the differential equation

$$\frac{1}{2}x^{2}y^{''} - xy^{\prime} + y = 0.$$

Although simply finding any solution to a differential equation is important, mathematicians and engineers often want to go beyond finding *one* solution to a differential equation to finding *all* solutions to a differential equation. In other words, we want to find a general solution. Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation. An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact as the following theorem.

Superposition Principle

If $y_{1}(x)$ and $y_{2}(x)$ are solutions to a linear homogeneous differential equation, then the function

$$y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x),$$

where $c_{1}$ and $c_{2}$ are constants, is also a solution.

The proof of this superposition principle theorem is left as an exercise.

Verifying the Superposition Principle

Consider the differential equation

$$y^{''} - 4y^{\prime} - 5y = 0.$$

Given that $e^{\text{−}x}$ and $e^{5x}$ are solutions to this differential equation, show that $4e^{\text{−}x} + e^{5x}$ is a solution.

Solution

We have

$$y(x) = 4e^{\text{−}x} + e^{5x},\ \text{so}\ y^{\prime}(x) = -4e^{\text{−}x} + 5e^{5x}\ \text{and}\ y^{''}(x) = 4e^{\text{−}x} + 25e^{5x}.$$

Then

$$\begin{array}{cl}

{y^{''} - 4y^{\prime} - 5y} & {= \left( {4e^{\text{−}x} + 25e^{5x}} \right) - 4\left( {-4e^{\text{−}x} + 5e^{5x}} \right) - 5\left( {4e^{\text{−}x} + e^{5x}} \right)} \\

& {= 4e^{\text{−}x} + 25e^{5x} + 16e^{\text{−}x} - 20e^{5x} - 20e^{\text{−}x} - 5e^{5x}} \\

& {= 0.}

\end{array}$$

Thus, $y(x) = 4e^{\text{−}x} + e^{5x}$ is a solution.

Consider the differential equation

$$y^{''} + 5y^{\prime} + 6y = 0.$$

Given that $e^{-2x}$ and $e^{-3x}$ are solutions to this differential equation, show that $3e^{-2x} + 6e^{-3x}$ is a solution.

Unfortunately, to find the general solution to a second-order differential equation, it is not enough to find any two solutions and then combine them. Consider the differential equation

$$x^{''} + 7x^{\prime} + 12x = 0.$$

Both $e^{-3t}$ and $2e^{-3t}$ are solutions (check this). However, $x(t) = c_{1}e^{-3t} + c_{2}\left( {2e^{-3t}} \right)$ is *not* the general solution. This expression does not account for all solutions to the differential equation. In particular, it fails to account for the function $e^{-4t},$ which is also a solution to the differential equation.

It turns out that to find the general solution to a second-order differential equation, we must find two linearly independent solutions. We define that terminology here.

A set of functions $f_{1}(x),f_{2}(x\text{),…,}f_{n}(x)$ is said to be linearly dependent if there are constants $c_{1},c_{2}\text{,…}c_{n},$ not all zero, such that $c_{1}f_{1}(x) + c_{2}f_{2}(x) + \text{⋯} + c_{n}f_{n}(x) = 0$ for all *x* over the interval of interest. A set of functions that is not linearly dependent is said to be linearly independent.

In this chapter, we usually test sets of only two functions for linear independence, which allows us to simplify this definition. From a practical perspective, we see that two functions are linearly dependent if either one of them is identically zero or if they are constant multiples of each other.

First we show that if the functions meet the conditions given previously, then they are linearly dependent. If one of the functions is identically zero—say, $f_{2}(x) \equiv 0$—then choose $c_{1} = 0$ and $c_{2} = 1,$ and the condition for linear dependence is satisfied. If, on the other hand, neither $f_{1}(x)$ nor $f_{2}(x)$ is identically zero, but $f_{1}(x) = Cf_{2}(x)$ for some constant $C,$ then choose $c_{1} = \frac{1}{C}$ and $c_{2} = -1,$ and again, the condition is satisfied.

Next, we show that if two functions are linearly dependent, then either one is identically zero or they are constant multiples of one another. Assume $f_{1}(x)$ and $f_{2}(x)$ are linearly independent. Then, there are constants, $c_{1}$ and $c_{2},$ not both zero, such that

$$c_{1}f_{1}(x) + c_{2}f_{2}(x) = 0$$

for all *x* over the interval of interest. Then,

$$c_{1}f_{1}(x) = \text{−}c_{2}f_{2}(x).$$

Now, since we stated that $c_{1}$ and $c_{2}$ can’t both be zero, assume $c_{2} \neq 0.$ Then, there are two cases: either $c_{1} = 0$ or $c_{1} \neq 0.$ If $c_{1} = 0,$ then

$$\begin{array}{l}

{0 = \text{−}c_{2}f_{2}(x)} \\

{0 = f_{2}(x),}

\end{array}$$

so one of the functions is identically zero. Now suppose $c_{1} \neq 0.$ Then,

$$f_{1}(x) = \left( {- \frac{c_{2}}{c_{1}}} \right)f_{2}(x)$$

and we see that the functions are constant multiples of one another.

Linear Dependence of Two Functions

Two functions, $f_{1}(x)$ and $f_{2}(x),$ are said to be linearly dependent if either one of them is identically zero or if $f_{1}(x) = Cf_{2}(x)$ for some constant *C* and for all *x* over the interval of interest. Functions that are not linearly dependent are said to be *linearly independent*.

Testing for Linear Dependence

Determine whether the following pairs of functions are linearly dependent or linearly independent.

1. $f_{1}(x) = x^{2},$ $f_{2}(x) = 5x^{2}$

2. $f_{1}(x) = \text{sin}\mspace{2mu} x,$ $f_{2}(x) = \text{cos}\mspace{2mu} x$

3. $f_{1}(x) = e^{3x},$ $f_{2}(x) = e^{-3x}$

4. $f_{1}(x) = 3x,$ $f_{2}(x) = 3x + 1$

Solution

1. $f_{2}(x) = 5f_{1}(x),$ so the functions are linearly dependent.

2. There is no constant *C* such that $f_{1}(x) = Cf_{2}(x),$ so the functions are linearly independent.

3. There is no constant *C* such that $f_{1}(x) = Cf_{2}(x),$ so the functions are linearly independent. Don’t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.

4. There is no constant *C* such that $f_{1}(x) = Cf_{2}(x),$ so the functions are linearly independent.

Determine whether the following pairs of functions are linearly dependent or linearly independent: $f_{1}(x) = e^{x},$ $f_{2}(x) = 3e^{3x}.$

If we are able to find two linearly independent solutions to a second-order differential equation, then we can combine them to find the general solution. This result is formally stated in the following theorem.

General Solution to a Homogeneous Equation

If $y_{1}(x)$ and $y_{2}(x)$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by

$$y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x),$$

where $c_{1}$ and $c_{2}$ are constants.

When we say a family of functions is the general solution to a differential equation, we mean that (1) every expression of that form is a solution and (2) every solution to the differential equation can be written in that form, which makes this theorem extremely powerful. If we can find two linearly independent solutions to a differential equation, we have, effectively, found *all* solutions to the differential equation—quite a remarkable statement. The proof of this theorem is beyond the scope of this text.

Writing the General Solution

If $y_{1}(t) = e^{3t}$ and $y_{2}(t) = e^{-3t}$ are solutions to $y^{''} - 9y = 0,$ what is the general solution?

Solution

Note that $y_{1}$ and $y_{2}$ are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is $y(t) = c_{1}e^{3t} + c_{2}e^{-3t}.$

If $y_{1}(x) = e^{3x}$ and $y_{2}(x) = xe^{3x}$ are solutions to $y^{''} - 6y^{\prime} + 9y = 0,$ what is the general solution?

Second-Order Equations with Constant Coefficients

Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form

$$ay^{''} + by^{\prime} + cy = 0,$$ (7.2)

where $a,$ $b,$ and $c$ are constants.

Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let’s see what happens when we try a solution of the form $y(x) = e^{\lambda x},$ where $\lambda$ (the lowercase Greek letter lambda) is some constant.

If $y(x) = e^{\lambda x},$ then $y^{\prime}(x) = \lambda e^{\lambda x}$ and $y^{''} = \lambda^{2}e^{\lambda x}.$ Substituting these expressions into Equation 7.1, we get

$$\begin{array}{cl}

{ay^{''} + by^{\prime} + cy} & {= a(\lambda^{2}e^{\lambda x}) + b(\lambda e^{\lambda x}) + ce^{\lambda x}} \\

& {= e^{\lambda x}(a\lambda^{2} + b\lambda + c).}

\end{array}$$

Since $e^{\lambda x}$ is never zero, this expression can be equal to zero for all *x* only if

$$a\lambda^{2} + b\lambda + c = 0.$$

We call this the characteristic equation of the differential equation.

The characteristic equation of the differential equation $ay^{''} + by^{\prime} + cy = 0$ is $a\lambda^{2} + b\lambda + c = 0.$

The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula

$$\lambda = \frac{\text{−}b \pm \sqrt{b^{2} - 4ac}}{2a}.$$

This gives three cases. The characteristic equation has (1) distinct real roots; (2) a single, repeated real root; or (3) complex conjugate roots. We consider each of these cases separately.

Distinct Real Roots

If the characteristic equation has distinct real roots $\lambda_{1}$ and $\lambda_{2},$ then $e^{\lambda_{1}x}$ and $e^{\lambda_{2}x}$ are linearly independent solutions to Example 7.1, and the general solution is given by

$$y(x) = c_{1}e^{\lambda_{1}x} + c_{2}e^{\lambda_{2}x},$$

where $c_{1}$ and $c_{2}$ are constants.

For example, the differential equation $y^{''} + 9y^{\prime} + 14y = 0$ has the associated characteristic equation $\lambda^{2} + 9\lambda + 14 = 0.$ This factors into $\left( {\lambda + 2} \right)\left( {\lambda + 7} \right) = 0,$ which has roots $\lambda_{1} = -2$ and $\lambda_{2} = -7.$ Therefore, the general solution to this differential equation is

$$y(x) = c_{1}e^{-2x} + c_{2}e^{-7x}.$$

Single Repeated Real Root

Things are a little more complicated if the characteristic equation has a repeated real root, $\lambda\text{.}$ In this case, we know $e^{\lambda x}$ is a solution to Equation 7.1, but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form $ke^{\lambda x},$ where *k* is some constant, but it would not be linearly independent of $e^{\lambda x}.$ Therefore, let’s try $xe^{\lambda x}$ as the second solution. First, note that by the quadratic formula,

$$\lambda = \frac{\text{−}b \pm \sqrt{b^{2} - 4ac}}{2a}.$$

But, $\lambda$ is a repeated root, so $b^{2} - 4ac = 0$ and $\lambda = \frac{\text{−}b}{2a}.$ Thus, if $y = xe^{\lambda x},$ we have

$$y^{\prime} = e^{\lambda x} + \lambda xe^{\lambda x}\ \text{and}\ y^{''} = 2\lambda e^{\lambda x} + \lambda^{2}xe^{\lambda x}.$$

Substituting these expressions into Equation 7.1, we see that

$$\begin{array}{cl}

{ay^{''} + by^{\prime} + cy} & {= a(2\lambda e^{\lambda x} + \lambda^{2}xe^{\lambda x}) + b(e^{\lambda x} + \lambda xe^{\lambda x}) + cxe^{\lambda x}} \\

& {= xe^{\lambda x}(a\lambda^{2} + b\lambda + c) + e^{\lambda x}(2a\lambda + b)} \\

& {= xe^{\lambda x}(0) + e^{\lambda x}\left( {2a\left( \frac{\text{−}b}{2a} \right) + b} \right)} \\

& {= 0 + e^{\lambda x}(0)} \\

& {= 0.}

\end{array}$$

This shows that $xe^{\lambda x}$ is a solution to Equation 7.1. Since $e^{\lambda x}$ and $xe^{\lambda x}$ are linearly independent, when the characteristic equation has a repeated root $\lambda,$ the general solution to Equation 7.1 is given by

$$y(x) = c_{1}e^{\lambda x} + c_{2}xe^{\lambda x},$$

where $c_{1}$ and $c_{2}$ are constants.

For example, the differential equation $y^{''} + 12y^{\prime} + 36y = 0$ has the associated characteristic equation $\lambda^{2} + 12\lambda + 36 = 0.$ This factors into $\left( {\lambda + 6} \right)^{2} = 0,$ which has a repeated root $\lambda = -6.$ Therefore, the general solution to this differential equation is

$$y(x) = c_{1}e^{-6x} + c_{2}xe^{-6x}.$$

Complex Conjugate Roots

The third case we must consider is when $b^{2} - 4ac < 0.$ In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number $i = \sqrt{-1}$ to find the roots, which take the form $\lambda_{1} = \alpha + \beta i$ and $\lambda_{2} = \alpha - \beta i\text{.}$ The complex number $\alpha + \beta i$ is called the *conjugate* of $\alpha - \beta i\text{.}$ Thus, we see that when $b^{2} - 4ac < 0,$ the roots of our characteristic equation are always complex conjugates.

This creates a little bit of a problem for us. If we follow the same process we used for distinct real roots—using the roots of the characteristic equation as the coefficients in the exponents of exponential functions—we get the functions $e^{(\alpha + \beta i)x}$ and $e^{(\alpha - \beta i)x}$ as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-value functions, in this course we do not address the idea of a derivative for such functions. So, if possible, we’d like to find two linearly independent *real-value* solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions $e^{(\alpha + \beta i)x}$ and $e^{(\alpha - \beta i)x}$ as if they were real-value functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-value functions do not follow the same rules as real-value functions. Those of you interested in a more in-depth discussion of complex-value functions should consult a complex analysis text.

Based on the roots $\alpha \pm \beta i$ of the characteristic equation, the functions $e^{(\alpha + \beta i)x}$ and $e^{(\alpha - \beta i)x}$ are linearly independent solutions to the differential equation. and the general solution is given by

$$y(x) = c_{1}e^{(\alpha + \beta i)x} + c_{2}e^{(\alpha - \beta i)x}.$$

Using some smart choices for $c_{1}$ and $c_{2},$ and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to Equation 7.1 and express our general solution in those terms.

We encountered exponential functions with complex exponents earlier. One of the key tools we used to express these exponential functions in terms of sines and cosines was Euler’s formula, which tells us that

$$e^{i\theta} = \text{cos}\mspace{2mu}\theta + i\mspace{2mu}\text{sin}\mspace{2mu}\theta$$

for all real numbers $\theta\text{.}$

Going back to the general solution, we have

$$\begin{array}{cl}

{y(x)} & {= c_{1}e^{(\alpha + \beta i)x} + c_{2}e^{(\alpha - \beta i)x}} \\

& {= c_{1}e^{\alpha x}e^{\beta ix} + c_{2}e^{\alpha x}e^{\text{−}\beta ix}} \\

& {= e^{\alpha x}\left( {c_{1}e^{\beta ix} + c_{2}e^{\text{−}\beta ix}} \right).}

\end{array}$$

Applying Euler’s formula together with the identities $\text{cos}(\text{−}x) = \text{cos}\mspace{2mu} x$ and $\text{sin}(\text{−}x) = \text{−}\text{sin}\mspace{2mu} x,$ we get

$$\begin{array}{cl}

{y(x)} & {= e^{\alpha x}\left\lbrack {c_{1}\left( {\text{cos}\mspace{2mu}\beta x + i\mspace{2mu}\text{sin}\mspace{2mu}\beta x} \right) + c_{2}\left( {\text{cos}(\text{−}\beta x) + i\mspace{2mu}\text{sin}(\text{−}\beta x)} \right)} \right\rbrack} \\

& {= e^{\alpha x}\left\lbrack {(c_{1} + c_{2})\text{cos}\mspace{2mu}\beta x + (c_{1} - c_{2})i\mspace{2mu}\text{sin}\mspace{2mu}\beta x} \right\rbrack.}

\end{array}$$

Now, if we choose $c_{1} = c_{2} = \frac{1}{2},$ the second term is zero and we get

$$y(x) = e^{\alpha x}\text{cos}\mspace{2mu}\beta x$$

as a real-value solution to Equation 7.1. Similarly, if we choose $c_{1} = \text{−}\frac{i}{2}$ and $c_{2} = \frac{i}{2},$ the first term is zero and we get

$$y(x) = e^{\alpha x}\text{sin}\mspace{2mu}\beta x$$

as a second, linearly independent, real-value solution to Equation 7.1.

Based on this, we see that if the characteristic equation has complex conjugate roots $\alpha \pm \beta i,$ then the general solution to Equation 7.1 is given by

$$\begin{array}{cl}

{y(x)} & {= c_{1}e^{\alpha x}\text{cos}\mspace{2mu}\beta x + c_{2}e^{\alpha x}\text{sin}\mspace{2mu}\beta x} \\

& {= e^{\alpha x}\left( {c_{1}\text{cos}\mspace{2mu}\beta x + c_{2}\text{sin}\mspace{2mu}\beta x} \right),}

\end{array}$$

where $c_{1}$ and $c_{2}$ are constants.

For example, the differential equation $y^{''} - 2y^{\prime} + 5y = 0$ has the associated characteristic equation $\lambda^{2} - 2\lambda + 5 = 0.$ By the quadratic formula, the roots of the characteristic equation are $1 \pm 2i\text{.}$ Therefore, the general solution to this differential equation is

$$y(x) = e^{x}\left( {c_{1}\text{cos}\mspace{2mu} 2x + c_{2}\text{sin}\mspace{2mu} 2x} \right).$$

Summary of Results

We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots. The three cases are summarized in Table 7.1.

| Characteristic Equation Roots | General Solution to the Differential Equation |

|------------------------------------------------------|---------------------------------------------------------------------------------------------------------------|

| Distinct real roots, $\lambda_{1}$ and $\lambda_{2}$ | $y(x) = c_{1}e^{\lambda_{1}x} + c_{2}e^{\lambda_{2}x}$ |

| A repeated real root, $\lambda$ | $y(x) = c_{1}e^{\lambda x} + c_{2}xe^{\lambda x}$ |

| Complex conjugate roots $\alpha \pm \beta i$ | $y(x) = e^{\alpha x}\left( {c_{1}\text{cos}\mspace{2mu}\beta x + c_{2}\text{sin}\mspace{2mu}\beta x} \right)$ |

Table 7.1 Summary of Characteristic Equation Cases

Using the Characteristic Equation to Solve Second-Order Differential Equations with Constant Coefficients

1. Write the differential equation in the form $ay^{''} + by^{\prime} + cy = 0.$

2. Find the corresponding characteristic equation $a\lambda^{2} + b\lambda + c = 0.$

3. Either factor the characteristic equation or use the quadratic formula to find the roots.

4. Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.

Solving Second-Order Equations with Constant Coefficients

Find the general solution to the following differential equations. Give your answers as functions of *x*.

1. $y^{''} + 3y^{\prime} - 4y = 0$

2. $y^{''} + 6y^{\prime} + 13y = 0$

3. $y^{''} + 2y^{\prime} + y = 0$

4. $y^{''} - 5y^{\prime} = 0$

5. $y^{''} - 16y = 0$

6. $y^{''} + 16y = 0$

Solution

Note that all these equations are already given in standard form (step 1).

1. The characteristic equation is $\lambda^{2} + 3\lambda - 4 = 0$ (step 2). This factors into $\left( {\lambda + 4} \right)\left( {\lambda - 1} \right) = 0,$ so the roots of the characteristic equation are $\lambda_{1} = -4$ and $\lambda_{2} = 1$ (step 3). Then the general solution to the differential equation is

$$y(x) = c_{1}e^{-4x} + c_{2}e^{x}\ \text{(step 4)}\text{.}$$

2. The characteristic equation is $\lambda^{2} + 6\lambda + 13 = 0$ (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots $-3 \pm 2i$ (step 3). Then the general solution to the differential equation is

$$y(t) = e^{-3t}\left( {c_{1}\text{cos}\mspace{2mu} 2t + c_{2}\text{sin}\mspace{2mu} 2t} \right)\ \text{(step 4)}\text{.}$$

3. The characteristic equation is $\lambda^{2} + 2\lambda + 1 = 0$ (step 2). This factors into $\left( {\lambda + 1} \right)^{2} = 0,$ so the characteristic equation has a repeated real root $\lambda = -1$ (step 3). Then the general solution to the differential equation is

$$y(t) = c_{1}e^{\text{−}t} + c_{2}te^{\text{−}t}\ \text{(step 4).}$$

4. The characteristic equation is $\lambda^{2} - 5\lambda$ (step 2). This factors into $\lambda\left( {\lambda - 5} \right) = 0,$ so the roots of the characteristic equation are $\lambda_{1} = 0$ and $\lambda_{2} = 5$ (step 3). Note that $e^{0x} = e^{0} = 1,$ so our first solution is just a constant. Then the general solution to the differential equation is

$$y(x) = c_{1} + c_{2}e^{5x}\ \text{(step 4)}\text{.}$$

5. The characteristic equation is $\lambda^{2} - 16 = 0$ (step 2). This factors into $(\lambda + 4)(\lambda - 4) = 0,$ so the roots of the characteristic equation are $\lambda_{1} = 4$ and $\lambda_{2} = -4$ (step 3). Then the general solution to the differential equation is

$$y(x) = c_{1}e^{4x} + c_{2}e^{-4x}\ \text{(step 4)}\text{.}$$

6. The characteristic equation is $\lambda^{2} + 16 = 0$ (step 2). This has complex conjugate roots $\pm 4i$ (step 3). Note that $e^{0x} = e^{0} = 1,$ so the exponential term in our solution is just a constant. Then the general solution to the differential equation is

$$y(t) = c_{1}\text{cos}\mspace{2mu} 4t + c_{2}\text{sin}\mspace{2mu} 4t\ \text{(step 4)}\text{.}$$

Find the general solution to the following differential equations:

1. $y^{''} - 2y^{\prime} + 10y = 0$

2. $y^{''} + 14y^{\prime} + 49y = 0$

Initial-Value Problems and Boundary-Value Problems

So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time. For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time $t = t_{0}.$ This means the system is at equilibrium, so $y(t_{0}) = 0,$ and the compression of the shock absorber is not changing, so $y^{\prime}(t_{0}) = 0.$ With these two initial conditions and the general solution to the differential equation, we can find the *specific* solution to the differential equation that satisfies both initial conditions. This process is known as *solving an initial-value problem*. (Recall that we discussed initial-value problems in Introduction to Differential Equations.) Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.

Sometimes we know the condition of the system at two different times. For example, we might know $y(t_{0}) = y_{0}$ and $y(t_{1}) = y_{1}.$ These conditions are called boundary conditions, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem.

Mathematicians, scientists, and engineers are interested in understanding the conditions under which an initial-value problem or a boundary-value problem has a unique solution. Although a complete treatment of this topic is beyond the scope of this text, it is useful to know that, within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.

Solving an Initial-Value Problem

Solve the following initial-value problem: $y^{''} + 3y^{\prime} - 4y = 0,$ $y(0) = 1,$ $y^{\prime}(0) = -9.$

Solution

We already solved this differential equation in Example 7.6a. and found the general solution to be

$$y(x) = c_{1}e^{-4x} + c_{2}e^{x}.$$

Then

$$y^{\prime}(x) = -4c_{1}e^{-4x} + c_{2}e^{x}.$$

When $x = 0,$ we have $y(0) = c_{1} + c_{2}$ and $y^{\prime}(0) = -4c_{1} + c_{2}.$ Applying the initial conditions, we have

$$\begin{array}{rll}

{c_{1} + c_{2}} & = & 1 \\

{-4c_{1} + c_{2}} & = & {-9.}

\end{array}$$

Then $c_{1} = 1 - c_{2}.$ Substituting this expression into the second equation, we see that

$$\begin{array}{rll}

{-4(1 - c_{2}) + c_{2}} & = & -9 \\

{-4 + 4c_{2} + c_{2}} & = & -9 \\

{5c_{2}} & = & -5 \\

c_{2} & = & {-1.}

\end{array}$$

So, $c_{1} = 2$ and the solution to the initial-value problem is

$$y(x) = 2e^{-4x} - e^{x}.$$

Solve the initial-value problem $y^{''} - 3y^{\prime} - 10y = 0,$ $y(0) = 0,$ $y^{\prime}(0) = 7.$

Solving an Initial-Value Problem and Graphing the Solution

Solve the following initial-value problem and graph the solution:

$$y^{''} + 6y^{\prime} + 13y = 0,\ y(0) = 0,\ y^{\prime}(0) = 2$$

Solution

We already solved this differential equation in Example 7.6b. and found the general solution to be

$$y(x) = e^{-3x}\left( {c_{1}\text{cos}\mspace{2mu} 2x + c_{2}\text{sin}\mspace{2mu} 2x} \right).$$

Then

$$y^{\prime}(x) = e^{-3x}\left( {-2c_{1}\text{sin}\mspace{2mu} 2x + 2c_{2}\text{cos}\mspace{2mu} 2x} \right) - 3e^{-3x}\left( {c_{1}\text{cos}\mspace{2mu} 2x + c_{2}\text{sin}\mspace{2mu} 2x} \right).$$

When $x = 0,$ we have $y(0) = c_{1}$ and $y^{\prime}(0) = 2c_{2} - 3c_{1}.$ Applying the initial conditions, we obtain

$$\begin{array}{rll}

c_{1} & = & 0 \\

{-3c_{1} + 2c_{2}} & = & {2.}

\end{array}$$

Therefore, $c_{1} = 0,$ $c_{2} = 1,$ and the solution to the initial value problem is shown in the following graph.

$$y = e^{-3x}\text{sin}\mspace{2mu} 2x\text{.}$$

Solve the following initial-value problem and graph the solution: $y^{''} - 2y^{\prime} + 10y = 0,y(0) = 2,y^{\prime}(0) = -1$

Initial-Value Problem Representing a Spring-Mass System

The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in Applications. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)

$$y^{''} + 2y^{\prime} + y = 0,y(0) = 1,y^{\prime}(0) = 0$$

Solve the initial-value problem and graph the solution. What is the position of the mass at time $t = 2$ sec? How fast is the mass moving at time $t = 1$ sec? In what direction?

Solution

In Example 7.6c. we found the general solution to this differential equation to be

$$y(t) = c_{1}e^{\text{−}t} + c_{2}te^{\text{−}t}.$$

Then

$$y^{\prime}(t) = \text{−}c_{1}e^{\text{−}t} + c_{2}\left( {\text{−}te^{\text{−}t} + e^{\text{−}t}} \right).$$

When $t = 0,$ we have $y(0) = c_{1}$ and $y^{\prime}(0) = \text{−}c_{1} + c_{2}.$ Applying the initial conditions, we obtain

$$\begin{array}{rll}

c_{1} & = & 1 \\

{{\text{−}c}_{1} + c_{2}} & = & {0.}

\end{array}$$

Thus, $c_{1} = 1,$ $c_{2} = 1,$ and the solution to the initial value problem is

$$y(t) = e^{\text{−}t} + te^{\text{−}t}.$$

This solution is represented in the following graph. At time $t = 2,$ the mass is at position $y(2) = e^{-2} + 2e^{-2} = 3e^{-2} \approx 0.406$ m below equilibrium.

To calculate the velocity at time $t = 1,$ we need to find the derivative. We have $y(t) = e^{\text{−}t} + te^{\text{−}t},$ so

$$y^{\prime}(t) = \text{−}e^{\text{−}t} + e^{\text{−}t} - te^{\text{−}t} = \text{−}te^{\text{−}t}.$$

Then $y^{\prime}(1) = \text{−}e^{-1} \approx - 0.3679.$ At time $t = 1,$ the mass is moving upward at 0.3679 m/sec.

Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time $t = 0.3$ sec? How fast is it moving at time $t = 0.1$ sec? In what direction?

$$y^{''} + 14y^{\prime} + 49y = 0,y(0) = 0,y^{\prime}(0) = 1$$

Solving a Boundary-Value Problem

In Example 7.6f. we solved the differential equation $y^{''} + 16y = 0$ and found the general solution to be $y(t) = c_{1}\text{cos}\mspace{2mu} 4t + c_{2}\text{sin}\mspace{2mu} 4t\text{.}$ If possible, solve the boundary-value problem if the boundary conditions are the following:

1. $y(0) = 0,$ $y\left( \frac{\pi}{4} \right) = 0$

2. $y(0) = 1,$ $y\left( \frac{\pi}{8} \right) = 0$

3. $y\left( \frac{\pi}{8} \right) = 0,$ $y\left( \frac{3\pi}{8} \right) = 2$

Solution

We have

$$y(t) = c_{1}\text{cos}\mspace{2mu} 4t + c_{2}\text{sin}\mspace{2mu} 4t\text{.}$$

1. Applying the first boundary condition given here, we get $y(0) = c_{1} = 0.$ So the solution is of the form $y(t) = c_{2}\text{sin}\mspace{2mu} 4t\text{.}$ When we apply the second boundary condition, though, we get $y\left( \frac{\pi}{4} \right) = c_{2}\text{sin}\mspace{2mu}\left( {4\left( \frac{\pi}{4} \right)} \right) = c_{2}\text{sin}\mspace{2mu}\pi = 0$ for all values of $c_{2}.$ The boundary conditions are not sufficient to determine a value for $c_{2},$ so this boundary-value problem has infinitely many solutions. Thus, $y(t) = c_{2}\text{sin}\mspace{2mu} 4t$ is a solution for any value of $c_{2}.$

2. Applying the first boundary condition given here, we get $y(0) = c_{1} = 1.$ Applying the second boundary condition gives $y\left( \frac{\pi}{8} \right) = c_{2} = 0,$ so $c_{2} = 0.$ In this case, we have a unique solution: $y(t) = \text{cos}\mspace{2mu} 4t\text{.}$

3. Applying the first boundary condition given here, we get $y\left( \frac{\pi}{8} \right) = c_{2} = 0.$ However, applying the second boundary condition gives $y\left( \frac{3\pi}{8} \right) = \text{−}c_{2} = 2,$ so $c_{2} = -2.$ We cannot have $c_{2} = 0 = -2,$ so this boundary value problem has no solution.

Section 7.1 Exercises

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine whether it is homogeneous or nonhomogeneous.

1.

$x^{3}y^{''} + \left( {x - 1} \right)y^{\prime} - 8y = 0$

2\.

$\left( {1 + y^{2}} \right)y^{''} + xy^{\prime} - 3y = \text{cos}\mspace{2mu} x$

3.

$xy^{''} + e^{y}y^{\prime} = x$

4\.

$y^{''} + \frac{4}{x}y^{\prime} - 8xy = 5x^{2} + 1$

5.

$y^{''} + \left( {\text{sin}\mspace{2mu} x} \right)y^{\prime} - xy = 4y$

6\.

$y^{''} + \left( \frac{x + 3}{y} \right)y^{\prime} = 0$

For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of *c*₁ and *c*₂. What do the solutions have in common?

7\.

\[T\]$y^{''} + 2y^{\prime} - 3y = 0;$ $y(x) = c_{1}e^{x} + c_{2}e^{-3x}$

8\.

\[T\]$x^{2}y^{''} - 2y - 3x^{2} + 1 = 0;$ $y(x) = c_{1}x^{2} + c_{2}x^{-1} + x^{2}\text{ln}\mspace{2mu}(x) + \frac{1}{2}$

9\.

\[T\]$y^{''} + 14y^{\prime} + 49y = 0;$ $y(x) = c_{1}e^{-7x} + c_{2}xe^{-7x}$

10\.

\[T\]$6y^{''} - 49y^{\prime} + 8y = 0;$ $y(x) = c_{1}e^{x\text{/}6} + c_{2}e^{8x}$

Find the general solution to the linear differential equation.

11.

$y^{''} - 3y^{\prime} - 10y = 0$

12\.

$y^{''} - 7y^{\prime} + 12y = 0$

13.

$y^{''} + 4y^{\prime} + 4y = 0$

14\.

$4y^{''} - 12y^{\prime} + 9y = 0$

15.

$2y^{''} - 3y^{\prime} - 5y = 0$

16\.

$3y^{''} - 14y^{\prime} + 8y = 0$

17.

$y^{''} + y^{\prime} + y = 0$

18\.

$5y^{''} + 2y^{\prime} + 4y = 0$

19.

$y^{''} - 121y = 0$

20\.

$8y^{''} + 14y^{\prime} - 15y = 0$

21.

$y^{''} + 81y = 0$

22\.

$y^{''} - y^{\prime} + 11y = 0$

23.

$2y^{''} = 0$

24\.

$y^{''} - 6y^{\prime} + 9y = 0$

25.

$3y^{''} - 2y^{\prime} - 7y = 0$

26\.

$4y^{''} - 10y^{\prime} = 0$

27.

$36\frac{d^{2}y}{dx^{2}} + 12\frac{dy}{dx} + y = 0$

28\.

$25\frac{d^{2}y}{dx^{2}} - 80\frac{dy}{dx} + 64y = 0$

29.

$\frac{d^{2}y}{dx^{2}} - 9\frac{dy}{dx} = 0$

30\.

$4\frac{d^{2}y}{dx^{2}} + 8y = 0$

Solve the initial-value problem.

31.

$y^{''} + 5y^{\prime} + 6y = 0,\qquad y(0) = 0,\ \ y^{\prime}(0) = -2$

32\.

$y^{''} + 2y^{\prime} - 8y = 0,\qquad y(0) = 5,\ \ y^{\prime}(0) = 4$

33.

$y^{''} + 4y = 0,\mspace{72mu} y(0) = 3,\ \ y^{\prime}(0) = 10$

34\.

$y^{''} - 18y^{\prime} + 81y = 0,\qquad y(0) = 1,\ \ y^{\prime}(0) = 5$

35.

$y^{''} - y^{\prime} - 30y = 0,\qquad y(0) = 1,\ \ y^{\prime}(0) = -16$

36\.

$4y^{''} + 4y^{\prime} - 8y = 0,\qquad y(0) = 2,\ \ y^{\prime}(0) = 1$

37.

$25y^{''} + 10y^{\prime} + y = 0,\qquad y(0) = 2,\ \ y^{\prime}(0) = 1$

38\.

$y^{''} + y = 0,\mspace{72mu} y(\pi) = 1,\ \ y^{\prime}(\pi) = -5$

Solve the boundary-value problem, if possible.

39.

$y^{''} + y^{\prime} - 42y = 0,\qquad y(0) = 0,\ \ y(1) = 2$

40\.

$9y^{''} + y = 0,\mspace{72mu} y(\frac{3\pi}{2}) = 6,\ \ y(0) = -8$

41.

$y^{''} + 10y^{\prime} + 34y = 0,\qquad y(0) = 6,\ \ y(\pi) = 2$

42\.

$y^{''} + 7y^{\prime} - 60y = 0,\qquad y(0) = 4,\ \ y(2) = 0$

43.

$y^{''} - 4y^{\prime} + 4y = 0,\qquad y(0) = 2,\ \ y(1) = -1$

44\.

$y^{''} - 5y^{\prime} = 0,\qquad y(0) = 3,\ \ y(-1) = 2$

45.

$y^{''} + 9y = 0,\qquad y(0) = 4,\ \ y\left( \frac{\pi}{3} \right) = -4$

46\.

$4y^{''} + 25y = 0,\qquad y(0) = 2,\ \ y(2\pi) = -2$

47.

Find a differential equation with a general solution that is $y = c_{1}e^{x\text{/}5} + c_{2}e^{-4x}.$

48\.

Find a differential equation with a general solution that is $y = c_{1}e^{x} + c_{2}e^{-4x\text{/}3}.$

For each of the following differential equations:

1. Solve the initial value problem.

2. \[T\] Use a graphing utility to graph the particular solution.

49.

$y^{''} + 64y = 0;\qquad y(0) = 3,\quad y^{\prime}(0) = –24$

50\.

$y^{''} - 2y^{\prime} + 10y = 0\qquad y(0) = 1,\quad y^{\prime}(0) = 13$

51.

$y^{''} + 5y^{\prime} + 15y = 0\qquad y(0) = -2,\quad y^{\prime}(0) = 7$

52\.

(Principle of superposition) Prove that if $y_{1}(x)$ and $y_{2}(x)$ are solutions to a linear homogeneous differential equation, $y^{''} + p(x)y^{\prime} + q(x)y = 0,$ then the function $y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x),$ where $c_{1}$ and $c_{2}$ are constants, is also a solution.

53\.

Prove that if *a, b,* and *c* are positive constants, then all solutions to the second-order linear differential equation $ay^{''} + by^{\prime} + cy = 0$ approach zero as $x\rightarrow\infty\text{.}$ (*Hint:* Consider three cases: two distinct roots, repeated real roots, and complex conjugate roots.)

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7.2 Nonhomogeneous Linear Equations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/7-2-nonhomogeneous-linear-equations

7.2 Nonhomogeneous Linear Equations

• 7.2.1 Write the general solution to a nonhomogeneous differential equation.
• 7.2.2 Solve a nonhomogeneous differential equation by the method of undetermined coefficients.
• 7.2.3 Solve a nonhomogeneous differential equation by the method of variation of parameters.

In this section, we examine how to solve nonhomogeneous differential equations. The terminology and methods are different from those we used for homogeneous equations, so let’s start by defining some new terms.

General Solution to a Nonhomogeneous Linear Equation

Consider the nonhomogeneous linear differential equation

$$a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x).$$

The associated homogeneous equation

$$a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = 0$$ (7.3)

is called the complementary equation. We will see that solving the complementary equation is an important step in solving a nonhomogeneous differential equation.

A solution $y_{p}(x)$ of a differential equation that contains no arbitrary constants is called a particular solution to the equation.

General Solution to a Nonhomogeneous Equation

Let $y_{p}(x)$ be any particular solution to the nonhomogeneous linear differential equation

$$a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x).$$

Also, let $c_{1}y_{1}(x) + c_{2}y_{2}(x)$ denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by

$$y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) + y_{p}(x\text{).}$$ (7.4)

Proof

To prove $y(x)$ is the general solution, we must first show that it solves the differential equation and, second, that any solution to the differential equation can be written in that form. Substituting $y(x)$ into the differential equation, we have

$$\begin{array}{cl}

{a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y} & {= a_{2}(x)\left( {c_{1}y_{1} + c_{2}y_{2} + y_{p}} \right)\text{''} + a_{1}(x)\left( {c_{1}y_{1} + c_{2}y_{2} + y_{p}} \right)^{\prime}} \\

& {\qquad + a_{0}(x)\left( {c_{1}y_{1} + c_{2}y_{2} + y_{p}} \right)} \\

& {= \left\lbrack {a_{2}(x)\left( {c_{1}y_{1} + c_{2}y_{2}} \right)\text{''} + a_{1}(x)\left( {c_{1}y_{1} + c_{2}y_{2}} \right)^{\prime} + a_{0}(x)\left( {c_{1}y_{1} + c_{2}y_{2}} \right)} \right\rbrack} \\

& {\qquad + a_{2}(x)y_{p}\text{''} + a_{1}(x)y_{p}{}^{\prime} + a_{0}(x)y_{p}} \\

& {= 0 + r(x)} \\

& {= r(x\text{).}}

\end{array}$$

So $y(x)$ is a solution.

Now, let $z(x)$ be any solution to $a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x).$ Then

$$\begin{array}{cl}

{a_{2}(x)\left( {z - y_{p}} \right)\text{''} + a_{1}(x)\left( {z - y_{p}} \right)^{\prime} + a_{0}(x)\left( {z - y_{p}} \right)} & {= \left( {a_{2}(x)z^{''} + a_{1}(x)z^{\prime} + a_{0}(x)z} \right)} \\

& {\qquad - \left( {a_{2}(x)y_{p}\text{''} + a_{1}(x)y_{p}{}^{\prime} + a_{0}(x)y_{p}} \right)} \\

& {= r(x) - r(x)} \\

& {= 0,}

\end{array}$$

so $z(x) - y_{p}(x)$ is a solution to the complementary equation. But, $c_{1}y_{1}(x) + c_{2}y_{2}(x)$ is the general solution to the complementary equation, so there are constants $c_{1}$ and $c_{2}$ such that

$$z(x) - y_{p}(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x\text{).}$$

Hence, we see that $z(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) + y_{p}(x\text{).}$

□

Verifying the General Solution

Given that $y_{p}(x) = x$ is a particular solution to the differential equation $y^{''} + y = x,$ write the general solution and check by verifying that the solution satisfies the equation.

Solution

The complementary equation is $y^{''} + y = 0,$ which has the general solution $c_{1}\text{cos}\mspace{2mu} x + c_{2}\text{sin}\mspace{2mu} x\text{.}$ So, the general solution to the nonhomogeneous equation is

$$y(x) = c_{1}\text{cos}\mspace{2mu} x + c_{2}\text{sin}\mspace{2mu} x + x\text{.}$$

To verify that this is a solution, substitute it into the differential equation. We have

$$y^{\prime}(x) = \text{−}c_{1}\text{sin}\mspace{2mu} x + c_{2}\text{cos}\mspace{2mu} x + 1\ \text{and}\ y^{''}(x) = \text{−}c_{1}\text{cos}\mspace{2mu} x - c_{2}\text{sin}\mspace{2mu} x\text{.}$$

Then

$$\begin{array}{cl}

{y^{''}(x) + y(x)} & {= \text{−}c_{1}\text{cos}\mspace{2mu} x - c_{2}\text{sin}\mspace{2mu} x + c_{1}\text{cos}\mspace{2mu} x + c_{2}\text{sin}\mspace{2mu} x + x} \\

& {= x\text{.}}

\end{array}$$

So, $y(x)$ is a solution to $y^{''} + y = x\text{.}$

Given that $y_{p}(x) = -2$ is a particular solution to $y^{''} - 3y^{\prime} - 4y = 8,$ write the general solution and verify that the general solution satisfies the equation.

In the preceding section, we learned how to solve homogeneous equations with constant coefficients. Therefore, for nonhomogeneous equations of the form $ay^{''} + by^{\prime} + cy = r(x),$ we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. We now examine two techniques for this: the method of undetermined coefficients and the method of variation of parameters.

Undetermined Coefficients

The method of undetermined coefficients involves making educated guesses about the form of the particular solution based on the form of $r(x\text{).}$ When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when $r(x)$ has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. Let’s look at some examples to see how this works.

Undetermined Coefficients When $r(x)$ Is a Polynomial

Find the general solution to $y^{''} + 4y^{\prime} + 3y = 3x\text{.}$

Solution

The complementary equation is $y^{''} + 4y^{\prime} + 3y = 0,$ with general solution $c_{1}e^{\text{−}x} + c_{2}e^{-3x}.$ Since $r(x) = 3x,$ the particular solution might have the form $y_{p}(x) = Ax + B\text{.}$ If this is the case, then we have $y_{p}{}^{\prime}(x) = A$ and $y_{p}{}\text{''}(x) = 0.$ For $y_{p}$ to be a solution to the differential equation, we must find values for $A$ and $B$ such that

$$\begin{array}{rll}

{y^{''} + 4y^{\prime} + 3y} & = & {3x} \\

{0 + 4(A) + 3(Ax + B)} & = & {3x} \\

{3Ax + (4A + 3B)} & = & {3x\text{.}}

\end{array}$$

Setting coefficients of like terms equal, we have

$$\begin{array}{rll}

{3A} & = & 3 \\

{4A + 3B} & = & {0.}

\end{array}$$

Then, $A = 1$ and $B = - \frac{4}{3},$ so $y_{p}(x) = x - \frac{4}{3}$ and the general solution is

$$y(x) = c_{1}e^{\text{−}x} + c_{2}e^{-3x} + x - \frac{4}{3}.$$

In Example 7.12, notice that even though $r(x)$ did not include a constant term, it was necessary for us to include the constant term in our guess. If we had assumed a solution of the form $y_{p} = Ax$ (with no constant term), we would not have been able to find a solution. (Verify this!) If the function $r(x)$ is a polynomial, our guess for the particular solution should be a polynomial of the same degree, and it must include all lower-order terms, regardless of whether they are present in $r(x\text{).}$

Undetermined Coefficients When $r(x)$ Is an Exponential

Find the general solution to $y^{''} - y^{\prime} - 2y = 2e^{3x}.$

Solution

The complementary equation is $y^{''} - y^{\prime} - 2y = 0,$ with the general solution $c_{1}e^{\text{−}x} + c_{2}e^{2x}.$ Since $r(x) = 2e^{3x},$ the particular solution might have the form $y_{p}(x) = Ae^{3x}.$ Then, we have $y_{p}{}^{\prime}(x) = 3Ae^{3x}$ and $y_{p}{}\text{''}(x) = 9Ae^{3x}.$ For $y_{p}$ to be a solution to the differential equation, we must find a value for $A$ such that

$$\begin{array}{rll}

{y^{''} - y^{\prime} - 2y} & = & {2e^{3x}} \\

{9Ae^{3x} - 3Ae^{3x} - 2Ae^{3x}} & = & {2e^{3x}} \\

{4Ae^{3x}} & = & {2e^{3x}.}

\end{array}$$

So, $4A = 2$ and $A = {1\text{/}2}.$ Then, $y_{p}(x) = \left( \frac{1}{2} \right)e^{3x},$ and the general solution is

$$y(x) = c_{1}e^{\text{−}x} + c_{2}e^{2x} + \frac{1}{2}e^{3x}.$$

Find the general solution to $y^{''} - 4y^{\prime} + 4y = 7\mspace{2mu}\text{sin}\mspace{2mu} t - \text{cos}\mspace{2mu} t\text{.}$

In the previous checkpoint, $r(x)$ included both sine and cosine terms. However, even if $r(x)$ included a sine term only or a cosine term only, both terms must be present in the guess. The method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines. Some of the key forms of $r(x)$ and the associated guesses for $y_{p}(x)$ are summarized in Table 7.2.

| $r(x)$ | Initial guess for $y_{p}(x)$ |

|--------------------------------------------------------------------------------------------|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| $k$ (a constant) | $A$ (a constant) |

| $ax + b$ | $Ax + B$ (*Note*: The guess must include both terms even if $b = 0.$) |

| $ax^{2} + bx + c$ | $Ax^{2} + Bx + C$ (*Note*: The guess must include all three terms even if $b$ or $c$ are zero.) |

| Higher-order polynomials | Polynomial of the same order as $r(x)$ |

| $ae^{\lambda x}$ | $Ae^{\lambda x}$ |

| $a\mspace{2mu}\text{cos}\mspace{2mu}\beta x + b\mspace{2mu}\text{sin}\mspace{2mu}\beta x$ | $A\mspace{2mu}\text{cos}\mspace{2mu}\beta x + B\mspace{2mu}\text{sin}\mspace{2mu}\beta x$ (*Note*: The guess must include both terms even if either $a = 0$ or $b = 0.$) |

| $ae^{\alpha x}\text{cos}\mspace{2mu}\beta x + be^{\alpha x}\text{sin}\mspace{2mu}\beta x$ | $Ae^{\alpha x}\text{cos}\mspace{2mu}\beta x + Be^{\alpha x}\text{sin}\mspace{2mu}\beta x$ |

| $\left( {ax^{2} + bx + c} \right)e^{\lambda x}$ | $\left( {Ax^{2} + Bx + C} \right)e^{\lambda x}$ |

| $\begin{array}{l}

{\left( {a_{2}x^{2} + a_{1}x + a_{0}} \right)\mspace{2mu}\text{cos}\mspace{2mu}\beta x} \\

{+ \left( {b_{2}x^{2} + b_{1}x + b_{0}} \right)\mspace{2mu}\text{sin}\mspace{2mu}\beta x}

\end{array}$ | $\begin{array}{l}

{\left( {A_{2}x^{2} + A_{1}x + A_{0}} \right)\mspace{2mu}\text{cos}\mspace{2mu}\beta x} \\

{+ \left( {B_{2}x^{2} + B_{1}x + B_{0}} \right)\mspace{2mu}\text{sin}\mspace{2mu}\beta x}

\end{array}$ |

| $\begin{array}{l}

{\left( {a_{2}x^{2} + a_{1}x + a_{0}} \right)e^{\alpha x}\text{cos}\mspace{2mu}\beta x} \\

{+ \left( {b_{2}x^{2} + b_{1}x + b_{0}} \right)e^{\alpha x}\text{sin}\mspace{2mu}\beta x}

\end{array}$ | $\begin{array}{l}

{\left( {A_{2}x^{2} + A_{1}x + A_{0}} \right)e^{\alpha x}\text{cos}\mspace{2mu}\beta x} \\

{+ \left( {B_{2}x^{2} + B_{1}x + B_{0}} \right)e^{\alpha x}\text{sin}\mspace{2mu}\beta x}

\end{array}$ |

Table 7.2 Key Forms for the Method of Undetermined Coefficients

Keep in mind that there is a key pitfall to this method. Consider the differential equation $y^{''} + 5y^{\prime} + 6y = 3e^{-2x}.$ Based on the form of $r(x),$ we guess a particular solution of the form $y_{p}(x) = Ae^{-2x}.$ But when we substitute this expression into the differential equation to find a value for $A,$ we run into a problem. We have

$$y_{p}{}^{\prime}(x) = -2Ae^{-2x}$$

and

$$y_{p}{}\text{''} = 4Ae^{-2x},$$

so we want

$$\begin{array}{rll}

{y^{''} + 5y^{\prime} + 6y} & = & {3e^{-2x}} \\

{4Ae^{-2x} + 5\left( {-2Ae^{-2x}} \right) + 6Ae^{-2x}} & = & {3e^{-2x}} \\

{4Ae^{-2x} - 10Ae^{-2x} + 6Ae^{-2x}} & = & {3e^{-2x}} \\

0 & = & {3e^{-2x},}

\end{array}$$

which is not possible.

Looking closely, we see that, in this case, the general solution to the complementary equation is $c_{1}e^{-2x} + c_{2}e^{-3x}.$ The exponential function in $r(x)$ is actually a solution to the complementary equation, so, as we just saw, all the terms on the left side of the equation cancel out. We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by$x\text{.}$ Using the new guess, $y_{p}(x) = Axe^{-2x},$ we have

$$y_{p}{}^{\prime}(x) = A\left( {e^{-2x} - 2xe^{-2x}} \right)$$

and

$$y_{p}{}\text{''}(x) = -4Ae^{-2x} + 4Axe^{-2x}.$$

Substitution gives

$$\begin{array}{rll}

{y^{''} + 5y^{\prime} + 6y} & = & {3e^{-2x}} \\

{\left( {-4Ae^{-2x} + 4Axe^{-2x}} \right) + 5\left( {Ae^{-2x} - 2Axe^{-2x}} \right) + 6Axe^{-2x}} & = & {3e^{-2x}} \\

{-4Ae^{-2x} + 4Axe^{-2x} + 5Ae^{-2x} - 10Axe^{-2x} + 6Axe^{-2x}} & = & {3e^{-2x}} \\

{Ae^{-2x}} & = & {3e^{-2x}.}

\end{array}$$

So, $A = 3$ and $y_{p}(x) = 3xe^{-2x}.$ This gives us the following general solution

$$y(x) = c_{1}e^{-2x} + c_{2}e^{-3x} + 3xe^{-2x}.$$

Note that if $xe^{-2x}$ were also a solution to the complementary equation, we would have to multiply by $x$ again, and we would try $y_{p}(x) = Ax^{2}e^{-2x}.$

Method of Undetermined Coefficients

1. Solve the complementary equation and write down the general solution.

2. Based on the form of $r(x),$ make an initial guess for $y_{p}(x\text{).}$

3. Check whether any term in the guess for $y_{p}(x)$ is a solution to the complementary equation. If so, multiply the guess by$x\text{.}$ Repeat this step until there are no terms in $y_{p}(x)$ that solve the complementary equation.

4. Substitute $y_{p}(x)$ into the differential equation and equate like terms to find values for the unknown coefficients in $y_{p}(x\text{).}$

5. Add the general solution to the complementary equation and the particular solution you just found to obtain the general solution to the nonhomogeneous equation.

Solving Nonhomogeneous Equations

Find the general solutions to the following differential equations.

1. $y^{''} - 9y = -6\mspace{2mu}\text{cos}\mspace{2mu} 3x$

2. $x^{''} + 2x^{\prime} + x = 4e^{\text{−}t}$

3. $y^{''} - 2y^{\prime} + 5y = 10x^{2} - 3x - 3$

4. $y^{''} - 3y^{\prime} = -12t$

Solution

1. The complementary equation is $y^{''} - 9y = 0,$ which has the general solution $c_{1}e^{3x} + c_{2}e^{-3x}$ (step 1). Based on the form of $r(x) = -6\mspace{2mu}\text{cos}\mspace{2mu} 3x,$ our initial guess for the particular solution is $y_{p}(x) = A\mspace{2mu}\text{cos}\mspace{2mu} 3x + B\mspace{2mu}\text{sin}\mspace{2mu} 3x$ (step 2). None of the terms in $y_{p}(x)$ solve the complementary equation, so this is a valid guess (step 3).

Now we want to find values for $A$ and $B,$ so substitute $y_{p}$ into the differential equation. We have

$$y_{p}{}^{\prime}(x) = -3A\mspace{2mu}\text{sin}\mspace{2mu} 3x + 3B\mspace{2mu}\text{cos}\mspace{2mu} 3x\ \text{and}\ y_{p}{}\text{''}(x) = -9A\mspace{2mu}\text{cos}\mspace{2mu} 3x - 9B\mspace{2mu}\text{sin}\mspace{2mu} 3x,$$

so we want to find values of $A$ and $B$ such that

$$\begin{array}{rll}

{y^{''} - 9y} & = & {-6\mspace{2mu}\text{cos}\mspace{2mu} 3x} \\

{-9A\mspace{2mu}\text{cos}\mspace{2mu} 3x - 9B\mspace{2mu}\text{sin}\mspace{2mu} 3x - 9\left( {A\mspace{2mu}\text{cos}\mspace{2mu} 3x + B\mspace{2mu}\text{sin}\mspace{2mu} 3x} \right)} & = & {-6\mspace{2mu}\text{cos}\mspace{2mu} 3x} \\

{-18A\mspace{2mu}\text{cos}\mspace{2mu} 3x - 18B\mspace{2mu}\text{sin}\mspace{2mu} 3x} & = & {-6\mspace{2mu}\text{cos}\mspace{2mu} 3x\text{.}}

\end{array}$$

Therefore,

$$\begin{array}{rll}

{-18A} & = & -6 \\

{-18B} & = & {0.}

\end{array}$$

This gives $A = \frac{1}{3}$ and $B = 0,$ so $y_{p}(x) = \left( \frac{1}{3} \right)\mspace{2mu}\text{cos}\mspace{2mu} 3x$ (step 4).

Putting everything together, we have the general solution

$$y(x) = c_{1}e^{3x} + c_{2}e^{-3x} + \frac{1}{3}\mspace{2mu}\text{cos}\mspace{2mu} 3x\text{.}$$

2. The complementary equation is $x^{''} + 2x^{\prime} + x = 0,$ which has the general solution $c_{1}e^{\text{−}t} + c_{2}te^{\text{−}t}$ (step 1). Based on the form $r(t) = 4e^{\text{−}t},$ our initial guess for the particular solution is $x_{p}(t) = Ae^{\text{−}t}$ (step 2). However, we see that this guess solves the complementary equation, so we must multiply by$t,$ which gives a new guess: $x_{p}(t) = Ate^{\text{−}t}$ (step 3). Checking this new guess, we see that it, too, solves the complementary equation, so we must multiply by *t* again, which gives $x_{p}(t) = At^{2}e^{\text{−}t}$ (step 3 again). Now, checking this guess, we see that $x_{p}(t)$ does not solve the complementary equation, so this is a valid guess (step 3 yet again).

We now want to find a value for $A,$ so we substitute $x_{p}$ into the differential equation. We have

$$\begin{array}{rll}

{x_{p}(t)} & = & {At^{2}e^{\text{−}t},\ \text{so}} \\

{x_{p}{}^{\prime}(t)} & = & {2Ate^{\text{−}t} - At^{2}e^{\text{−}t}}

\end{array}$$

and $x_{p}{}\text{''}(t) = 2Ae^{\text{−}t} - 2Ate^{\text{−}t} - \left( {2Ate^{\text{−}t} - At^{2}e^{\text{−}t}} \right) = 2Ae^{\text{−}t} - 4Ate^{\text{−}t} + At^{2}e^{\text{−}t}.$

Substituting into the differential equation, we want to find a value of $A$ so that

$$\begin{array}{rll}

{x^{''} + 2x^{\prime} + x} & = & {4e^{\text{−}t}} \\

{2Ae^{\text{−}t} - 4Ate^{\text{−}t} + At^{2}e^{\text{−}t} + 2\left( {2Ate^{\text{−}t} - At^{2}e^{\text{−}t}} \right) + At^{2}e^{\text{−}t}} & = & {4e^{\text{−}t}} \\

{2Ae^{\text{−}t}} & = & {4e^{\text{−}t}.}

\end{array}$$

This gives $A = 2,$ so $x_{p}(t) = 2t^{2}e^{\text{−}t}$ (step 4). Putting everything together, we have the general solution

$$x(t) = c_{1}e^{\text{−}t} + c_{2}te^{\text{−}t} + 2t^{2}e^{\text{−}t}.$$

3. The complementary equation is $y^{''} - 2y^{\prime} + 5y = 0,$ which has the general solution $c_{1}e^{x}\text{cos}\mspace{2mu} 2x + c_{2}e^{x}\text{sin}\mspace{2mu} 2x$ (step 1). Based on the form $r(x) = 10x^{2} - 3x - 3,$ our initial guess for the particular solution is $y_{p}(x) = Ax^{2} + Bx + C$ (step 2). None of the terms in $y_{p}(x)$ solve the complementary equation, so this is a valid guess (step 3). We now want to find values for $A,$ $B,$ and $C,$ so we substitute $y_{p}$ into the differential equation. We have $y_{p}{}^{\prime}(x) = 2Ax + B$ and $y_{p}{}\text{''}(x) = 2A,$ so we want to find values of $A,$ $B,$ and $C$ such that

$$\begin{array}{rll}

{y^{''} - 2y^{\prime} + 5y} & = & {10x^{2} - 3x - 3} \\

{2A - 2\left( {2Ax + B} \right) + 5\left( {Ax^{2} + Bx + C} \right)} & = & {10x^{2} - 3x - 3} \\

{5Ax^{2} + \left( {5B - 4A} \right)x + \left( {5C - 2B + 2A} \right)} & = & {10x^{2} - 3x - 3.}

\end{array}$$

Therefore,

$$\begin{array}{rll}

{5A} & = & 10 \\

{5B - 4A} & = & -3 \\

{5C - 2B + 2A} & = & {-3.}

\end{array}$$

This gives $A = 2,$ $B = 1,$ and $C = -1,$ so $y_{p}(x) = 2x^{2} + x - 1$ (step 4). Putting everything together, we have the general solution

$$y(x) = c_{1}e^{x}\text{cos}\mspace{2mu} 2x + c_{2}e^{x}\text{sin}\mspace{2mu} 2x + 2x^{2} + x - 1.$$

4. The complementary equation is $y^{''} - 3y^{\prime} = 0,$ which has the general solution $c_{1}e^{3t} + c_{2}$ (step 1). Based on the form $r(t) = -12t,$ our initial guess for the particular solution is $y_{p}(t) = At + B$ (step 2). However, we see that the constant term in this guess solves the complementary equation, so we must multiply by $t,$ which gives a new guess: $y_{p}(t) = At^{2} + Bt$ (step 3). Checking this new guess, we see that none of the terms in $y_{p}(t)$ solve the complementary equation, so this is a valid guess (step 3 again). We now want to find values for $A$ and $B,$ so we substitute $y_{p}$ into the differential equation. We have $y_{p}{}^{\prime}(t) = 2At + B$ and $y_{p}{}\text{''}(t) = 2A,$ so we want to find values of $A$ and $B$ such that

$$\begin{array}{rll}

{y^{''} - 3y^{\prime}} & = & {-12t} \\

{2A - 3\left( {2At + B} \right)} & = & {-12t} \\

{- 6At + \left( {2A - 3B} \right)} & = & {-12t\text{.}}

\end{array}$$

Therefore,

$$\begin{array}{rll}

{-6A} & = & -12 \\

{2A - 3B} & = & {0.}

\end{array}$$

This gives $A = 2$ and $B = {4\text{/}3},$ so $y_{p}(t) = 2t^{2} + \left( {4\text{/}3} \right)t$ (step 4). Putting everything together, we have the general solution

$$y(t) = c_{1}e^{3t} + c_{2} + 2t^{2} + \frac{4}{3}t\text{.}$$

Find the general solution to the following differential equations.

1. $y^{''} - 5y^{\prime} + 4y = 3e^{x}$

2. $y^{''} + y^{\prime} - 6y = 52\mspace{2mu}\text{cos}\mspace{2mu} 2t$

Variation of Parameters

Sometimes, $r(x)$ is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the method of variation of parameters.

To simplify our calculations a little, we are going to divide the differential equation through by $a,$ so we have a leading coefficient of 1. Then the differential equation has the form

$$y^{''} + py^{\prime} + qy = r(x),$$

where $p$ and $q$ are constants.

If the general solution to the complementary equation is given by $c_{1}y_{1}(x) + c_{2}y_{2}(x),$ we are going to look for a particular solution of the form $y_{p}(x) = u(x)y_{1}(x) + v(x)y_{2}(x\text{).}$ In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of *x*, rather than constants. We want to find functions $u(x)$ and $v(x)$ such that $y_{p}(x)$ satisfies the differential equation. We have

$$\begin{array}{rll}

y_{p} & = & {uy_{1} + vy_{2}} \\

{y_{p}{}^{\prime}} & = & {u^{\prime}y_{1} + uy_{1}{}^{\prime} + v^{\prime}y_{2} + vy_{2}{}^{\prime}} \\

{y_{p}\text{''}} & = & {\left( {u^{\prime}y_{1} + v^{\prime}y_{2}} \right)^{\prime} + u^{\prime}y_{1}{}^{\prime} + uy_{1}\text{''} + v^{\prime}y_{2}{}^{\prime} + vy_{2}{\text{''}\text{.}}}

\end{array}$$

Substituting into the differential equation, we obtain

$$\begin{array}{cl}

{y_{p}\text{''} + py_{p}{}^{\prime} + qy_{p}} & {= \left\lbrack {\left( {u^{\prime}y_{1} + v^{\prime}y_{2}} \right)^{\prime} + u^{\prime}y_{1}{}^{\prime} + uy_{1}\text{''} + v^{\prime}y_{2}{}^{\prime} + vy_{2}\text{''}} \right\rbrack} \\

& {\qquad + p\left\lbrack {u^{\prime}y_{1} + uy_{1}{}^{\prime} + v^{\prime}y_{2} + vy_{2}{}^{\prime}} \right\rbrack + q\left\lbrack {uy_{1} + vy_{2}} \right\rbrack} \\

& {= u\left\lbrack {y_{1}\text{''} + py_{1}{}^{\prime} + qy_{1}} \right\rbrack + v\left\lbrack {y_{2}\text{''} + py_{2}{}^{\prime} + qy_{2}} \right\rbrack} \\

& {\qquad + \left( {u^{\prime}y_{1} + v^{\prime}y_{2}} \right)^{\prime} + p\left( {u^{\prime}y_{1} + v^{\prime}y_{2}} \right) + {\left( {u^{\prime}y_{1}{}^{\prime} + v^{\prime}y_{2}{}^{\prime}} \right).}}

\end{array}$$

Note that $y_{1}$ and $y_{2}$ are solutions to the complementary equation, so the first two terms are zero. Thus, we have

$$\left( {u^{\prime}y_{1} + v^{\prime}y_{2}} \right)^{\prime} + p\left( {u^{\prime}y_{1} + v^{\prime}y_{2}} \right) + \left( {u^{\prime}y_{1}{}^{\prime} + v^{\prime}y_{2}{}^{\prime}} \right) = r(x\text{).}$$

If we simplify this equation by imposing the additional condition $u^{\prime}y_{1} + v^{\prime}y_{2} = 0,$ the first two terms are zero, and this reduces to $u^{\prime}y_{1}{}^{\prime} + v^{\prime}y_{2}{}^{\prime} = r(x\text{).}$ So, with this additional condition, we have a system of two equations in two unknowns:

$$\begin{array}{rll}

{u^{\prime}y_{1} + v^{\prime}y_{2}} & = & 0 \\

{u^{\prime}y_{1}{}^{\prime} + v^{\prime}y_{2}{}^{\prime}} & = & {r(x\text{).}}

\end{array}$$

Solving this system gives us $u^{\prime}$ and $v^{\prime},$ which we can integrate to find *u* and *v*.

Then, $y_{p}(x) = u(x)y_{1}(x) + v(x)y_{2}(x)$ is a particular solution to the differential equation. Solving this system of equations is sometimes challenging, so let’s take this opportunity to review Cramer’s rule, which allows us to solve the system of equations using determinants.

The system of equations

$$\begin{array}{rll}

{a_{1}z_{1} + b_{1}z_{2}} & = & r_{1} \\

{a_{2}z_{1} + b_{2}z_{2}} & = & r_{2}

\end{array}$$

has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by

$$z_{1} = \frac{\left| \begin{array}{ll}

r_{1} & b_{1} \\

r_{2} & b_{2}

\end{array} \right|}{\left| \begin{array}{ll}

a_{1} & b_{1} \\

a_{2} & b_{2}

\end{array} \right|}\qquad\text{and}\qquad z_{2} = \frac{\left| \begin{array}{ll}

a_{1} & r_{1} \\

a_{2} & r_{2}

\end{array} \right|}{\left| \begin{array}{ll}

a_{1} & b_{1} \\

a_{2} & b_{2}

\end{array} \right|}.$$

Using Cramer’s Rule

Use Cramer’s rule to solve the following system of equations.

$$\begin{array}{rll}

{x^{2}z_{1} + 2xz_{2}} & = & 0 \\

{z_{1} - 3x^{2}z_{2}} & = & {2x}

\end{array}$$

Solution

We have

$$\begin{array}{rll}

& & \\

& & \\

{a_{1}(x)} & = & x^{2} \\

{a_{2}(x)} & = & 1 \\

{b_{1}(x)} & = & {2x} \\

{b_{2}(x)} & = & {-3x^{2}} \\

{r_{1}(x)} & = & 0 \\

{r_{2}(x)} & = & {2x\text{.}}

\end{array}$$

Then,

$$\left| \begin{array}{ll}

a_{1} & b_{1} \\

a_{2} & b_{2}

\end{array} \right| = \left| \begin{array}{ll}

x^{2} & {\mspace{14mu}{2x}} \\

1 & {-3x^{2}}

\end{array} \right| = -3x^{4} - 2x$$

and

$$\left| \begin{array}{ll}

r_{1} & b_{1} \\

r_{2} & b_{2}

\end{array} \right| = \left| \begin{array}{ll}

0 & {\mspace{14mu}{2x}} \\

{2x} & {-3x^{2}}

\end{array} \right| = 0 - 4x^{2} = -4x^{2}.$$

Thus,

$$z_{1} = \frac{\left| \begin{array}{ll}

r_{1} & b_{1} \\

r_{2} & b_{2}

\end{array} \right|}{\left| \begin{array}{ll}

a_{1} & b_{1} \\

a_{2} & b_{2}

\end{array} \right|} = \frac{-4x^{2}}{-3x^{4} - 2x} = \frac{4x}{3x^{3} + 2}.$$

In addition,

$$\left| \begin{array}{ll}

a_{1} & r_{1} \\

a_{2} & r_{2}

\end{array} \right| = \left| \begin{array}{ll}

x^{2} & 0 \\

1 & {2x}

\end{array} \right| = 2x^{3} - 0 = 2x^{3}.$$

Thus,

$$z_{2} = \frac{\left| \begin{array}{ll}

a_{1} & r_{1} \\

a_{2} & r_{2}

\end{array} \right|}{\left| \begin{array}{ll}

a_{1} & b_{1} \\

a_{2} & b_{2}

\end{array} \right|} = \frac{2x^{3}}{-3x^{4} - 2x} = \frac{-2x^{2}}{3x^{3} + 2}.$$

Use Cramer’s rule to solve the following system of equations.

$$\begin{array}{rll}

{2xz_{1} - 3z_{2}} & = & 0 \\

{x^{2}z_{1} + 4xz_{2}} & = & {x + 1}

\end{array}$$

Method of Variation of Parameters

1. Solve the complementary equation and write down the general solution

$$c_{1}y_{1}(x) + c_{2}y_{2}(x\text{).}$$

2. Use Cramer’s rule or another suitable technique to find functions $u^{\prime}(x)$ and $v^{\prime}(x)$ satisfying

$$\begin{array}{rll}

{u^{\prime}y_{1} + v^{\prime}y_{2}} & = & 0 \\

{u^{\prime}y_{1}{}^{\prime} + v^{\prime}y_{2}{}^{\prime}} & = & {r(x\text{).}}

\end{array}$$

3. Integrate $u^{\prime}$ and $v^{\prime}$ to find $u(x)$ and $v(x\text{).}$ Then, $y_{p}(x) = u(x)y_{1}(x) + v(x)y_{2}(x)$ is a particular solution to the equation.

4. Add the general solution to the complementary equation and the particular solution found in step 3 to obtain the general solution to the nonhomogeneous equation.

Using the Method of Variation of Parameters

Find the general solution to the following differential equations.

1. $y^{''} - 2y^{\prime} + y = \frac{e^{t}}{t^{2}}$

2. $y^{''} + y = 3\mspace{2mu}\text{sin}^{2}x$

Solution

1. The complementary equation is $y^{''} - 2y^{\prime} + y = 0$ with associated general solution $c_{1}e^{t} + c_{2}te^{t}.$ Therefore, $y_{1}(t) = e^{t}$ and $y_{2}(t) = te^{t}.$ Calculating the derivatives, we get $y_{1}{}^{\prime}(t) = e^{t}$ and $y_{2}{}^{\prime}(t) = e^{t} + te^{t}$ (step 1). Then, we want to find functions $u^{\prime}(t)$ and $v^{\prime}(t)$ so that

$$\begin{array}{rll}

{u^{\prime}e^{t} + v^{\prime}te^{t}} & = & 0 \\

{u^{\prime}e^{t} + v^{\prime}\left( {e^{t} + te^{t}} \right)} & = & {\frac{e^{t}}{t^{2}}.}

\end{array}$$

Applying Cramer’s rule, we have

$$u^{\prime} = \frac{\left| \begin{array}{lc}

0 & {te^{t}} \\

\frac{e^{t}}{t^{2}} & {e^{t} + te^{t}}

\end{array} \right|}{\left| \begin{array}{lc}

e^{t} & {te^{t}} \\

e^{t} & {e^{t} + te^{t}}

\end{array} \right|} = \frac{0 - te^{t}\left( \frac{e^{t}}{t^{2}} \right)}{e^{t}\left( {e^{t} + te^{t}} \right) - e^{t}te^{t}} = \frac{- \frac{e^{2t}}{t}}{e^{2t}} = - \frac{1}{t}$$

and

$$v^{\prime} = \frac{\left| \begin{array}{ll}

e^{t} & 0 \\

e^{t} & \frac{e^{t}}{t^{2}}

\end{array} \right|}{\left| \begin{array}{lc}

e^{t} & {te^{t}} \\

e^{t} & {e^{t} + te^{t}}

\end{array} \right|} = \frac{e^{t}\left( \frac{e^{t}}{t^{2}} \right)}{e^{2t}} = \frac{1}{t^{2}}\ \text{(step 2).}$$

Integrating, we get

$$\begin{array}{l}

\\

\\

{u = \text{−}\left. \int{\frac{1}{t}dt} \right. = \text{−}\text{ln}|t|} \\

{v = \left. \int\frac{1}{t^{2}} \right.dt = - \frac{1}{t}\ \text{(step 3).}}

\end{array}$$

Then we have

$$\begin{array}{cl}

y_{p} & {= \text{−}e^{t}\text{ln}|t| - \frac{1}{t}te^{t}} \\

& {= \text{−}e^{t}\text{ln}|t| - e^{t}\ \text{(step 4).}}

\end{array}$$

The $e^{t}$ term is a solution to the complementary equation, so we don’t need to carry that term into our general solution explicitly. The general solution is

$$y(t) = c_{1}e^{t} + c_{2}te^{t} - e^{t}\text{ln}|t|\ \text{(step 5).}$$

2. The complementary equation is $y^{''} + y = 0$ with associated general solution $c_{1}\text{cos}\mspace{2mu} x + c_{2}\text{sin}\mspace{2mu} x\text{.}$ So, $y_{1}(x) = \text{cos}\mspace{2mu} x$ and $y_{2}(x) = \text{sin}\mspace{2mu} x$ (step 1). Then, we want to find functions $u^{\prime}(x)$ and $v^{\prime}(x)$ such that

$$\begin{array}{cll}

{u^{\prime}\text{cos}\mspace{2mu} x + v^{\prime}\text{sin}\mspace{2mu} x} & = & 0 \\

{\text{−}u^{\prime}\text{sin}\mspace{2mu} x + v^{\prime}\text{cos}\mspace{2mu} x} & = & {3\mspace{2mu}\text{sin}^{2}x\text{.}}

\end{array}$$

Applying Cramer’s rule, we have

$$u^{\prime} = \frac{\left| \begin{array}{cl}

0 & {\text{sin}\mspace{2mu} x} \\

{3\mspace{2mu}\text{sin}^{2}x} & {\text{cos}\mspace{2mu} x}

\end{array} \right|}{\left| \begin{array}{ll}

{\text{cos}\mspace{2mu} x} & {\text{sin}\mspace{2mu} x} \\

{\text{−}\text{sin}\mspace{2mu} x} & {\text{cos}\mspace{2mu} x}

\end{array} \right|} = \frac{0 - 3\mspace{2mu}\text{sin}^{3}x}{\text{cos}^{2}x + \text{sin}^{2}x} = -3\mspace{2mu}\text{sin}^{3}x$$

and

$$v^{\prime} = \frac{\left| \begin{array}{lc}

{\text{cos}\mspace{2mu} x} & 0 \\

{\text{−}\text{sin}\mspace{2mu} x} & {3\mspace{2mu}\text{sin}^{2}x}

\end{array} \right|}{\left| \begin{array}{ll}

{\text{cos}\mspace{2mu} x} & {\text{sin}\mspace{2mu} x} \\

{\text{−}\text{sin}\mspace{2mu} x} & {\text{cos}\mspace{2mu} x}

\end{array} \right|} = \frac{3\mspace{2mu}\text{sin}^{2}x\mspace{2mu}\text{cos}\mspace{2mu} x}{1} = 3\mspace{2mu}\text{sin}^{2}x\mspace{2mu}\text{cos}\mspace{2mu} x\ \text{(step 2).}$$

Integrating first to find *u*, we get

$$u = {\int{-3\mspace{2mu}\text{sin}^{3}x}}dx = -3\left\lbrack {\text{−}\frac{1}{3}\text{sin}^{2}x\mspace{2mu}\text{cos}\mspace{2mu} x + \frac{2}{3}{\int{\text{sin}\mspace{2mu} x}}dx} \right\rbrack = \text{sin}^{2}x\mspace{2mu}\text{cos}\mspace{2mu} x + 2\mspace{2mu}\text{cos}\mspace{2mu} x\text{.}$$

Now, we integrate to find *v*. Using substitution (with $w = \text{sin}\mspace{2mu} x$), we get

$$v = {\int{3\mspace{2mu}\text{sin}^{2}x\mspace{2mu}\text{cos}\mspace{2mu} xdx}} = {\int 3}w^{2}dw = w^{3} = \text{sin}^{3}x\text{.}$$

Then,

$$\begin{array}{cl}

y_{p} & {= \left( {\text{sin}^{2}x\mspace{2mu}\text{cos}\mspace{2mu} x + 2\mspace{2mu}\text{cos}\mspace{2mu} x} \right)\mspace{2mu}\text{cos}\mspace{2mu} x + \left( {\text{sin}^{3}x} \right)\mspace{2mu}\text{sin}\mspace{2mu} x} \\

& {= \text{sin}^{2}x\mspace{2mu}\text{cos}^{2}x + 2\mspace{2mu}\text{cos}^{2}x + \text{sin}^{4}x} \\

& {= 2\mspace{2mu}\text{cos}^{2}x + \text{sin}^{2}x\left( {\text{cos}^{2}x + \text{sin}^{2}x} \right)\qquad\text{(step 3).}} \\

& {= 2\mspace{2mu}\text{cos}^{2}x + \text{sin}^{2}x} \\

& {= \text{cos}^{2}x + 1}

\end{array}$$

The general solution is

$$y(x) = c_{1}\text{cos}\mspace{2mu} x + c_{2}\text{sin}\mspace{2mu} x + 1 + \text{cos}^{2}x\ \text{(step 4).}$$

Find the general solution to the following differential equations.

1. $y^{''} + y = \text{sec}\mspace{2mu} x$

2. $x^{''} - 2x^{\prime} + x = \frac{e^{t}}{t}$

Section 7.2 Exercises

Solve the following equations using the method of undetermined coefficients.

54\.

$2y^{''} - 5y^{\prime} - 12y = 6$

55.

$3y^{''} + y^{\prime} - 4y = 8$

56\.

$y^{''} - 6y^{\prime} + 5y = e^{\text{−}x}$

57.

$y^{''} + 16y = e^{-2x}$

58\.

$y^{''} - 4y = x^{2} + 1$

59.

$y^{''} - 4y^{\prime} + 4y = 8x^{2} + 4x$

60\.

$y^{''} - 2y^{\prime} - 3y = \text{sin}\mspace{2mu} 2x$

61.

$y^{''} + 2y^{\prime} + y = \text{sin}\mspace{2mu} x + \text{cos}\mspace{2mu} x$

62\.

$y^{''} + 9y = e^{x}\text{cos}\mspace{2mu} x$

63.

$y^{''} + y = 3\mspace{2mu}\text{sin}\mspace{2mu} 2x + x\mspace{2mu}\text{cos}\mspace{2mu} 2x$

64\.

$y^{''} + 3y^{\prime} - 28y = 10e^{4x}$

65.

$y^{''} + 10y^{\prime} + 25y = xe^{-5x} + 4$

In each of the following problems,

1. Write the form for the particular solution $y_{p}(x)$ for the method of undetermined coefficients.

2. \[T\] Use a computer algebra system to find a particular solution to the given equation.

66\.

$y^{''} - y^{\prime} - y = x + e^{\text{−}x}$

67.

$y^{''} - 3y = x^{2} - 4x + 11$

68\.

$y^{''} - y^{\prime} - 4y = e^{x}\text{cos}\mspace{2mu} 3x$

69.

$2y^{''} - y^{\prime} + y = \left( {x^{2} - 5x} \right)e^{\text{−}x}$

70\.

$4y^{''} + 5y^{\prime} - 2y = e^{2x} + x\mspace{2mu}\text{sin}\mspace{2mu} x$

71.

$y^{''} - y^{\prime} - 2y = x^{2}e^{x}\text{sin}\mspace{2mu} x$

Solve the differential equation using either the method of undetermined coefficients or the variation of parameters.

72\.

$y^{''} + 3y^{\prime} - 4y = 2e^{x}$

73.

$y^{''} + 2y^{\prime} = e^{3x}$

74\.

$y^{''} + 6y^{\prime} + 9y = e^{\text{−}x}$

75.

$y^{''} + 2y^{\prime} - 8y = 6e^{2x}$

Solve the differential equation using the method of variation of parameters.

76\.

$4y^{''} + y = 2\mspace{2mu}\text{sin}\mspace{2mu} x$

77.

$y^{''} - 9y = 8x$

78\.

$y^{''} + y = \text{sec}\mspace{2mu} x,\qquad 0 < x < \pi\text{/}2$

79.

$y^{''} + 4y = 3\mspace{2mu}\text{csc}\mspace{2mu} 2x,\quad 0 < x < \pi\text{/}2$

Find the unique solution satisfying the differential equation and the initial conditions given, where $y_{p}(x)$ is the particular solution.

80\.

$y^{''} - 2y^{\prime} + y = 12e^{x},$ $\text{y}_{p}(x) = 6x^{2}e^{x},$ $y(0) = 6,\ \ y\prime(0) = 0$

81.

$y^{''} - 7y^{\prime} = 4xe^{7x},$ $y_{p}(x) = \frac{2}{7}x^{2}e^{7x} - \frac{4}{49}xe^{7x},$ $y(0) = -1,\ \ y\prime(0) = 0$

82\.

$y^{''} + y = \text{cos}\mspace{2mu} x - 4\mspace{2mu}\text{sin}\mspace{2mu} x,$ $y_{p}(x) = 2x\mspace{2mu}\text{cos}\mspace{2mu} x + \frac{1}{2}x\mspace{2mu}\text{sin}\mspace{2mu} x,$ $y(0) = 8,\ \ y\prime(0) = -4$

83.

$y^{''} - 5y^{\prime} = e^{5x} + 8e^{-5x},$ $y_{p}(x) = \frac{1}{5}xe^{5x} + \frac{4}{25}e^{-5x},$ $y(0) = -2,\ \ y\prime(0) = 0$

In each of the following problems, two linearly independent solutions—$y_{1}$ and $y_{2}$—are given that satisfy the corresponding homogeneous equation. Use the method of variation of parameters to find a particular solution to the given nonhomogeneous equation. Assume *x* \> 0 in each exercise.

84\.

$x^{2}y^{''} + 2xy^{\prime} - 2y = 3x,$ $y_{1}(x) = x,\quad y_{2}(x) = x^{-2}$

85.

$x^{2}y^{''} - 2y = 10x^{2} - 1,$ $y_{1}(x) = x^{2},\quad y_{2}(x) = x^{-1}$

---

7.3 Applications

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/7-3-applications

7.3 Applications

• 7.3.1 Solve a second-order differential equation representing simple harmonic motion.
• 7.3.2 Solve a second-order differential equation representing damped simple harmonic motion.
• 7.3.3 Solve a second-order differential equation representing forced simple harmonic motion.
• 7.3.4 Solve a second-order differential equation representing charge and current in an RLC series circuit.

We saw in the chapter introduction that second-order linear differential equations are used to model many situations in physics and engineering. In this section, we look at how this works for systems of an object with mass attached to a vertical spring and an electric circuit containing a resistor, an inductor, and a capacitor connected in series. Models such as these can be used to approximate other more complicated situations; for example, bonds between atoms or molecules are often modeled as springs that vibrate, as described by these same differential equations.

Simple Harmonic Motion

Consider a mass suspended from a spring attached to a rigid support. (This is commonly called a spring-mass system.) Gravity is pulling the mass downward and the restoring force of the spring is pulling the mass upward. As shown in Figure 7.2, when these two forces are equal, the mass is said to be at the equilibrium position. If the mass is displaced from equilibrium, it oscillates up and down. This behavior can be modeled by a second-order constant-coefficient differential equation.

Let $x(t)$ denote the displacement of the mass from equilibrium. Note that for spring-mass systems of this type, it is customary to adopt the convention that down is positive. Thus, a positive displacement indicates the mass is *below* the equilibrium point, whereas a negative displacement indicates the mass is *above* equilibrium. Displacement is usually given in feet in the English system or meters in the metric system.

Consider the forces acting on the mass. The force of gravity is given by $mg\text{.}$ In the English system, mass is in slugs and the acceleration resulting from gravity is in feet per second squared. The acceleration resulting from gravity is constant, so in the English system, $g = 32$ ft/sec². Recall that 1 slug-foot/sec² is a pound, so the expression *mg* can be expressed in pounds. Metric system units are kilograms for mass and m/sec² for gravitational acceleration. In the metric system, we have $g = 9.8$ m/sec².

According to Hooke’s law, the restoring force of the spring is proportional to the displacement and acts in the opposite direction from the displacement, so the restoring force is given by $\text{−}k\left( {s + x} \right).$ The spring constant is given in pounds per foot in the English system and in newtons per meter in the metric system.

Now, by Newton’s second law, the sum of the forces on the system (gravity plus the restoring force) is equal to mass times acceleration, so we have

$$\begin{array}{cl}

{mx^{''}} & {= \text{−}k\left( {s + x} \right) + mg} \\

& {= \text{−}ks - kx + mg\text{.}}

\end{array}$$

However, by the way we have defined our equilibrium position, $mg = ks,$ the differential equation becomes

$$mx^{''} + kx = 0.$$

It is convenient to rearrange this equation and introduce a new variable, called the angular frequency, $\omega\text{.}$ Letting $\omega = \sqrt{k\text{/}m},$ we can write the equation as

$$x^{''} + \omega^{2}x = 0.$$ (7.5)

This differential equation has the general solution

$$x(t) = c_{1}\text{cos}\mspace{2mu}\omega t + c_{2}\text{sin}\mspace{2mu}\omega t,$$ (7.6)

which gives the position of the mass at any point in time. The motion of the mass is called simple harmonic motion. The period of this motion (the time it takes to complete one oscillation) is $T = \frac{2\pi}{\omega}$ and the frequency is $f = \frac{1}{T} = \frac{\omega}{2\pi}$ (Figure 7.3).

Simple Harmonic Motion

Assume an object weighing 2 lb stretches a spring 6 in. Find the equation of motion if the spring is released from the equilibrium position with an upward velocity of 16 ft/sec. What is the period of the motion?

Solution

We first need to find the spring constant. We have

$$\begin{array}{rll}

{mg} & = & {ks} \\

2 & = & {k\left( \frac{1}{2} \right)} \\

k & = & {4.}

\end{array}$$

We also know that weight *W* equals the product of mass *m* and the acceleration due to gravity *g*. In English units, the acceleration due to gravity is 32 ft/sec².

$$\begin{array}{rll}

W & = & {mg} \\

2 & = & {m(32)} \\

m & = & \frac{1}{16}

\end{array}$$

Thus, the differential equation representing this system is

$$\frac{1}{16}x^{''} + 4x = 0.$$

Multiplying through by 16, we get $x^{''} + 64x = 0,$ which can also be written in the form $x^{''} + {(8^{2})}x = 0.$ This equation has the general solution

$$x(t) = c_{1}\text{cos}\mspace{2mu}\left( {8t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {8t} \right).$$

The mass was released from the equilibrium position, so $x(0) = 0,$ and it had an initial upward velocity of 16 ft/sec, so $x^{\prime}(0) = -16.$ Applying these initial conditions to solve for $c_{1}$ and $c_{2}.$ gives

$$x(t) = -2\mspace{2mu}\text{sin}\mspace{2mu} 8t\text{.}$$

The period of this motion is $\frac{2\pi}{8} = \frac{\pi}{4}$ sec.

A 200-g mass stretches a spring 5 cm. Find the equation of motion of the mass if it is released from rest from a position 10 cm below the equilibrium position. What is the frequency of this motion?

Writing the general solution in the form $x(t) = c_{1}\text{cos}\mspace{2mu}\left( {\omega t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {\omega t} \right)$ has some advantages. It is easy to see the link between the differential equation and the solution, and the period and frequency of motion are evident. This form of the function tells us very little about the amplitude of the motion, however. In some situations, we may prefer to write the solution in the form

$$x(t) = A\mspace{2mu}\text{sin}\mspace{2mu}\left( {\omega t + \phi} \right).$$ (7.7)

Although the link to the differential equation is not as explicit in this case, the period and frequency of motion are still evident. Furthermore, the amplitude of the motion, *A*, is obvious in this form of the function. The constant $\phi$ is called a *phase shift* and has the effect of shifting the graph of the function to the left or right.

To convert the solution to this form, we want to find the values of *A* and $\phi$ such that

$$c_{1}\text{cos}\mspace{2mu}\left( {\omega t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {\omega t} \right) = A\mspace{2mu}\text{sin}\mspace{2mu}\left( {\omega t + \phi} \right).$$

We first apply the trigonometric identity

$$\text{sin}(\alpha + \beta) = \text{sin}\mspace{2mu}\alpha\mspace{2mu}\text{cos}\mspace{2mu}\beta + \text{cos}\mspace{2mu}\alpha\mspace{2mu}\text{sin}\mspace{2mu}\beta$$

to get

$$\begin{matrix}

{c_{1}\text{cos}\mspace{2mu}\left( {\omega t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {\omega t} \right)} & {= A\left( {\text{sin}\mspace{2mu}\left( {\omega t} \right)\mspace{2mu}\text{cos}\mspace{2mu}\phi + \text{cos}\mspace{2mu}\left( {\omega t} \right)\mspace{2mu}\text{sin}\mspace{2mu}\phi} \right)} \\

& {= A\mspace{2mu}\text{sin}\mspace{2mu}\phi\left( {\text{cos}\mspace{2mu}\left( {\omega t} \right)} \right) + A\mspace{2mu}\text{cos}\mspace{2mu}\phi\left( {\text{sin}\mspace{2mu}\left( {\omega t} \right)} \right).}

\end{matrix}$$

Thus,

$$c_{1} = A\mspace{2mu}\text{sin}\mspace{2mu}\phi\ \text{and}\ c_{2} = A\mspace{2mu}\text{cos}\mspace{2mu}\phi\text{.}$$

If we square both of these equations and add them together, we get

$$\begin{array}{cl}

{c_{1}^{2} + c_{2}^{2}} & {= A^{2}\text{sin}^{2}\phi + A^{2}\text{cos}^{2}\phi} \\

& {= A^{2}\left( {\text{sin}^{2}\phi + \text{cos}^{2}\phi} \right)} \\

& {= A^{2}.}

\end{array}$$

Thus,

$$A = \sqrt{c_{1}^{2} + c_{2}^{2}}.$$

Now, to find $\phi,$ go back to the equations for $c_{1}$ and $c_{2},$ but this time, divide the first equation by the second equation to get

$$\begin{array}{cl}

\frac{c_{1}}{c_{2}} & {= \frac{A\mspace{2mu}\text{sin}\mspace{2mu}\phi}{A\mspace{2mu}\text{cos}\mspace{2mu}\phi}} \\

& {= \text{tan}\mspace{2mu}\phi\text{.}}

\end{array}$$

Then,

$$\text{tan}\mspace{2mu}\phi = \frac{c_{1}}{c_{2}}.$$

We summarize this finding in the following theorem.

Solution to the Equation for Simple Harmonic Motion

The function $x(t) = c_{1}\text{cos}\mspace{2mu}\left( {\omega t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {\omega t} \right)$ can be written in the form $x(t) = A\mspace{2mu}\text{sin}\mspace{2mu}\left( {\omega t + \phi} \right),$ where $A = \sqrt{c_{1}^{2} + c_{2}^{2}}$ and $\text{tan}\mspace{2mu}\phi = \frac{c_{1}}{c_{2}}.$

Note that when using the formula $\text{tan}\mspace{2mu}\phi = \frac{c_{1}}{c_{2}}$ to find $\phi,$ we must take care to ensure $\phi$ is in the right quadrant (Figure 7.4).

Expressing the Solution with a Phase Shift

Express the following functions in the form $A\mspace{2mu}\text{sin}\mspace{2mu}\left( {\omega t + \phi} \right).$ What is the frequency of motion? The amplitude?

1. $x(t) = 2\mspace{2mu}\text{cos}\mspace{2mu}\left( {3t} \right) + \text{sin}\mspace{2mu}\left( {3t} \right)$

2. $x(t) = 3\mspace{2mu}\text{cos}\mspace{2mu}\left( {2t} \right) - 2\mspace{2mu}\text{sin}\mspace{2mu}\left( {2t} \right)$

Solution

1. We have

$$A = \sqrt{c_{1}^{2} + c_{2}^{2}} = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$$

and

$$\text{tan}\mspace{2mu}\phi = \frac{c_{1}}{c_{2}} = \frac{2}{1} = 2.$$

Note that both $c_{1}$ and $c_{2}$ are positive, so $\phi$ is in the first quadrant. Thus,

$$\phi \approx 1.107\ \text{rad,}$$

so we have

$$x(t) = 2\mspace{2mu}\text{cos}\mspace{2mu}\left( {3t} \right) + \text{sin}\mspace{2mu}\left( {3t} \right) = \sqrt{5}\text{sin}\mspace{2mu}\left( {3t + 1.107} \right).$$

The frequency is $\frac{\omega}{2\pi} = \frac{3}{2\pi} \approx 0.477.$ The amplitude is $\sqrt{5}.$

2. We have

$$A = \sqrt{c_{1}^{2} + c_{2}^{2}} = \sqrt{3^{2} + 2^{2}} = \sqrt{13}$$

and

$$\text{tan}\mspace{2mu}\phi = \frac{c_{1}}{c_{2}} = \frac{3}{-2} = - \frac{3}{2}.$$

Note that $c_{1}$ is positive but $c_{2}$ is negative, so $\phi$ is in the fourth quadrant. Thus,

$$\phi \approx - 0.983\ \text{rad,}$$

so we have

$$\begin{array}{cl}

{x(t)} & {= 3\mspace{2mu}\text{cos}\mspace{2mu}\left( {2t} \right) - 2\mspace{2mu}\text{sin}\mspace{2mu}\left( {2t} \right)} \\

& {= \sqrt{13}\text{sin}\mspace{2mu}\left( {2t - 0.983} \right).}

\end{array}$$

The frequency is $\frac{\omega}{2\pi} = \frac{2}{2\pi} \approx 0.318.$ The amplitude is $\sqrt{13}.$

Express the function $x(t) = \text{cos}\mspace{2mu}\left( {4t} \right) + 4\mspace{2mu}\text{sin}\mspace{2mu}\left( {4t} \right)$ in the form $A\mspace{2mu}\text{sin}\mspace{2mu}\left( {\omega t + \phi} \right).$ What is the frequency of motion? The amplitude?

Damped Vibrations

With the model just described, the motion of the mass continues indefinitely. Clearly, this doesn’t happen in the real world. In the real world, there is almost always some friction in the system, which causes the oscillations to die off slowly—an effect called *damping*. So now let’s look at how to incorporate that damping force into our differential equation.

Physical spring-mass systems almost always have some damping as a result of friction, air resistance, or a physical damper, called a *dashpot* (a pneumatic cylinder; see Figure 7.5).

Because damping is primarily a friction force, we assume it is proportional to the velocity of the mass and acts in the opposite direction. So the damping force is given by $\text{−}bx^{\prime}$ for some constant $b > 0.$ Again applying Newton’s second law, the differential equation becomes

$$mx^{''} + bx^{\prime} + kx = 0.$$

Then the associated characteristic equation is

$$m\lambda^{2} + b\lambda + k = 0.$$

Applying the quadratic formula, we have

$$\lambda = \frac{\text{−}b \pm \sqrt{b^{2} - 4mk}}{2m}.$$

Just as in Second-Order Linear Equations we consider three cases, based on whether the characteristic equation has distinct real roots, a repeated real root, or complex conjugate roots.

Case 1: $b^{2} > 4mk$

In this case, we say the system is *overdamped*. The general solution has the form

$$x(t) = c_{1}e^{\lambda_{1}t} + c_{2}e^{\lambda_{2}t},$$

where both $\lambda_{1}$ and $\lambda_{2}$ are less than zero. Because the exponents are negative, the displacement decays to zero over time, usually quite quickly. Overdamped systems do not oscillate (no more than one change of direction), but simply move back toward the equilibrium position. Figure 7.6 shows what typical overdamped behavior looks like.

Overdamped Spring-Mass System

A 16-lb mass is attached to a 10-ft spring. When the mass comes to rest in the equilibrium position, the spring measures 15 ft 4 in. The system is immersed in a medium that imparts a damping force equal to $\frac{5}{2}$ times the instantaneous velocity of the mass. Find the equation of motion if the mass is pushed upward from the equilibrium position with an initial upward velocity of 5 ft/sec. What is the position of the mass after 10 sec? Its velocity?

Solution

The mass stretches the spring 5 ft 4 in., or $\frac{16}{3}$ ft. Thus, $16 = \left( \frac{16}{3} \right)k,$ so $k = 3.$ We also have $m = \frac{16}{32} = \frac{1}{2},$ so the differential equation is

$$\frac{1}{2}x^{''} + \frac{5}{2}x^{\prime} + 3x = 0.$$

Multiplying through by 2 gives $x^{''} + 5x^{\prime} + 6x = 0,$ which has the general solution

$$x(t) = c_{1}e^{-2t} + c_{2}e^{-3t}.$$

Applying the initial conditions, $x(0) = 0$ and $x^{\prime}(0) = -5,$ we get

$$x(t) = -5e^{-2t} + 5e^{-3t}.$$

After 10 sec the mass is at position

$$x(10) = -5e^{-20} + 5e^{-30} \approx - 1.0305\ \times \ 10^{-8} \approx 0,$$

so it is, effectively, at the equilibrium position. We have $x^{\prime}(t) = 10e^{-2t} - 15e^{-3t},$ so after 10 sec the mass is moving at a velocity of

$$x^{\prime}(10) = 10e^{-20} - 15e^{-30} \approx 2.061\ \times \ 10^{-8} \approx 0.$$

After only 10 sec, the mass is barely moving.

A 2-kg mass is attached to a spring with spring constant 24 N/m. The system is then immersed in a medium imparting a damping force equal to 16 times the instantaneous velocity of the mass. Find the equation of motion if it is released from rest at a point 40 cm below equilibrium.

Case 2: $b^{2} = 4mk$

In this case, we say the system is *critically damped*. The general solution has the form

$$x(t) = c_{1}e^{\lambda_{1}t} + c_{2}te^{\lambda_{1}t},$$

where $\lambda_{1}$ is less than zero. The motion of a critically damped system is very similar to that of an overdamped system. It does not oscillate. However, with a critically damped system, if the damping is reduced even a little, oscillatory behavior results. From a practical perspective, physical systems are almost always either overdamped or underdamped (case 3, which we consider next). It is impossible to fine-tune the characteristics of a physical system so that $b^{2}$ and $4mk$ are exactly equal. Figure 7.7 shows what typical critically damped behavior looks like.

Critically Damped Spring-Mass System

A 1-kg mass stretches a spring 20 cm. The system is attached to a dashpot that imparts a damping force equal to 14 times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from equilibrium with an upward velocity of 3 m/sec.

Solution

We have $mg = 1(9.8) = 0.2k,$ so $k = 49.$ Then, the differential equation is

$$x^{''} + 14x^{\prime} + 49x = 0,$$

which has general solution

$$x(t) = c_{1}e^{-7t} + c_{2}te^{-7t}.$$

Applying the initial conditions $x(0) = 0$ and $x^{\prime}(0) = -3$ gives

$$x(t) = -3te^{-7t}.$$

A 1-lb weight stretches a spring 6 in., and the system is attached to a dashpot that imparts a damping force equal to half the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 6 in. below equilibrium.

Case 3: $b^{2} < 4mk$

In this case, we say the system is *underdamped*. The general solution has the form

$$x(t) = e^{\alpha t}\left( {c_{1}\text{cos}\mspace{2mu}\left( {\beta t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {\beta t} \right)} \right),$$

where $\alpha$ is less than zero. Underdamped systems do oscillate because of the sine and cosine terms in the solution. However, the exponential term dominates eventually, so the amplitude of the oscillations decreases over time. Figure 7.8 shows what typical underdamped behavior looks like.

Note that for all damped systems, $\underset{t\rightarrow\infty}{\text{lim}}x(t) = 0.$ The system always approaches the equilibrium position over time.

Underdamped Spring-Mass System

A 16-lb weight stretches a spring 3.2 ft. Assume the damping force on the system is equal to the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 9 in. below equilibrium.

Solution

We have $k = \frac{16}{3.2} = 5$ and $m = \frac{16}{32} = \frac{1}{2},$ so the differential equation is

$$\frac{1}{2}x^{''} + x^{\prime} + 5x = 0,\ \text{or}\ x^{''} + 2x^{\prime} + 10x = 0.$$

This equation has the general solution

$$x(t) = e^{\text{−}t}\left( {c_{1}\text{cos}\mspace{2mu}\left( {3t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {3t} \right)} \right).$$

Applying the initial conditions, $x(0) = \frac{3}{4}$ and $x^{\prime}(0) = 0,$ we get

$$x(t) = e^{\text{−}t}\left( {\frac{3}{4}\mspace{2mu}\text{cos}\mspace{2mu}\left( {3t} \right) + \frac{1}{4}\mspace{2mu}\text{sin}\mspace{2mu}\left( {3t} \right)} \right).$$

A 1-kg mass stretches a spring 49 cm. The system is immersed in a medium that imparts a damping force equal to four times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 24 cm above equilibrium.

Chapter Opener: Modeling a Motorcycle Suspension System

For motocross riders, the suspension systems on their motorcycles are very important. The off-road courses on which they ride often include jumps, and losing control of the motorcycle when they land could cost them the race.

This suspension system can be modeled as a damped spring-mass system. We define our frame of reference with respect to the frame of the motorcycle. Assume the end of the shock absorber attached to the motorcycle frame is fixed. Then, the “mass” in our spring-mass system is the motorcycle wheel. We measure the position of the wheel with respect to the motorcycle frame. This may seem counterintuitive, since, in many cases, it is actually the motorcycle frame that moves, but this frame of reference preserves the development of the differential equation that was done earlier. As with earlier development, we define the downward direction to be positive.

When the motorcycle is lifted by its frame, the wheel hangs freely and the spring is uncompressed. This is the spring’s natural position. When the motorcycle is placed on the ground and the rider mounts the motorcycle, the spring compresses and the system is in the equilibrium position (Figure 7.10).

This system can be modeled using the same differential equation we used before:

$$mx^{''} + bx^{\prime} + kx = 0.$$

A motocross motorcycle weighs 204 lb, and we assume a rider weight of 180 lb. When the rider mounts the motorcycle, the suspension compresses 4 in., then comes to rest at equilibrium. The suspension system provides damping equal to 240 times the instantaneous vertical velocity of the motorcycle (and rider).

1. Set up the differential equation that models the behavior of the motorcycle suspension system.

2. We are interested in what happens when the motorcycle lands after taking a jump. Let time $t = 0$ denote the time when the motorcycle first contacts the ground. If the motorcycle hits the ground with a velocity of 10 ft/sec downward, find the equation of motion of the motorcycle after the jump.

3. Graph the equation of motion over the first second after the motorcycle hits the ground.

Solution

1. We have defined equilibrium to be the point where $mg = ks,$ so we have

$$\begin{array}{rll}

{mg} & = & {ks} \\

384 & = & {k\left( \frac{1}{3} \right)} \\

k & = & {1152.}

\end{array}$$

We also have

$$\begin{array}{rll}

W & = & {mg} \\

384 & = & {m(32)} \\

m & = & {12.}

\end{array}$$

Therefore, the differential equation that models the behavior of the motorcycle suspension is

$$12x^{''} + 240x^{\prime} + 1152x = 0.$$

Dividing through by 12, we get

$$x^{''} + 20x^{\prime} + 96x = 0.$$

2. The differential equation found in part a. has the general solution

$$x(t) = c_{1}e^{-8t} + c_{2}e^{-12t}.$$

Now, to determine our initial conditions, we consider the position and velocity of the motorcycle wheel when the wheel first contacts the ground. Since the motorcycle was in the air prior to contacting the ground, the wheel was hanging freely and the spring was uncompressed. Therefore the wheel is 4 in. $\left( \frac{1}{3}\ \text{ft} \right)$ below the equilibrium position (with respect to the motorcycle frame), and we have $x(0) = \frac{1}{3}.$ According to the problem statement, the motorcycle has a velocity of 10 ft/sec downward when the motorcycle contacts the ground, so $x^{\prime}(0) = 10.$ Applying these initial conditions, we get $c_{1} = \frac{7}{2}$ and $c_{2} = \text{−}\left( \frac{19}{6} \right),$ so the equation of motion is

$$x(t) = \frac{7}{2}e^{-8t} - \frac{19}{6}e^{-12t}.$$

3. The graph is shown in Figure 7.11.

Landing Vehicle

NASA is planning a mission to Mars. To save money, engineers have decided to adapt one of the moon landing vehicles for the new mission. However, they are concerned about how the different gravitational forces will affect the suspension system that cushions the craft when it touches down. The acceleration resulting from gravity on the moon is 1.6 m/sec², whereas on Mars it is 3.7 m/sec².

The suspension system on the craft can be modeled as a damped spring-mass system. In this case, the spring is below the moon lander, so the spring is slightly compressed at equilibrium, as shown in Figure 7.12.

We retain the convention that down is positive. Despite the new orientation, an examination of the forces affecting the lander shows that the same differential equation can be used to model the position of the landing craft relative to equilibrium:

$$mx^{''} + bx^{\prime} + kx = 0,$$

where *m* is the mass of the lander, *b* is the damping coefficient, and *k* is the spring constant.

1. The lander has a mass of 15,000 kg and the spring is 2 m long when uncompressed. The lander is designed to compress the spring 0.5 m to reach the equilibrium position under lunar gravity. The dashpot imparts a damping force equal to 48,000 times the instantaneous velocity of the lander. Set up the differential equation that models the motion of the lander when the craft lands on the moon.

2. Let time $t = 0$ denote the instant the lander touches down. The rate of descent of the lander can be controlled by the crew, so that it is descending at a rate of 2 m/sec when it touches down. Find the equation of motion of the lander on the moon.

3. If the lander is traveling too fast when it touches down, it could fully compress the spring and “bottom out.” Bottoming out could damage the landing craft and must be avoided at all costs. Graph the equation of motion found in part 2. If the spring is 0.5 m long when fully compressed, will the lander be in danger of bottoming out?

4. Assuming NASA engineers make no adjustments to the spring or the damper, how far does the lander compress the spring to reach the equilibrium position under Martian gravity?

5. If the lander crew uses the same procedures on Mars as on the moon, and keeps the rate of descent to 2 m/sec, will the lander bottom out when it lands on Mars?

6. What adjustments, if any, should the NASA engineers make to use the lander safely on Mars?

Forced Vibrations

The last case we consider is when an external force acts on the system. In the case of the motorcycle suspension system, for example, the bumps in the road act as an external force acting on the system. Another example is a spring hanging from a support; if the support is set in motion, that motion would be considered an external force on the system. We model these forced systems with the nonhomogeneous differential equation

$$mx^{''} + bx^{\prime} + kx = f(t),$$ (7.8)

where the external force is represented by the $f(t)$ term. As we saw in Nonhomogeneous Linear Equations, differential equations such as this have solutions of the form

$$x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t) + x_{p}(t),$$

where $c_{1}x_{1}(t) + c_{2}x_{2}(t)$ is the general solution to the complementary equation and $x_{p}(t)$ is a particular solution to the nonhomogeneous equation. If the system is damped, $\underset{t\rightarrow\infty}{\text{lim}}c_{1}x_{1}(t) + c_{2}x_{2}(t) = 0.$ Since these terms do not affect the long-term behavior of the system, we call this part of the solution the *transient solution*. The long-term behavior of the system is determined by $x_{p}(t),$ so we call this part of the solution the steady-state solution.

This website shows a simulation of forced vibrations.

Forced Vibrations

A mass of 1 slug stretches a spring 2 ft and comes to rest at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. Find the equation of motion if an external force equal to $f(t) = 8\mspace{2mu}\text{sin}\mspace{2mu}\left( {4t} \right)$ is applied to the system beginning at time $t = 0.$ What is the transient solution? What is the steady-state solution?

Solution

We have $mg = 1(32) = 2k,$ so $k = 16$ and the differential equation is

$$x^{''} + 8x^{\prime} + 16x = 8\mspace{2mu}\text{sin}\mspace{2mu}\left( {4t} \right).$$

The general solution to the complementary equation is

$$c_{1}e^{-4t} + c_{2}te^{-4t}.$$

Assuming a particular solution of the form $x_{p}(t) = A\mspace{2mu}\text{cos}\mspace{2mu}\left( {4t} \right) + B\mspace{2mu}\text{sin}\mspace{2mu}\left( {4t} \right)$ and using the method of undetermined coefficients, we find $x_{p}(t) = - \frac{1}{4}\mspace{2mu}\text{cos}\mspace{2mu}\left( {4t} \right),$ so

$$x(t) = c_{1}e^{-4t} + c_{2}te^{-4t} - \frac{1}{4}\mspace{2mu}\text{cos}\mspace{2mu}\left( {4t} \right).$$

At $t = 0,$ the mass is at rest in the equilibrium position, so $x(0) = x^{\prime}(0) = 0.$ Applying these initial conditions to solve for $c_{1}$ and $c_{2},$ we get

$$x(t) = \frac{1}{4}e^{-4t} + te^{-4t} - \frac{1}{4}\mspace{2mu}\text{cos}\mspace{2mu}\left( {4t} \right).$$

The transient solution is $\frac{1}{4}e^{-4t} + te^{-4t}.$ The steady-state solution is $- \frac{1}{4}\mspace{2mu}\text{cos}\mspace{2mu}\left( {4t} \right).$

A mass of 2 kg is attached to a spring with constant 32 N/m and comes to rest in the equilibrium position. Beginning at time $t = 0,$ an external force equal to$f(t) = 68e^{-2t}\text{cos}\mspace{2mu}\left( {4t} \right)$ is applied to the system. Find the equation of motion if there is no damping. What is the transient solution? What is the steady-state solution?

Resonance

Consider an undamped system exhibiting simple harmonic motion. In the real world, we never truly have an undamped system; –some damping always occurs. For theoretical purposes, however, we could imagine a spring-mass system contained in a vacuum chamber. With no air resistance, the mass would continue to move up and down indefinitely.

The frequency of the resulting motion, given by $f = \frac{1}{T} = \frac{\omega}{2\pi},$ is called the *natural frequency of the system*. If an external force acting on the system has a frequency close to the natural frequency of the system, a phenomenon called *resonance* results. The external force reinforces and amplifies the natural motion of the system.

1. Consider the differential equation $x^{''} + x = 0.$ Find the general solution. What is the natural frequency of the system?

2. Now suppose this system is subjected to an external force given by $f(t) = 5\mspace{2mu}\text{cos}\mspace{2mu} t\text{.}$ Solve the initial-value problem $x^{''} + x = 5\mspace{2mu}\text{cos}\mspace{2mu} t,$ $x(0) = 0,$ $x^{\prime}(0) = 1.$

3. Graph the solution. What happens to the behavior of the system over time?

4. In the real world, there is always some damping. However, if the damping force is weak, and the external force is strong enough, real-world systems can still exhibit resonance. One of the most famous examples of resonance is the collapse of the Tacoma Narrows Bridge on November 7, 1940. The bridge had exhibited strange behavior ever since it was built. The roadway had a strange “bounce” to it. On the day it collapsed, a strong windstorm caused the roadway to twist and ripple violently. The bridge was unable to withstand these forces and it ultimately collapsed. Experts believe the windstorm exerted forces on the bridge that were very close to its natural frequency, and the resulting resonance ultimately shook the bridge apart.

This website contains more information about the collapse of the Tacoma Narrows Bridge.

During the short time the Tacoma Narrows Bridge stood, it became quite a tourist attraction. Several people were on site the day the bridge collapsed, and one of them caught the collapse on film. Watch the video to see the collapse.

5. Another real-world example of resonance is a singer shattering a crystal wineglass when she sings just the right note. When someone taps a crystal wineglass or wets a finger and runs it around the rim, a tone can be heard. That note is created by the wineglass vibrating at its natural frequency. If a singer then sings that same note at a high enough volume, the glass shatters as a result of resonance.

The TV show *Mythbusters* aired an episode on this phenomenon. Adam Savage described the experience. Watch this video for his account.

The *RLC* Series Circuit

Consider an electrical circuit containing a resistor, an inductor, and a capacitor, as shown in Figure 7.10. Such a circuit is called an *RLC* series circuit. *RLC* circuits are used in many electronic systems, most notably as tuners in AM/FM radios. The tuning knob varies the capacitance of the capacitor, which in turn tunes the radio. Such circuits can be modeled by second-order, constant-coefficient differential equations.

Let $I(t)$ denote the current in the *RLC* circuit and $q(t)$ denote the charge on the capacitor. Furthermore, let *L* denote inductance in henrys (H), *R* denote resistance in ohms $\left( \text{Ω} \right),$ and *C* denote capacitance in farads (F). Last, let $E(t)$ denote electric potential in volts (V).

Kirchhoff’s voltage rule states that the sum of the voltage drops around any closed loop must be zero. So, we need to consider the voltage drops across the inductor (denoted $E_{L}$), the resistor (denoted $E_{R}$), and the capacitor (denoted $E_{C}$). Because the *RLC* circuit shown in Figure 7.10 includes a voltage source, $E(t),$ which adds voltage to the circuit, we have $E_{L} + E_{R} + E_{C} = E(t).$

We present the formulas below without further development. Those of you interested in the derivation of these formulas should consult a physics text. Using Faraday’s law and Lenz’s law, the voltage drop across an inductor can be shown to be proportional to the instantaneous rate of change of current, with proportionality constant *L*. Thus,

$$E_{L} = L\frac{dI}{dt}.$$

Next, according to Ohm’s law, the voltage drop across a resistor is proportional to the current passing through the resistor, with proportionality constant *R*. Therefore,

$$E_{R} = RI\text{.}$$

Last, the voltage drop across a capacitor is proportional to the charge, *q*, on the capacitor, with proportionality constant ${1\text{/}C}.$ Thus,

$$E_{C} = \frac{1}{C}q\text{.}$$

Adding these terms together, we get

$$L\frac{dI}{dt} + RI + \frac{1}{C}q = E(t).$$

Noting that $I = {\left( {dq} \right)\text{/}\left( {dt} \right)},$ this becomes

$$L\frac{d^{2}q}{dt^{2}} + R\frac{dq}{dt} + \frac{1}{C}q = E(t).$$ (7.9)

Mathematically, this system is analogous to the spring-mass systems we have been examining in this section.

The *RLC* Series Circuit

Find the charge on the capacitor in an *RLC* series circuit where $L = {5\text{/}3}$ H, $R = 10\text{Ω},$ $C = {1\text{/}30}$ F, and $E(t) = 300$ V. Assume the initial charge on the capacitor is 0 C and the initial current is 9 A. What happens to the charge on the capacitor over time?

Solution

We have

$$\begin{array}{rll}

& & \\

& & \\

{L\frac{d^{2}q}{dt^{2}} + R\frac{dq}{dt} + \frac{1}{C}q} & = & {E(t)} \\

{\frac{5}{3}\ \frac{d^{2}q}{dt^{2}} + 10\frac{dq}{dt} + 30q} & = & 300 \\

{\frac{d^{2}q}{dt^{2}} + 6\frac{dq}{dt} + 18q} & = & {180.}

\end{array}$$

The general solution to the complementary equation is

$$e^{-3t}\left( {c_{1}\text{cos}\mspace{2mu}\left( {3t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {3t} \right)} \right).$$

Assume a particular solution of the form $q_{p} = A,$ where $A$ is a constant. Using the method of undetermined coefficients, we find $A = 10.$ So,

$$q(t) = e^{-3t}\left( {c_{1}\text{cos}\mspace{2mu}\left( {3t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {3t} \right)} \right) + 10.$$

Applying the initial conditions $q(0) = 0$ and $i(0) = \left( {\left( {dq} \right)\text{/}\left( {dt} \right)} \right)(0) = 9,$ we find $c_{1} = -10$ and $c_{2} = -7.$ So the charge on the capacitor is

$$q(t) = -10e^{-3t}\text{cos}\mspace{2mu}\left( {3t} \right) - 7e^{-3t}\text{sin}\mspace{2mu}\left( {3t} \right) + 10.$$

Looking closely at this function, we see the first two terms will decay over time (as a result of the negative exponent in the exponential function). Therefore, the capacitor eventually approaches a steady-state charge of 10 C.

Find the charge on the capacitor in an *RLC* series circuit where $L = {1\text{/}5}$ H, $R = {2\text{/}5}\text{Ω},$ $C = {1\text{/}2}$ F, and $E(t) = 50$ V. Assume the initial charge on the capacitor is 0 C and the initial current is 4 A.

Section 7.3 Exercises

86\.

A mass weighing 4 lb stretches a spring 8 in. Find the equation of motion if the spring is released from the equilibrium position with a downward velocity of 12 ft/sec. What is the period and frequency of the motion?

87.

A mass weighing 2 lb stretches a spring 2 ft. Find the equation of motion if the spring is released from 2 in. below the equilibrium position with an upward velocity of 8 ft/sec. What is the period and frequency of the motion?

88\.

A 100-g mass stretches a spring 0.1 m. Find the equation of motion of the mass if it is released from rest from a position 20 cm below the equilibrium position. What is the frequency of this motion?

89.

A 400-g mass stretches a spring 5 cm. Find the equation of motion of the mass if it is released from rest from a position 15 cm below the equilibrium position. What is the frequency of this motion?

90\.

A block has a mass of 9 kg and is attached to a vertical spring with a spring constant of 0.25 N/m. The block is stretched 0.75 m below its equilibrium position and released.

1. Find the position function $x(t)$ of the block.

2. Find the period and frequency of the vibration.

3. Sketch a graph of $x(t).$

4. At what time does the block first pass through the equilibrium position?

91.

A block has a mass of 5 kg and is attached to a vertical spring with a spring constant of 20 N/m. The block is released from the equilibrium position with a downward velocity of 10 m/sec.

1. Find the position function $x(t)$ of the block.

2. Find the period and frequency of the vibration.

3. Sketch a graph of $x(t).$

4. At what time does the block first pass through the equilibrium position?

92\.

A 1-kg mass is attached to a vertical spring with a spring constant of 21 N/m. The resistance in the spring-mass system is equal to 10 times the instantaneous velocity of the mass.

1. Find the equation of motion if the mass is released from a position 2 m below its equilibrium position with a downward velocity of 2 m/sec.

2. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.

93.

An 800-lb weight (25 slugs) is attached to a vertical spring with a spring constant of 226 lb/ft. The system is immersed in a medium that imparts a damping force equal to 10 times the instantaneous velocity of the mass.

1. Find the equation of motion if it is released from a position 20 ft below its equilibrium position with a downward velocity of 41 ft/sec.

2. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.

94\.

A 9-kg mass is attached to a vertical spring with a spring constant of 16 N/m. The system is immersed in a medium that imparts a damping force equal to 24 times the instantaneous velocity of the mass.

1. Find the equation of motion if it is released from its equilibrium position with an upward velocity of 4 m/sec.

2. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.

95.

A 1-kg mass stretches a spring 61.25 cm. The resistance in the spring-mass system is equal to eight times the instantaneous velocity of the mass.

1. Find the equation of motion if the mass is released from a position 5 m below its equilibrium position with an upward velocity of 10 m/sec.

2. Determine whether the motion is overdamped, critically damped, or underdamped.

96\.

A 32-lb weight (1 slug) stretches a vertical spring 128 in. The resistance in the spring-mass system is equal to four times the instantaneous velocity of the mass.

1. Find the equation of motion if it is released from its equilibrium position with a downward velocity of 12 ft/sec.

2. Determine whether the motion is overdamped, critically damped, or underdamped.

97.

A 64-lb weight is attached to a vertical spring with a spring constant of 4.625 lb/ft. The resistance in the spring-mass system is equal to the instantaneous velocity. The weight is set in motion from a position 1 ft below its equilibrium position with an upward velocity of 2 ft/sec. Is the mass above or below the equilibrium position at the end of $\pi$ sec? By what distance?

98\.

A mass that weighs 8 lb stretches a spring 6 inches. The system is acted on by an external force of $8\mspace{2mu}\text{sin}\mspace{2mu} 8t$ lb. If the mass is pulled down 3 inches and then released, determine the position of the mass at any time.

99.

A mass that weighs 6 lb stretches a spring 3 in. The system is acted on by an external force of $8\mspace{2mu}\text{sin}\mspace{2mu}\left( {4t} \right)$ lb. If the mass is pulled down 1 inch and then released, determine the position of the mass at any time.

100\.

Find the charge on the capacitor in an *RLC* series circuit where $L = 40$ H, $R = 30\text{Ω},$ $C = {1\text{/}200}$ F, and $E(t) = 200$ V. Assume the initial charge on the capacitor is 7 C and the initial current is 0 A.

101.

Find the charge on the capacitor in an *RLC* series circuit where $L = 2$ H, $R = 24\text{Ω},$ $C = 0.005$ F, and $E(t) = 12\mspace{2mu}\text{sin}\mspace{2mu} 10t$ V. Assume the initial charge on the capacitor is 0.001 C and the initial current is 0 A.

102\.

A series circuit consists of a device where $L = 1$ H, $R = 20\text{Ω},$ $C = 0.002$ F, and $E(t) = 12$ V. If the initial charge and current are both zero, find the charge and current at time *t*.

103.

A series circuit consists of a device where $L = \frac{1}{2}$ H, $R = 10\text{Ω},$ $C = \frac{1}{50}$ F, and $E(t) = 250$ V. If the initial charge on the capacitor is 0 C and the initial current is 18 A, find the charge and current at time *t*.

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7.4 Series Solutions of Differential Equations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/7-4-series-solutions-of-differential-equations

7.4 Series Solutions of Differential Equations

• 7.4.1 Use power series to solve first-order and second-order differential equations.

In Introduction to Power Series, we studied how functions can be represented as power series, $y(x) = \sum\limits_{n = 0}^{\infty}a_{n}x^{n}.$ We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives $y^{\prime}(x) = {\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1}$ and $y^{''}(x) = {\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2}.$ In some cases, these power series representations can be used to find solutions to differential equations.

Be aware that this subject is given only a very brief treatment in this text. Most introductory differential equations textbooks include an entire chapter on power series solutions. This text has only a single section on the topic, so several important issues are not addressed here, particularly issues related to existence of solutions. The examples and exercises in this section were chosen for which power solutions exist. However, it is not always the case that power solutions exist. Those of you interested in a more rigorous treatment of this topic should consult a differential equations text.

Finding Power Series Solutions to Differential Equations

1. Assume the differential equation has a solution of the form $y(x) = \sum\limits_{n = 0}^{\infty}a_{n}x^{n}.$

2. Differentiate the power series term by term to get $y^{\prime}(x) = {\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1}$ and $y^{''}(x) = {\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2}.$

3. Substitute the power series expressions into the differential equation.

4. Re-index sums as necessary to combine terms and simplify the expression.

5. Equate coefficients of like powers of $x$ to determine values for the coefficients $a_{n}$ in the power series.

6. Substitute the coefficients back into the power series and write the solution.

Series Solutions to Differential Equations

Find a power series solution for the following differential equations.

1. $y^{''} - y = 0$

2. $(x^{2} - 1)y^{''} + 6xy^{\prime} + 4y = -4$

Solution

1. Assume $y(x) = \sum\limits_{n = 0}^{\infty}a_{n}x^{n}$ (step 1). Then, $y^{\prime}(x) = {\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1}$ and $y^{''}(x) = {\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2}$ (step 2). We want to find values for the coefficients $a_{n}$ such that

$$\begin{array}{rll}

{y^{''} - y} & = & 0 \\

{{\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} - \sum\limits_{n = 0}^{\infty}a_{n}x^{n}} & = & {0\ \text{(step 3).}}

\end{array}$$

We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with $n = 0.$

To re-index the first term, replace *n* with $n + 2$ inside the sum, and change the lower summation limit to $n = 0.$ We get

$${\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} = {\sum\limits_{n = 0}^{\infty}{(n + 2)}}(n + 1)a_{n + 2}x^{n}.$$

This gives

$$\begin{array}{rll}

{{\sum\limits_{n = 0}^{\infty}{(n + 2)}}(n + 1)a_{n + 2}x^{n} - \sum\limits_{n = 0}^{\infty}a_{n}x^{n}} & = & 0 \\

{\sum\limits_{n = 0}^{\infty}\left\lbrack {(n + 2)(n + 1)a_{n + 2} - a_{n}} \right\rbrack x^{n}} & = & {0\ \text{(step 4).}}

\end{array}$$

Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of *x* are zero. So we have

$$(n + 2)(n + 1)a_{n + 2} - a_{n} = 0\ \text{for}\ n = 0,1,2\text{,….}$$

This recurrence relationship allows us to express each coefficient $a_{n}$ in terms of the coefficient two terms earlier. This yields one expression for even values of *n* and another expression for odd values of *n*. Looking first at the equations involving even values of *n*, we see that

$$\begin{array}{clc}

a_{2} & = & \frac{a_{0}}{2} \\

a_{4} & = & {\frac{a_{2}}{4 \cdot 3} = \frac{a_{0}}{4!}} \\

a_{6} & = & {\frac{a_{4}}{6 \cdot 5} = \frac{a_{0}}{6!}} \\

& \text{⋮.} &

\end{array}$$

Thus, in general, when *n* is even, $a_{n} = \frac{a_{0}}{n!}$ (step 5).

For the equations involving odd values of *n*, we see that

$$\begin{array}{clc}

a_{3} & = & {\frac{a_{1}}{3 \cdot 2} = \frac{a_{1}}{3!}} \\

a_{5} & = & {\frac{a_{3}}{5 \cdot 4} = \frac{a_{1}}{5!}} \\

a_{7} & = & {\frac{a_{5}}{7 \cdot 6} = \frac{a_{1}}{7!}} \\

& \text{⋮.} &

\end{array}$$

Therefore, in general, when *n* is odd, $a_{n} = \frac{a_{1}}{n!}$ (step 5 continued).

Putting this together, we have

$$\begin{array}{cl}

{y(x)} & {= {\sum\limits_{n = 0}^{\infty}a_{n}}x^{n}} \\

& {= a_{0} + a_{1}x + \frac{a_{0}}{2}x^{2} + \frac{a_{1}}{3!}x^{3} + \frac{a_{0}}{4!}x^{4} + \frac{a_{1}}{5!}x^{5} + \text{⋯.}}

\end{array}$$

Re-indexing the sums to account for the even and odd values of *n* separately, we obtain

$$y(x) = a_{0}\sum\limits_{k = 0}^{\infty}\frac{1}{(2k)!}x^{2k} + a_{1}\sum\limits_{k = 0}^{\infty}\frac{1}{(2k + 1)!}x^{2k + 1}\ \text{(step 6).}$$

Analysis for part a.

As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since $y(x) = c_{1}e^{x} + c_{2}e^{\text{−}x}$ is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.

Fortunately, after writing the power series representations of $e^{x}$ and $e^{\text{−}x},$ and doing some algebra, we find that if we choose

$$c_{0} = \frac{(a_{0} + a_{1})}{2},\quad c_{1} = \frac{(a_{0} - a_{1})}{2},$$

we then have $a_{0} = c_{0} + c_{1}$ and$a_{1} = c_{0} - c_{1},$ and

$$\begin{array}{cl}

{y(x)} & {= a_{0} + a_{1}x + \frac{a_{0}}{2}x^{2} + \frac{a_{1}}{3!}x^{3} + \frac{a_{0}}{4!}x^{4} + \frac{a_{1}}{5!}x^{5} + \text{⋯}} \\

& {= (c_{0} + c_{1}) + (c_{0} - c_{1})x + \frac{(c_{0} + c_{1})}{2}x^{2} + \frac{(c_{0} - c_{1})}{3!}x^{3} + \frac{(c_{0} + c_{1})}{4!}x^{4} + \frac{(c_{0} - c_{1})}{5!}x^{5} + \text{⋯}} \\

& {= c_{0}{\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n!}} + c_{1}{\sum\limits_{n = 0}^{\infty}\frac{{(\text{−}x)}^{n}}{n!}}} \\

& {= c_{0}e^{x} + c_{1}e^{\text{−}x}.}

\end{array}$$

So we have, in fact, found the same general solution. Note that this choice of $c_{1}$ and $c_{2}$ is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.

2. Assume $y(x) = \sum\limits_{n = 0}^{\infty}a_{n}x^{n}$ (step 1). Then, $y^{\prime}(x) = {\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1}$ and $y^{''}(x) = {\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2}$ (step 2). We want to find values for the coefficients $a_{n}$ such that

$$\begin{array}{rll}

& & \\

& & \\

{(x^{2} - 1)y^{''} + 6xy^{\prime} + 4y} & = & -4 \\

{\left( {x^{2} - 1} \right){\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} + 6x{\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1} + 4{\sum\limits_{n = 0}^{\infty}a_{n}}x^{n}} & = & -4 \\

{x^{2}{\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} - {\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} + 6x{\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1} + 4{\sum\limits_{n = 0}^{\infty}a_{n}}x^{n}} & = & {-4.}

\end{array}$$

Taking the external factors inside the summations, we get

$${\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n} - {\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} + {\sum\limits_{n = 1}^{\infty}{6n}}a_{n}x^{n} + {\sum\limits_{n = 0}^{\infty}{4a_{n}}}x^{n} = -4\ \text{(step 3).}$$

Now, in the first summation, we see that when $n = 0$ or $n = 1,$ the term evaluates to zero, so we can add these terms back into our sum to get

$$\sum\limits_{n = 2}^{\infty}n\left( {n - 1} \right)a_{n}x^{n} = \sum\limits_{n = 0}^{\infty}n\left( {n - 1} \right)a_{n}x^{n}.$$

Similarly, in the third term, we see that when $n = 0,$ the expression evaluates to zero, so we can add that term back in as well. We have

$${\sum\limits_{n = 1}^{\infty}6}na_{n}x^{n} = {\sum\limits_{n = 0}^{\infty}6}na_{n}x^{n}.$$

Then, we need only shift the indices in our second term. We get

$${\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} = {\sum\limits_{n = 0}^{\infty}{(n + 2)}}(n + 1)a_{n + 2}x^{n}.$$

Thus, we have

$$\begin{array}{rll}

{{\sum\limits_{n = 0}^{\infty}n}(n - 1)a_{n}x^{n} - {\sum\limits_{n = 0}^{\infty}{(n + 2)}}(n + 1)a_{n + 2}x^{n} + {\sum\limits_{n = 0}^{\infty}{6n}}a_{n}x^{n} + {\sum\limits_{n = 0}^{\infty}{4a_{n}}}x^{n}} & = & {-4\ \text{(step 4).}} \\

{\sum\limits_{n = 0}^{\infty}\left\lbrack {n(n - 1)a_{n} - (n + 2)(n + 1)a_{n + 2} + 6na_{n} + 4a_{n}} \right\rbrack x^{n}} & = & -4 \\

{\sum\limits_{n = 0}^{\infty}\left\lbrack {(n^{2} - n)a_{n} + 6na_{n} + 4a_{n} - (n + 2)(n + 1)a_{n + 2}} \right\rbrack x^{n}} & = & -4 \\

{\sum\limits_{n = 0}^{\infty}\left\lbrack {n^{2}a_{n} + 5na_{n} + 4a_{n} - (n + 2)(n + 1)a_{n + 2}} \right\rbrack x^{n}} & = & -4 \\

{\sum\limits_{n = 0}^{\infty}\left\lbrack {(n^{2} + 5n + 4)a_{n} - (n + 2)(n + 1)a_{n + 2}} \right\rbrack x^{n}} & = & -4 \\

{\sum\limits_{n = 0}^{\infty}\left\lbrack {(n + 4)(n + 1)a_{n} - (n + 2)(n + 1)a_{n + 2}} \right\rbrack x^{n}} & = & -4

\end{array}$$

Looking at the coefficients of each power of *x*, we see that the constant term must be equal to $-4,$ and the coefficients of all other powers of *x* must be zero. Then, looking first at the constant term,

$$\begin{array}{rll}

{4a_{0} - 2a_{2}} & = & -4 \\

a_{2} & = & {2a_{0} + 2\ \text{(step 3).}}

\end{array}$$

For $n \geq 1,$ we have

$$\begin{array}{rll}

{(n + 4)(n + 1)a_{n} - (n + 2)(n + 1)a_{n + 2}} & = & 0 \\

{(n + 1)\left\lbrack {(n + 4)a_{n} - (n + 2)a_{n + 2}} \right\rbrack} & = & {0.}

\end{array}$$

Since $n \geq 1,$ $n + 1 \neq 0,$ we see that

$$(n + 4)a_{n} - (n + 2)a_{n + 2} = 0$$

and thus

$$a_{n + 2} = \frac{n + 4}{n + 2}\mspace{2mu} a_{n}.$$

For even values of *n*, we have

$$\begin{array}{clc}

a_{4} & = & {\frac{6}{4}\left( {2a_{0} + 2} \right) = 3a_{0} + 3} \\

a_{6} & = & {\frac{8}{6}\left( {3a_{0} + 3} \right) = 4a_{0} + 4} \\

& \text{⋮.} &

\end{array}$$

In general, $a_{2k} = \left( {k + 1} \right)\left( {a_{0} + 1} \right)$ (step 5).

For odd values of *n*, we have

$$\begin{array}{clc}

a_{3} & = & {\frac{5}{3}\mspace{2mu} a_{1}} \\

a_{5} & = & {\frac{7}{5}\mspace{2mu} a_{3} = \frac{7}{3}\mspace{2mu} a_{1}} \\

a_{7} & = & {\frac{9}{7}\mspace{2mu} a_{5} = \frac{9}{3}\mspace{2mu} a_{1} = 3a_{1}} \\

& \text{⋮.} &

\end{array}$$

In general, $a_{2k + 1} = \frac{2k + 3}{3}\mspace{2mu} a_{1}$ (step 5 continued).

Putting this together, we have

$$y(x) = {\sum\limits_{k = 0}^{\infty}{(k + 1)(a_{0} + 1)}}x^{2k} + {\sum\limits_{k = 0}^{\infty}{\left( \frac{2k + 3}{3} \right)a_{1}}}x^{2k + 1}\ \text{(step 6).}$$

Find a power series solution for the following differential equations.

1. $y^{\prime} + 2xy = 0$

2. $(x + 1)y^{\prime} = 3y$

We close this section with a brief introduction to Bessel functions. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a little taste of the topic here so we can see how series solutions to differential equations are used in real-world applications. The Bessel equation of order *n* is given by

$$x^{2}y^{''} + xy^{\prime} + (x^{2} - n^{2})y = 0.$$

This equation arises in many physical applications, particularly those involving cylindrical coordinates, such as the vibration of a circular drum head and transient heating or cooling of a cylinder. In the next example, we find a power series solution to the Bessel equation of order 0.

Power Series Solution to the Bessel Equation

Find a power series solution to the Bessel equation of order 0 and graph the solution.

Solution

The Bessel equation of order 0 is given by

$$x^{2}y^{''} + xy^{\prime} + x^{2}y = 0.$$

We assume a solution of the form $y = \sum\limits_{n = 0}^{\infty}a_{n}x^{n}.$ Then $y^{\prime}(x) = {\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1}$ and $y^{''}(x) = {\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2}.$ Substituting this into the differential equation, we get

$$\begin{array}{lcl}

& & \\

& & \\

{x^{2}{\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n - 2} + x{\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n - 1} + x^{2}\sum\limits_{n = 0}^{\infty}a_{n}x^{n} = 0} & & \text{Substitution.} \\

{{\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n} + {\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n} + \sum\limits_{n = 0}^{\infty}a_{n}x^{n + 2} = 0} & & \text{Bring external factors within sums.} \\

{{\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n} + {\sum\limits_{n = 1}^{\infty}n}a_{n}x^{n} + \sum\limits_{n = 2}^{\infty}a_{n - 2}x^{n} = 0} & & \text{Re-index third sum.} \\

{{\sum\limits_{n = 2}^{\infty}n}(n - 1)a_{n}x^{n} + a_{1}x + {\sum\limits_{n = 2}^{\infty}n}a_{n}x^{n} + \sum\limits_{n = 2}^{\infty}a_{n - 2}x^{n} = 0} & & {\text{Separate}\ n = 1\ \text{term from second sum.}} \\

{a_{1}x + {\sum\limits_{n = 2}^{\infty}\left\lbrack {n(n - 1)a_{n} + na_{n} + a_{n - 2}} \right\rbrack}x^{n} = 0} & & \text{Collect summation terms.} \\

{a_{1}x + {\sum\limits_{n = 2}^{\infty}\left\lbrack {(n^{2} - n)a_{n} + na_{n} + a_{n - 2}} \right\rbrack}x^{n} = 0} & & \text{Multiply through in first term.} \\

{a_{1}x + {\sum\limits_{n = 2}^{\infty}\left\lbrack {n^{2}a_{n} + a_{n - 2}} \right\rbrack}x^{n} = 0.} & & \text{Simplify.}

\end{array}$$

Then, $a_{1} = 0,$ and for $n \geq 2,$

$$\begin{array}{l}

{n^{2}a_{n} + a_{n - 2} = 0} \\

\\

{a_{n} = - \frac{1}{n^{2}}\mspace{2mu} a_{n - 2}.}

\end{array}$$

Because $a_{1} = 0,$ all odd terms are zero. Then, for even values of *n*, we have

$$\begin{array}{rll}

a_{2} & = & {- \frac{1}{2^{2}}\mspace{2mu} a_{0}} \\

a_{4} & = & {- \frac{1}{4^{2}}\mspace{2mu} a_{2} = \frac{1}{4^{2} \cdot 2^{2}}\mspace{2mu} a_{0}.} \\

a_{6} & = & {- \frac{1}{6^{2}}\mspace{2mu} a_{4} = - \frac{1}{6^{2} \cdot 4^{2} \cdot 2^{2}}\mspace{2mu} a_{0}}

\end{array}$$

In general,

$$a_{2k} = \frac{{(-1)}^{k}}{{(2)}^{2k}{(k!)}^{2}}\mspace{2mu} a_{0}.$$

Thus, we have

$$y(x) = a_{0}\sum\limits_{k = 0}^{\infty}\frac{{(-1)}^{k}}{{(2)}^{2k}{(k!)}^{2}}x^{2k}.$$ (7.10)

The graph appears below for $a_{0} = 1$.

Verify that the expression found in Example 7.26 is a solution to the Bessel equation of order 0.

Section 7.4 Exercises

Find a power series solution for the following differential equations.

104\.

$y^{''} + 6y^{\prime} = 0$

105.

$5y^{''} + y^{\prime} = 0$

106\.

$y^{''} + 25y = 0$

107.

$y^{''} - y = 0$

108\.

$2y^{\prime} + y = 0$

109.

$y^{\prime} - 2xy = 0$

110\.

$\left( {x - 7} \right)y^{\prime} + 2y = 0$

111.

$y^{''} - xy^{\prime} - y = 0$

112\.

$\left( {1 + x^{2}} \right)y^{''} - 4xy^{\prime} + 6y = 0$

113.

$x^{2}y^{''} - xy^{\prime} - 3y = 0$

114\.

$y^{''} - 8y^{\prime} = 0,\qquad y(0) = -2,\ \ y^{\prime}(0) = 10$

115.

$y^{''} - 2xy = 0,\qquad y(0) = 1,\ \ y^{\prime}(0) = -3$

116\.

The differential equation $x^{2}y^{''} + xy^{\prime} + \left( {x^{2} - 1} \right)y = 0$ is a Bessel equation of order 1. Use a power series of the form $y = {\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}}$ to find the solution.

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Key Terms

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Key Terms

boundary conditions

the conditions that give the state of a system at different times, such as the position of a spring-mass system at two different times

boundary-value problem

a differential equation with associated boundary conditions

characteristic equation

the equation $a\lambda^{2} + b\lambda + c = 0$ for the differential equation $ay^{''} + by^{\prime} + cy = 0$

complementary equation

for the nonhomogeneous linear differential equation

$$a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x),$$

the associated homogeneous equation, called the *complementary equation*, is

$$a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = 0$$

homogeneous linear equation

a second-order differential equation that can be written in the form $a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x),$ but $r(x) = 0$ for every value of $x$

linearly dependent

a set of functions $f_{1}(x),f_{2}(x)\text{,…,}f_{n}(x)$ for which there are constants $c_{1},c_{2}\text{,…}c_{n},$ not all zero, such that $c_{1}f_{1}(x) + c_{2}f_{2}(x) + \text{⋯} + c_{n}f_{n}(x) = 0$ for all *x* in the interval of interest

linearly independent

a set of functions $f_{1}(x),f_{2}(x)\text{,…,}f_{n}(x)$ for which there are no constants $c_{1},c_{2}\text{,…}c_{n},$ such that $c_{1}f_{1}(x) + c_{2}f_{2}(x) + \text{⋯} + c_{n}f_{n}(x) = 0$ for all *x* in the interval of interest

method of undetermined coefficients

a method that involves making a guess about the form of the particular solution, then solving for the coefficients in the guess

method of variation of parameters

a method that involves looking for particular solutions in the form $y_{p}(x) = u(x)y_{1}(x) + v(x)y_{2}(x),$ where $y_{1}$ and $y_{2}$ are linearly independent solutions to the complementary equations, and then solving a system of equations to find $u(x)$ and $v(x)$

nonhomogeneous linear equation

a second-order differential equation that can be written in the form $a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x),$ but $r(x) \neq 0$ for some value of $x$

particular solution

a solution $y_{p}(x)$ of a differential equation that contains no arbitrary constants

*RLC* series circuit

a complete electrical path consisting of a resistor, an inductor, and a capacitor; a second-order, constant-coefficient differential equation can be used to model the charge on the capacitor in an *RLC* series circuit

simple harmonic motion

motion described by the equation $x(t) = c_{1}\text{cos}\mspace{2mu}\left( {\omega t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {\omega t} \right),$ as exhibited by an undamped spring-mass system in which the mass continues to oscillate indefinitely

steady-state solution

a solution to a nonhomogeneous differential equation related to the forcing function; in the long term, the solution approaches the steady-state solution

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
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Key Equations

| | |

|------------------------------------------------------|----------------------------------------------------------|

| Linear second-order differential equation | $a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x)$ |

| Second-order equation with constant coefficients | $ay^{''} + by^{\prime} + cy = 0$ |

| | |

|-----------------------------------------------------------------------|-------------------------------------------------------|

| Complementary equation | $a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = 0$ |

| General solution to a nonhomogeneous linear differential equation | $y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) + y_{p}(x)$ |

| | |

|----------------------------------------------|----------------------------------------------------------------------------------------------------------------------|

| Equation of simple harmonic motion | $x^{''} + \omega^{2}x = 0$ |

| Solution for simple harmonic motion | $x(t) = c_{1}\text{cos}\mspace{2mu}\left( {\omega t} \right) + c_{2}\text{sin}\mspace{2mu}\left( {\omega t} \right)$ |

| Alternative form of solution for SHM | $x(t) = A\mspace{2mu}\text{sin}\mspace{2mu}\left( {\omega t + \phi} \right)$ |

| Forced harmonic motion | $mx^{''} + bx^{\prime} + kx = f(t)$ |

| Charge in a *RLC* series circuit | $L\frac{d^{2}q}{dt^{2}} + R\frac{dq}{dt} + \frac{1}{C}q = E(t)$ |

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7.1 Second-Order Linear Equations

• Second-order differential equations can be classified as linear or nonlinear, homogeneous or nonhomogeneous.
• To find a general solution for a homogeneous second-order differential equation, we must find two linearly independent solutions. If $y_{1}(x)$ and $y_{2}(x)$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by

$$y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x).$$

• To solve homogeneous second-order differential equations with constant coefficients, find the roots of the characteristic equation. The form of the general solution varies depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.
• Initial conditions or boundary conditions can then be used to find the specific solution to a differential equation that satisfies those conditions, except when there is no solution or infinitely many solutions.

7.2 Nonhomogeneous Linear Equations

• To solve a nonhomogeneous linear second-order differential equation, first find the general solution to the complementary equation, then find a particular solution to the nonhomogeneous equation.
• Let $y_{p}(x)$ be any particular solution to the nonhomogeneous linear differential equation

$$a_{2}(x)y^{''} + a_{1}(x)y^{\prime} + a_{0}(x)y = r(x),$$

and let $c_{1}y_{1}(x) + c_{2}y_{2}(x)$ denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by

$$y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) + y_{p}(x\text{).}$$

• When $r(x)$ is a combination of polynomials, exponential functions, sines, and cosines, use the method of undetermined coefficients to find the particular solution. To use this method, assume a solution in the same form as $r(x),$ multiplying by *x* as necessary until the assumed solution is linearly independent of the general solution to the complementary equation. Then, substitute the assumed solution into the differential equation to find values for the coefficients.
• When $r(x)$ is *not* a combination of polynomials, exponential functions, or sines and cosines, use the method of variation of parameters to find the particular solution. This method involves using Cramer’s rule or another suitable technique to find functions $u^{\prime}(x)$ and $v^{\prime}(x)$ satisfying

$$\begin{array}{rll}

{u^{\prime}y_{1} + v^{\prime}y_{2}} & = & 0 \\

{u^{\prime}y_{1}{}^{\prime} + v^{\prime}y_{2}{}^{\prime}} & = & {r(x\text{).}}

\end{array}$$

Then, $y_{p}(x) = u(x)y_{1}(x) + v(x)y_{2}(x)$ is a particular solution to the differential equation.

7.3 Applications

• Second-order constant-coefficient differential equations can be used to model spring-mass systems.
• An examination of the forces on a spring-mass system results in a differential equation of the form

$$mx^{''} + bx^{\prime} + kx = f(t),$$

where $m$ represents the mass, $b$ is the coefficient of the damping force, $k$ is the spring constant, and $f(t)$ represents any net external forces on the system.

• If $b = 0,$ there is no damping force acting on the system, and simple harmonic motion results. If $b \neq 0,$ the behavior of the system depends on whether $b^{2} - 4mk > 0,$ $b^{2} - 4mk = 0,$ or $b^{2} - 4mk < 0.$
• If $b^{2} - 4mk > 0,$ the system is overdamped and does not exhibit oscillatory behavior.
• If $b^{2} - 4mk = 0,$ the system is critically damped. It does not exhibit oscillatory behavior, but any slight reduction in the damping would result in oscillatory behavior.
• If $b^{2} - 4mk < 0,$ the system is underdamped. It exhibits oscillatory behavior, but the amplitude of the oscillations decreases over time.
• If $f(t) \neq 0,$ the solution to the differential equation is the sum of a transient solution and a steady-state solution. The steady-state solution governs the long-term behavior of the system.
• The charge on the capacitor in an *RLC* series circuit can also be modeled with a second-order constant-coefficient differential equation of the form

$$L\frac{d^{2}q}{dt^{2}} + R\frac{dq}{dt} + \frac{1}{C}q = E(t),$$

where *L* is the inductance, *R* is the resistance, *C* is the capacitance, and $E(t)$ is the voltage source.

7.4 Series Solutions of Differential Equations

• Power series representations of functions can sometimes be used to find solutions to differential equations.
• Differentiate the power series term by term and substitute into the differential equation to find relationships between the power series coefficients.

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/7-key-concepts

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Review Exercises

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/7-review-exercises

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Review Exercises

*True or False*? Justify your answer with a proof or a counterexample.

117.

If $y$ and $z$ are both solutions to $y^{''} + 2y^{\prime} + y = 0,$ then $y + z$ is also a solution.

118\.

The following system of algebraic equations has a unique solution:

$\begin{array}{l}

{6z_{1} + 3z_{2} = 8} \\

{4z_{1} + 2z_{2} = 4.}

\end{array}$

119.

$y = e^{x}\text{cos}(3x) + e^{x}\text{sin}(2x)$ is a solution to the second-order differential equation $y^{''} + 2y\prime + 10 = 0.$

120\.

To find the particular solution to a second-order differential equation, you need one initial condition.

Classify the differential equation. Determine the order, whether it is linear and, if linear, whether the differential equation is homogeneous or nonhomogeneous. If the equation is second-order homogeneous and linear, find the characteristic equation.

121.

$y^{''} - 2y = 0$

122\.

$y^{''} - 3y + 2y = \text{cos}\mspace{2mu}(t)$

123.

$\left( \frac{dy}{dt} \right)^{2} + yy^{\prime} = 1$

124\.

$\frac{d^{2}y}{dt^{2}} + t\frac{dy}{dt} + \text{sin}^{2}(t)y = e^{t}$

For the following problems, find the general solution.

125.

$y^{''} + 9y = 0$

126\.

$y^{''} + 2y^{\prime} + y = 0$

127.

$y^{''} - 2y^{\prime} + 10y = 4x$

128\.

$y^{''} = \text{cos}\mspace{2mu}(x) + 2y^{\prime} + y$

129.

$y^{\operatorname{''}} + 5y\prime + 4y = x + e^{2x}$

130\.

$y^{''} = 3y^{\prime} + xe^{\text{−}x}$

131.

$y^{''} - x^{2} = -3y^{\prime} - \frac{9}{4}y + 3x$

132\.

$y^{''} = 2\mspace{2mu}\text{cos}\mspace{2mu} x + y^{\prime} - y$

For the following problems, find the solution to the initial-value problem, if possible.

133.

$y^{''} + 4y^{\prime} + 6y = 0,$ $y(0) = 0,$ $y^{\prime}(0) = \sqrt{2}$

134\.

$y^{''} = 3y - \text{cos}\mspace{2mu}(x),$ $y(0) = \frac{9}{4},$ $y^{\prime}(0) = 0$

For the following problems, find the solution to the boundary-value problem, if possible.

135.

$4y^{\prime} = -6y + 2y^{''},$ $y(0) = 0,$ $y(1) = 1$

136\.

$y^{''} = 3x - y - y^{\prime},$ $y(0) = -3,$ $y(1) = 0$

For the following problem, set up and solve the differential equation.

137.

The motion of a swinging pendulum for small angles $\theta$ can be approximated by $\frac{d^{2}\theta}{dt^{2}} + \frac{g}{L}\theta = 0,$ where $\theta$ is the angle the pendulum makes with respect to a vertical line, *g* is the acceleration resulting from gravity, and *L* is the length of the pendulum. Find the equation describing the angle of the pendulum at time $t,$ assuming an initial displacement of $\theta_{0}$ and an initial velocity of zero.

The following problems consider the “beats” that occur when the forcing term of a differential equation causes “slow” and “fast” amplitudes. Consider the general differential equation$ay^{''} + by = \text{cos}(\omega t)$ that governs undamped motion. Assume that $\sqrt{\frac{b}{a}} \neq \omega\text{.}$

138\.

Find the general solution to this equation (*Hint:* call $\omega_{0} = \sqrt{b\text{/}a}$).

139\.

Assuming the system starts from rest, show that the particular solution can be written as $y = \frac{2}{a\left( {\omega_{0}{}^{2} - \omega^{2}} \right)}\mspace{2mu}\text{sin}\mspace{2mu}\left( \frac{\omega_{0} - \omega t}{2} \right)\mspace{2mu}\text{sin}\mspace{2mu}\left( \frac{\omega_{0} + \omega t}{2} \right).$

140\.

\[T\] Using your solutions derived earlier, plot the solution to the system $2y^{''} + 9y = \text{cos}(2t)$ over the interval $t = \lbrack-50,50\rbrack.$ Find, analytically, the period of the fast and slow amplitudes.

For the following problem, set up and solve the differential equations.

141.

An opera singer is attempting to shatter a glass by singing a particular note. The vibrations of the glass can be modeled by ${y^{''} + ay = 0},$ where $a > 0$ represents the natural frequency of the glass and the singer is forcing the vibrations at $\text{cos}(bt\text{).}$ For what value $b$ would the singer be able to break that glass? (*Note*: in order for the glass to break, the oscillations would need to get higher and higher.)

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Citation information

• Use the information below to generate a citation. We recommend using a citation tool such as this one.
• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/7-review-exercises

© Jul 15, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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A | Table of Integrals

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/a-table-of-integrals

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A \| Table of Integrals

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Basic Integrals

1\. ${\int{u^{n}\ du =}}\frac{u^{n + 1}}{n + 1} + C,n \neq \text{−}1$

2\. ${\int{\frac{du}{u} =}}\text{ln}\mspace{2mu}|u| + C$

3\. ${\int{e^{u}\ du}} = e^{u} + C$

4\. ${\int{a^{u}\ du =}}\frac{a^{u}}{\text{ln}\mspace{2mu} a} + C$

5\. $\int{\text{sin}\ u\ du = \text{−cos}\ u + C}$

6\. $\int{\text{cos}\ u\ du = \text{sin}\ u + C}$

7\. $\int{\text{sec}^{2}u\ du = \text{tan}\ u + C}$

8\. $\int{\text{csc}^{2}u\ du = \text{−cot}\ u + C}$

9\. $\int{\text{sec}\ u\ \text{tan}\ u\ du = \text{sec}\ u + C}$

10\. $\int{\text{csc}\ u\ \text{cot}\ u\ du = \text{−csc}\ u + C}$

11\. $\int{\text{tan}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sec}\ u} \right| + C}$

12\. $\int{\text{cot}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sin}\ u} \right| + C}$

13\. $\int{\text{sec}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sec}\ u + \text{tan}\ u} \right| + C}$

14\. $\int{\text{csc}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{csc}\ u - \text{cot}\ u} \right| + C}$

15\. ${\int\frac{\ du}{\sqrt{a^{2} - u^{2}}}} = \text{sin}^{-1}\frac{u}{a} + C$

16\. ${\int\frac{\ du}{a^{2} + u^{2}}} = \frac{1}{a}\text{tan}^{-1}\frac{u}{a} + C$

17\. ${\int\frac{\ du}{u\sqrt{u^{2} - a^{2}}}} = \frac{1}{a}\text{sec}^{-1}\frac{u}{a} + C$

Trigonometric Integrals

18\. ${\int{\text{sin}^{2}u\ du = \frac{1}{2}u -}}\frac{1}{4}\text{sin}\ 2u + C$

19\. ${\int{\text{cos}^{2}u\ du = \frac{1}{2}u +}}\frac{1}{4}\text{sin}\ 2u + C$

20\. ${\int{\text{tan}^{2}u\ du = \text{tan}\ u - u}} + C$

21\. ${\int{\text{cot}^{2}u\ du = \text{−}\text{cot}\ u - u}} + C$

22\. ${\int{\text{sin}^{3}u\ du = - \frac{1}{3}\left( {2 + \text{sin}^{2}u} \right)\mspace{2mu}\text{cos}\ u}} + C$

23\. ${\int{\text{cos}^{3}u\ du = \frac{1}{3}\left( {2 + \text{cos}^{2}u} \right)\mspace{2mu}\text{sin}\ u}} + C$

24\. ${\int{\text{tan}^{3}u\ du = \frac{1}{2}\text{tan}^{2}u +}}\text{ln}\mspace{2mu}\left| {\text{cos}\ u} \right| + C$

25\. ${\int{\text{cot}^{3}u\ du = - \frac{1}{2}\text{cot}^{2}u -}}\text{ln}\mspace{2mu}\left| {\text{sin}\ u} \right| + C$

26\. ${\int{\text{sec}^{3}u\ du = \frac{1}{2}\text{sec}\ u\ \text{tan}\ u + \frac{1}{2}}}\text{ln}\mspace{2mu}\left| {\text{sec}\ u + \text{tan}\ u} \right| + C$

27\. ${\int{\text{csc}^{3}u\ du = - \frac{1}{2}\text{csc}\ u\ \text{cot}\ u + \frac{1}{2}}}\text{ln}\mspace{2mu}\left| {\text{csc}\ u - \text{cot}\ u} \right| + C$

28\. ${\int{\text{sin}^{n}u\ du = - \frac{1}{n}\text{sin}^{n - 1}u\ \text{cos}\ u + \frac{n - 1}{n}}}{\int{\text{sin}^{n - 2}u}}\ du$

29\. ${\int{\text{cos}^{n}u\ du = \frac{1}{n}\text{cos}^{n - 1}u\ \text{sin}\ u + \frac{n - 1}{n}}}{\int{\text{cos}^{n - 2}u}}\ du$

30\. ${\int{\text{tan}^{n}u\ du = \frac{1}{n - 1}\text{tan}^{n - 1}u -}}{\int{\text{tan}^{n - 2}u}}\ du$

31\. ${\int{\text{cot}^{n}u\ du = \frac{-1}{n - 1}\text{cot}^{n - 1}u -}}{\int{\text{cot}^{n - 2}u}}\ du$

32\. ${\int{\text{sec}^{n}u\ du = \frac{1}{n - 1}\text{tan}\ u\ \text{sec}^{n - 2}u + \frac{n - 2}{n - 1}}}{\int{\text{sec}^{n - 2}u}}\ du$

33\. ${\int{\text{csc}^{n}u\ du = \frac{-1}{n - 1}\text{cot}\ u\ \text{csc}^{n - 2}u + \frac{n - 2}{n - 1}}}{\int{\text{csc}^{n - 2}u}}\ du$

34\. ${\int{\text{sin}\ au\ \text{sin}\ bu\ du = \frac{\left. \text{sin}\left( \left( {a - b} \right) \right.u \right)}{2\left( {a - b} \right)}}} - \frac{\left. \text{sin}\left( \left( {a + b} \right) \right.u \right)}{2\left( {a + b} \right)} + C$

35\. ${\int{\text{cos}\ au\ \text{cos}\ bu\ du = \frac{\left. \text{sin}\left( \left( {a - b} \right) \right.u \right)}{2\left( {a - b} \right)}}} + \frac{\left. \text{sin}\left( \left( {a + b} \right) \right.u \right)}{2\left( {a + b} \right)} + C$

36\. ${\int{\text{sin}\ au\ \text{cos}\ bu\ du = - \frac{\left. \text{cos}\left( \left( {a - b} \right) \right.u \right)}{2\left( {a - b} \right)}}} - \frac{\left. \text{cos}\left( \left( {a + b} \right) \right.u \right)}{2\left( {a + b} \right)} + C$

37\. $\int{u\ \text{sin}\ u\ du = \text{sin}\ u - u\ \text{cos}\ u + C}$

38\. $\int{u\ \text{cos}\ u\ du = \text{cos}\ u + u\ \text{sin}\ u + C}$

39\. $\int{u^{n}\text{sin}\ u\ du = \text{−}u^{n}\text{cos}\ u + n{\int{u^{n - 1}\text{cos}\ u}}\ du}$

40\. $\int{u^{n}\text{cos}\ u\ du = u^{n}\text{sin}\ u - n{\int{u^{n - 1}\text{sin}\ u}}\ du}$

41\. $\begin{array}{cl}

{{\int\text{sin}^{n}}u\ \text{cos}^{m}\ u\ du} & {= - \frac{\text{sin}^{n - 1}\ u\ \text{cos}^{m + 1}\ u}{n + m} + \frac{n - 1}{n + m}{\int{\text{sin}^{n - 2}u\ \text{cos}^{m}u}}\ du} \\

& {= \frac{\text{sin}^{n + 1}u\ \text{cos}^{m - 1}u}{n + m} + \frac{m - 1}{n + m}{\int{\text{sin}^{n}u\ \text{cos}^{m - 2}u}}\ du}

\end{array}$

Exponential and Logarithmic Integrals

42\. ${\int{ue^{au}\ du =}}\frac{1}{a^{2}}\left( {au - 1} \right)e^{au} + C$

43\. ${\int{u^{n}e^{au}\ du}} = \frac{1}{a}u^{n}e^{au} - \frac{n}{a}{\int{u^{n - 1}e^{au}\ du}}$

44\. ${\int{e^{au}\text{sin}\ bu\ du}} = \frac{e^{au}}{a^{2} + b^{2}}\left( {a\mspace{2mu}\text{sin}\ bu - b\ \text{cos}\ bu} \right) + C$

45\. ${\int{e^{au}\text{cos}\ bu\ du}} = \frac{e^{au}}{a^{2} + b^{2}}\left( {a\ \text{cos}\ bu + b\ \text{sin}\ bu} \right) + C$

46\. $\int{\text{ln}\mspace{2mu} u\ du = u\ \text{ln}\mspace{2mu} u - u + C}$

47\. ${\int u^{n}}\text{ln}\mspace{2mu} u\ du = \frac{u^{n + 1}}{\left( {n + 1} \right)^{2}}\left\lbrack {\left( {n + 1} \right)\text{ln}\mspace{2mu} u - 1} \right\rbrack + C$

48\. ${\int{\frac{1}{u\ \text{ln}\mspace{2mu} u}\ du = \text{ln}\mspace{2mu}\left| {\text{ln}\mspace{2mu} u} \right|}} + C$

Hyperbolic Integrals

49\. $\int{\text{sinh}\ u\ du = \text{cosh}\ u + C}$

50\. $\int{\text{cosh}\ u\ du = \text{sinh}\ u + C}$

51\. $\int{\text{tanh}\ u\ du = \text{ln}\mspace{2mu}\text{cosh}\ u + C}$

52\. $\int{\text{coth}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sinh}\ u} \right| + C}$

53\. $\int{\text{sech}\ u\ du = \text{tan}^{-1}\left| {\text{sinh}\ u} \right| + C}$

54\. ${\int{\text{csch}\ u\ du =}}\text{ln}\mspace{2mu}\left| {\text{tanh}\mspace{2mu}\frac{1}{2}u} \right| + C$

55\. $\int{\text{sech}^{2}u\ du = \text{tanh}\ u + C}$

56\. $\int{\text{csch}^{2}u\ du = \text{−}\text{coth}\ u + C}$

57\. $\int{\text{sech}\ u\ \text{tanh}\ u\ du = \text{−}\text{sech}\ u + C}$

58\. $\int{\text{csch}\ u\ \text{coth}\ u\ du = \text{−}\text{csch}\ u + C}$

Inverse Trigonometric Integrals

59\. ${\int{\text{sin}^{-1}u\ du =}}u\ \text{sin}^{-1}u + \sqrt{1 - u^{2}} + C$

60\. ${\int{\text{cos}^{-1}u\ du =}}u\ \text{cos}^{-1}u - \sqrt{1 - u^{2}} + C$

61\. ${\int{\text{tan}^{-1}u\ du =}}u\ \text{tan}^{-1}u - \frac{1}{2}\text{ln}\mspace{2mu}\left( {1 + u^{2}} \right) + C$

62\. ${\int{u\ \text{sin}^{-1}u\ du =}}\frac{2u^{2} - 1}{4}\text{sin}^{-1}u + \frac{u\sqrt{1 - u^{2}}}{4} + C$

63\. ${\int{u\ \text{cos}^{-1}u\ du =}}\frac{2u^{2} - 1}{4}\text{cos}^{-1}u - \frac{u\sqrt{1 - u^{2}}}{4} + C$

64\. ${\int{u\ \text{tan}^{-1}u\ du =}}\frac{u^{2} + 1}{2}\text{tan}^{-1}u - \frac{u}{2} + C$

65\. ${\int{u^{n}\text{sin}^{-1}u\ du =}}\frac{1}{n + 1}\left\lbrack {u^{n + 1}\text{sin}^{-1}u - {\int\frac{u^{n + 1}\ du}{\sqrt{1 - u^{2}}}}} \right\rbrack,n \neq \text{−}1$

66\. ${\int{u^{n}\text{cos}^{-1}u\ du =}}\frac{1}{n + 1}\left\lbrack {u^{n + 1}\text{cos}^{-1}u + {\int\frac{u^{n + 1}\ du}{\sqrt{1 - u^{2}}}}} \right\rbrack,n \neq \text{−}1$

67\. ${\int{u^{n}\text{tan}^{-1}u\ du =}}\frac{1}{n + 1}\left\lbrack {u^{n + 1}\text{tan}^{-1}u - {\int\frac{u^{n + 1}\ du}{1 + u^{2}}}} \right\rbrack,n \neq \text{−}1$

Integrals Involving *a*² + *u*², *a* \> 0

68\. ${\int\sqrt{a^{2} + u^{2}}}\ du = \frac{u}{2}\sqrt{a^{2} + u^{2}} + \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

69\. ${\int{u^{2}\sqrt{a^{2} + u^{2}}}}\ du = \frac{u}{8}\left( {a^{2} + 2u^{2}} \right)\sqrt{a^{2} + u^{2}} - \frac{a^{4}}{8}\text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

70\. ${\int\frac{\sqrt{a^{2} + u^{2}}}{u}}\ du = \sqrt{a^{2} + u^{2}} - a\ \text{ln}\mspace{2mu}\left| \frac{a + \sqrt{a^{2} + u^{2}}}{u} \right| + C$

71\. ${\int\frac{\sqrt{a^{2} + u^{2}}}{u^{2}}}\ du = - \frac{\sqrt{a^{2} + u^{2}}}{u} + \text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

72\. ${\int\frac{\ du}{\sqrt{a^{2} + u^{2}}}} = \text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

73\. ${\int\frac{u^{2}\ du}{\sqrt{a^{2} + u^{2}}}} = \frac{u}{2}\left( \sqrt{a^{2} + u^{2}} \right) - \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

74\. ${\int\frac{\ du}{u\sqrt{a^{2} + u^{2}}}} = - \frac{1}{a}\text{ln}\mspace{2mu}\left| \frac{\sqrt{a^{2} + u^{2}} + a}{u} \right| + C$

75\. ${\int\frac{\ du}{u^{2}\sqrt{a^{2} + u^{2}}}} = - \frac{\sqrt{a^{2} + u^{2}}}{a^{2}u} + C$

76\. ${\int\frac{\ du}{\left( {a^{2} + u^{2}} \right)^{3\text{/}2}}} = \frac{u}{a^{2}\sqrt{a^{2} + u^{2}}} + C$

Integrals Involving *u*² − *a*², *a* \> 0

77\. ${\int{\sqrt{u^{2} - a^{2}}\ du = \frac{u}{2}}}\sqrt{u^{2} - a^{2}} - \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

78\. ${\int{u^{2}\sqrt{u^{2} - a^{2}}\ du = \frac{u}{8}}}\left( {2u^{2} - a^{2}} \right)\sqrt{u^{2} - a^{2}} - \frac{a^{4}}{8}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

79\. ${\int{\frac{\sqrt{u^{2} - a^{2}}}{u}\ du =}}\sqrt{u^{2} - a^{2}} - a\text{cos}^{-1}\frac{a}{|u|} + C$

80\. ${\int{\frac{\sqrt{u^{2} - a^{2}}}{u^{2}}\ du =}} - \frac{\sqrt{u^{2} - a^{2}}}{u} + \text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

81\. ${\int{\frac{\ du}{\sqrt{u^{2} - a^{2}}} =}}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

82\. ${\int{\frac{u^{2}\ du}{\sqrt{u^{2} - a^{2}}} =}}\frac{u}{2}\sqrt{u^{2} - a^{2}} + \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

83\. ${\int{\frac{\ du}{u^{2}\sqrt{u^{2} - a^{2}}} =}}\frac{\sqrt{u^{2} - a^{2}}}{a^{2}u} + C$

84a. ${\int{\frac{\ du}{\left( {u^{2} - a^{2}} \right)^{3\text{/}2}} = \text{−}}}\frac{u}{a^{2}\sqrt{u^{2} - a^{2}}} + C$

84b. $\int\frac{du}{u^{2} - a^{2}} = \frac{1}{2a}\ln\left| \frac{u - a}{u + a} \right| + C$

Integrals Involving *a*² − *u*², *a* \> 0

85\. ${\int{\sqrt{a^{2} - u^{2}}\ du = \frac{u}{2}}}\sqrt{a^{2} - u^{2}} + \frac{a^{2}}{2}\text{sin}^{-1}\frac{u}{a} + C$

86\. ${\int{u^{2}\sqrt{a^{2} - u^{2}}\ du = \frac{u}{8}}}\left( {2u^{2} - a^{2}} \right)\sqrt{a^{2} - u^{2}} + \frac{a^{4}}{8}\text{sin}^{-1}\frac{u}{a} + C$

87\. ${\int{\frac{\sqrt{a^{2} - u^{2}}}{u}\ du =}}\sqrt{a^{2} - u^{2}} - a\text{ln}\mspace{2mu}\left| \frac{a + \sqrt{a^{2} - u^{2}}}{u} \right| + C$

88\. ${\int{\frac{\sqrt{a^{2} - u^{2}}}{u^{2}}\ du = - \frac{1}{u}}}\sqrt{a^{2} - u^{2}} - \text{sin}^{-1}\frac{u}{a} + C$

89\. $\int\frac{u^{2}\ du}{\sqrt{a^{2} - u^{2}}} = - \frac{u}{2}\sqrt{a^{2} - u^{2}} + \frac{a^{2}}{2}\text{sin}^{-1}\frac{u}{a} + C$

90\. ${\int{\frac{\ du}{u\sqrt{a^{2} - u^{2}}} =}} - \frac{1}{a}\text{ln}\mspace{2mu}\left| \frac{a + \sqrt{a^{2} - u^{2}}}{u} \right| + C$

91\. ${\int{\frac{\ du}{u^{2}\sqrt{a^{2} - u^{2}}} =}} - \frac{1}{a^{2}u}\sqrt{a^{2} - u^{2}} + C$

92\. ${\int{\left( {a^{2} - u^{2}} \right)^{3\text{/}2}\ du = - \frac{u}{8}\left( {2u^{2} - 5a^{2}} \right)}}\sqrt{a^{2} - u^{2}} + \frac{3a^{4}}{8}\text{sin}^{-1}\frac{u}{a} + C$

93a. ${\int{\frac{\ du}{\left( {a^{2} - u^{2}} \right)^{3\text{/}2}} = \frac{u}{a^{2}\sqrt{a^{2} - u^{2}}} +}}C$

93b. $\int\frac{du}{a^{2} - u^{2}} = \frac{1}{2a}\ln\left| \frac{u + a}{u - a} \right| + C$

Integrals Involving 2*au* − *u*², *a* \> 0

94\. ${\int\sqrt{2au - u^{2}}}\ du = \frac{u - a}{2}\sqrt{2au - u^{2}} + \frac{a^{2}}{2}\text{cos}^{-1}\left( \frac{a - u}{a} \right) + C$

95\. ${\int\frac{\ du}{\sqrt{2au - u^{2}}}} = \text{cos}^{-1}\left( \frac{a - u}{a} \right) + C$

96\. ${\int{u\sqrt{2au - u^{2}}}}\ du = \frac{2u^{2} - au - 3a^{2}}{6}\sqrt{2au - u^{2}} + \frac{a^{3}}{2}\text{cos}^{-1}\left( \frac{a - u}{a} \right) + C$

97\. ${\int\frac{\ du}{u\sqrt{2au - u^{2}}}} = - \frac{\sqrt{2au - u^{2}}}{au} + C$

Integrals Involving *a* + *bu*, *a* ≠ 0

98\. ${\int\frac{u\ du}{a + bu}} = \frac{1}{b^{2}}\left( {a + bu - a\text{ln}\mspace{2mu}\left| {a + bu} \right|} \right) + C$

99\. ${\int\frac{u^{2}\ du}{a + bu}} = \frac{1}{2b^{3}}\left\lbrack {\left( {a + bu} \right)^{2} - 4a\left( {a + bu} \right) + 2a^{2}\text{ln}\mspace{2mu}\left| {a + bu} \right|} \right\rbrack + C$

100\. ${\int\frac{\ du}{u\left( {a + bu} \right)}} = \frac{1}{a}\text{ln}\mspace{2mu}\left| \frac{u}{a + bu} \right| + C$

101\. ${\int\frac{\ du}{u^{2}\left( {a + bu} \right)}} = - \frac{1}{au} + \frac{b}{a^{2}}\text{ln}\mspace{2mu}\left| \frac{a + bu}{u} \right| + C$

102\. ${\int\frac{u\ du}{\left( {a + bu} \right)^{2}}} = \frac{a}{b^{2}\left( {a + bu} \right)} + \frac{1}{b^{2}}\text{ln}\mspace{2mu}\left| {a + bu} \right| + C$

103\. ${\int\frac{u\ du}{u\ \left( {a + bu} \right)^{2}}} = \frac{1}{a\left( {a + bu} \right)} - \frac{1}{a^{2}}\text{ln}\mspace{2mu}\left| \frac{a + bu}{u} \right| + C$

104\. ${\int\frac{u^{2}\ du}{\left( {a + bu} \right)^{2}}} = \frac{1}{b^{3}}\left( {a + bu - \frac{a^{2}}{a + bu} - 2a\text{ln}\mspace{2mu}\left| {a + bu} \right|} \right) + C$

105\. ${\int{u\sqrt{a + bu}}}\ du = \frac{2}{15b^{2}}\left( {3bu - 2a} \right)\left( {a + bu} \right)^{3\text{/}2} + C$

106\. ${\int\frac{u\ du}{\sqrt{a + bu}}} = \frac{2}{3b^{2}}\left( {bu - 2a} \right)\sqrt{a + bu} + C$

107\. ${\int\frac{u^{2}\ du}{\sqrt{a + bu}}} = \frac{2}{15b^{3}}\left( {8a^{2} + 3b^{2}u^{2} - 4abu} \right)\sqrt{a + bu} + C$

108\. $\begin{array}{cll}

{\int\frac{\ du}{u\sqrt{a + bu}}} & {= \frac{1}{\sqrt{a}}\text{ln}\mspace{2mu}\left| \frac{\sqrt{a + bu} - \sqrt{a}}{\sqrt{a + bu} + \sqrt{a}} \right| + C,} & {\text{if}\ a > 0} \\

& {= \frac{2}{\sqrt{\text{−}a}}\text{tan} - 1\sqrt{\frac{a + bu}{\text{−}a}} + C,} & {\text{if}\ a < 0}

\end{array}$

109\. ${\int\frac{\sqrt{a + bu}}{u}}\ du = 2\sqrt{a + bu} + a{\int\frac{\ du}{u\sqrt{a + bu}}}$

110\. ${\int\frac{\sqrt{a + bu}}{u^{2}}}\ du = - \frac{\sqrt{a + bu}}{u} + \frac{b}{2}{\int\frac{\ du}{u\sqrt{a + bu}}}$

111\. ${\int{u^{n}\sqrt{a + bu}}}\ du = \frac{2}{b\left( {2n + 3} \right)}\left\lbrack {u^{n}\left( {a + bu} \right)^{3\text{/}2} - na{\int{u^{n - 1}\sqrt{a + bu}\ du}}} \right\rbrack$

112\. ${\int\frac{u^{n}\ du}{\sqrt{a + bu}}} = \frac{2u^{n}\sqrt{a + bu}}{b\left( {2n + 1} \right)} - \frac{2na}{b\left( {2n + 1} \right)}{\int\frac{u^{n - 1}\ du}{\sqrt{a + bu}}}$

113\. ${\int\frac{\ du}{u^{n}\sqrt{a + bu}}} = - \frac{\sqrt{a + bu}}{a\left( {n - 1} \right)u^{n - 1}} - \frac{b\left( {2n - 3} \right)}{2a\left( {n - 1} \right)}{\int\frac{\ du}{u^{n - 1}\sqrt{a + bu}}}$

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/a-table-of-integrals

© Jul 15, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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B | Table of Derivatives

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/b-table-of-derivatives

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General Formulas

1\. $\frac{d}{dx}(c) = 0$

2\. $\frac{d}{dx}\left( {f(x) + g(x)} \right) = f^{\prime}(x) + g^{\prime}(x)$

3\. $\frac{d}{dx}\left( {f(x)g(x)} \right) = f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$

4\. $\frac{d}{dx}\left( x^{n} \right) = nx^{n - 1},\ \text{for real numbers}\ n$

5\. $\frac{d}{dx}\left( {cf(x)} \right) = cf^{\prime}(x)$

6\. $\frac{d}{dx}\left( {f(x) - g(x)} \right) = f^{\prime}(x) - g^{\prime}(x)$

7\. $\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f^{\prime}(x) - f(x)g^{\prime}(x)}{\left( {g(x)} \right)^{2}}$

8\. $\frac{d}{dx}\left\lbrack {f\left( {g(x)} \right)} \right\rbrack = f^{\prime}\left( {g(x)} \right) \cdot g^{\prime}(x)$

Trigonometric Functions

9\. $\frac{d}{dx}\left( {\text{sin}\mspace{2mu} x} \right) = \text{cos}\mspace{2mu} x$

10\. $\frac{d}{dx}\left( {\text{tan}x} \right) = \text{sec}^{2}x$

11\. $\frac{d}{dx}\left( {\text{sec}x} \right) = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$

12\. $\frac{d}{dx}\left( {\text{cos}\mspace{2mu} x} \right) = \text{−}\text{sin}\mspace{2mu} x$

13\. $\frac{d}{dx}\left( {\text{cot}x} \right) = \text{−}\text{csc}^{2}x$

14\. $\frac{d}{dx}\left( {\text{csc}x} \right) = \text{−csc}\mspace{2mu} x\mspace{2mu}\text{cot}\mspace{2mu} x$

Inverse Trigonometric Functions

15\. $\frac{d}{dx}\left( {\text{sin}^{-1}x} \right) = \frac{1}{\sqrt{1 - x^{2}}}$

16\. $\frac{d}{dx}\left( {\text{tan}^{-1}x} \right) = \frac{1}{1 + x^{2}}$

17\. $\frac{d}{dx}\left( {\text{sec}^{-1}x} \right) = \frac{1}{|x|\sqrt{x^{2} - 1}}$

18\. $\frac{d}{dx}\left( {\text{cos}^{-1}x} \right) = - \frac{1}{\sqrt{1 - x^{2}}}$

19\. $\frac{d}{dx}\left( {\text{cot}^{-1}x} \right) = - \frac{1}{1 + x^{2}}$

20\. $\frac{d}{dx}\left( {\text{csc}^{-1}x} \right) = - \frac{1}{|x|\sqrt{x^{2} - 1}}$

Exponential and Logarithmic Functions

21\. $\frac{d}{dx}\left( e^{x} \right) = e^{x}$

22\. $\frac{d}{dx}\left( {\text{ln}\mspace{2mu}|x|} \right) = \frac{1}{x}$

23\. $\frac{d}{dx}\left( b^{x} \right) = b^{x}\text{ln}\mspace{2mu} b$

24\. $\frac{d}{dx}\left( {\text{log}_{b}x} \right) = \frac{1}{x\mspace{2mu}\text{ln}\mspace{2mu} b}$

Hyperbolic Functions

25\. $\frac{d}{dx}\left( {\text{sinh}\mspace{2mu} x} \right) = \text{cosh}\mspace{2mu} x$

26\. $\frac{d}{dx}\left( {\text{tanh}\mspace{2mu} x} \right) = \text{sech}^{2}\mspace{2mu} x$

27\. $\frac{d}{dx}\left( {\text{sech}\mspace{2mu} x} \right) = \text{−sech}\mspace{2mu} x\ \text{tanh}\mspace{2mu} x$

28\. $\frac{d}{dx}\left( {\text{cosh}\mspace{2mu} x} \right) = \text{sinh}\mspace{2mu} x$

29\. $\frac{d}{dx}\left( {\text{coth}\mspace{2mu} x} \right) = \text{−}\text{csch}^{2}\mspace{2mu} x$

30\. $\frac{d}{dx}\left( {\text{csch}\mspace{2mu} x} \right) = \text{−csch}\mspace{2mu} x\ \text{coth}\mspace{2mu} x$

Inverse Hyperbolic Functions

31\. $\frac{d}{dx}\left( {\text{sinh}^{-1}x} \right) = \frac{1}{\sqrt{x^{2} + 1}}$

32\. $\frac{d}{dx}\left( {\text{tanh}^{-1}x} \right) = \frac{1}{1 - x^{2}}\left( {|x| < 1} \right)$

33\. $\frac{d}{dx}\left( {\text{sech}^{-1}x} \right) = - \frac{1}{x\sqrt{1 - x^{2}}}\quad\left( {0 < x < 1} \right)$

34\. $\frac{d}{dx}\left( {\text{cosh}^{-1}x} \right) = \frac{1}{\sqrt{x^{2} - 1}}\quad\left( {x > 1} \right)$

35\. $\frac{d}{dx}\left( {\text{coth}^{-1}x} \right) = \frac{1}{1 - x^{2}}\quad\left( {|x| > 1} \right)$

36\. $\frac{d}{dx}\left( {\text{csch}^{-1}x} \right) = - \frac{1}{|x|\sqrt{1 + x^{2}}}\ \left( {x \neq 0} \right)$

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Citation information

• Use the information below to generate a citation. We recommend using a citation tool such as this one.
• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/b-table-of-derivatives

© Jul 15, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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C | Review of Pre-Calculus

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/c-review-of-pre-calculus

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Calculus Volume 3C \| Review of Pre-Calculus

------------------------------------------------------------------------

Formulas from Geometry

$A = \text{area},$ $V = \text{Volume},\ \text{and}$ $S = \text{lateral surface area}$

Formulas from Algebra

Laws of Exponents

$\begin{array}{rllcccccccccc}

{x^{m}x^{n}} & = & x^{m + n} & & & \frac{x^{m}}{x^{n}} & = & x^{m - n} & & & \left( x^{m} \right)^{n} & = & x^{mn} \\

x^{\text{−}n} & = & \frac{1}{x^{n}} & & & \left( {xy} \right)^{n} & = & {x^{n}y^{n}} & & & \left( \frac{x}{y} \right)^{n} & = & \frac{x^{n}}{y^{n}} \\

x^{1\text{/}n} & = & \sqrt[n]{x} & & & \sqrt[n]{xy} & = & {\sqrt[n]{x}\sqrt[n]{y}} & & & \sqrt[n]{\frac{x}{y}} & = & \frac{\sqrt[n]{x}}{\sqrt[n]{y}} \\

x^{m\text{/}n} & = & {\sqrt[n]{x^{m}} = \left( \sqrt[n]{x} \right)^{m}} & & & & & & & & & &

\end{array}$

Special Factorizations

$\begin{array}{rll}

{x^{2} - y^{2}} & = & {\left( {x + y} \right)\left( {x - y} \right)} \\

{x^{3} + y^{3}} & = & {\left( {x + y} \right)\left( {x^{2} - xy + y^{2}} \right)} \\

{x^{3} - y^{3}} & = & {\left( {x - y} \right)\left( {x^{2} + xy + y^{2}} \right)}

\end{array}$

Quadratic Formula

If $ax^{2} + bx + c = 0,$ then $x = \frac{\text{−}b \pm \sqrt{b^{2} - 4ac}}{2a}.$

Binomial Theorem

$\left( {a + b} \right)^{n} = a^{n} + \left( \begin{array}{l}

n \\

1

\end{array} \right)a^{n - 1}b + \left( \begin{array}{l}

n \\

2

\end{array} \right)a^{n - 2}b^{2} + \cdots + \begin{pmatrix}

n \\

{n - 1}

\end{pmatrix}ab^{n - 1} + b^{n},$

where $\left( \begin{array}{l}

n \\

k

\end{array} \right) = \frac{n\left( {n - 1} \right)\left( {n - 2} \right)\cdots\left( {n - k + 1} \right)}{k\left( {k - 1} \right)\left( {k - 2} \right)\cdots 3 \cdot 2 \cdot 1} = \frac{n!}{k!\left( {n - k} \right)!}$

Formulas from Trigonometry

Right-Angle Trigonometry

$\begin{array}{lccl}

{\text{sin}\mspace{2mu}\theta = \frac{\text{opp}}{\text{hyp}}} & & & {\text{csc}\mspace{2mu}\theta = \frac{\text{hyp}}{\text{opp}}} \\

{\text{cos}\mspace{2mu}\theta = \frac{\text{adj}}{\text{hyp}}} & & & {\text{sec}\mspace{2mu}\theta = \frac{\text{hyp}}{\text{adj}}} \\

{\text{tan}\mspace{2mu}\theta = \frac{\text{opp}}{\text{adj}}} & & & {\text{cot}\mspace{2mu}\theta = \frac{\text{adj}}{\text{opp}}}

\end{array}$

Trigonometric Functions of Important Angles

| | | | | |

|--------------|----------------------------|--------------------------------|--------------------------------|--------------------------------|

| $\theta$ | $\text{Radians}$ | $\text{sin}\mspace{2mu}\theta$ | $\text{cos}\mspace{2mu}\theta$ | $\text{tan}\mspace{2mu}\theta$ |

| $0\text{°}$ | $0$ | $0$ | $1$ | $0$ |

| $30\text{°}$ | $\text{π}\text{/}\text{6}$ | $1\text{/}2$ | $\sqrt{3}\text{/}2$ | $\sqrt{3}\text{/}3$ |

| $45\text{°}$ | $\text{π}\text{/}\text{4}$ | $\sqrt{2}\text{/}2$ | $\sqrt{2}\text{/}2$ | $1$ |

| $60\text{°}$ | $\text{π}\text{/}\text{3}$ | $\sqrt{3}\text{/}2$ | $1\text{/}2$ | $\sqrt{3}$ |

| $90\text{°}$ | $\text{π}\text{/}2$ | $1$ | $0$ | — |

Fundamental Identities

$\begin{array}{rllccrll}

{\text{sin}^{2}\theta + \text{cos}^{2}\theta} & = & 1 & & & {\text{sin}\left( {\text{−}\mspace{2mu}\theta} \right)} & = & {\text{−}\text{sin}\mspace{2mu}\theta} \\

{1 + \text{tan}^{2}\theta} & = & {\text{sec}^{2}\theta} & & & {\text{cos}\left( {\text{−}\mspace{2mu}\theta} \right)} & = & {\text{cos}\mspace{2mu}\theta} \\

{1 + \text{cot}^{2}\theta} & = & {\text{csc}^{2}\theta} & & & {\text{tan}\left( {\text{−}\mspace{2mu}\theta} \right)} & = & {\text{−}\text{tan}\mspace{2mu}\theta} \\

{\text{sin}\left( {\frac{\pi}{2} - \theta} \right)} & = & {\text{cos}\mspace{2mu}\theta} & & & {\text{sin}\left( {\theta + 2\pi} \right)} & = & {\text{sin}\mspace{2mu}\theta} \\

{\text{cos}\left( {\frac{\pi}{2} - \theta} \right)} & = & {\text{sin}\mspace{2mu}\theta} & & & {\text{cos}\left( {\theta + 2\pi} \right)} & = & {\text{cos}\mspace{2mu}\theta} \\

{\text{tan}\left( {\frac{\pi}{2} - \theta} \right)} & = & {\text{cot}\mspace{2mu}\theta} & & & {\text{tan}\left( {\theta + \pi} \right)} & = & {\text{tan}\mspace{2mu}\theta}

\end{array}$

Law of Sines

$\frac{\text{sin}\mspace{2mu} A}{a} = \frac{\text{sin}\mspace{2mu} B}{b} = \frac{\text{sin}\mspace{2mu} C}{c}$

Law of Cosines

$\begin{array}{rll}

a^{2} & = & {b^{2} + c^{2} - 2bc\ \text{cos}\ A} \\

b^{2} & = & {a^{2} + c^{2} - 2ac\ \text{cos}\ B} \\

c^{2} & = & {a^{2} + b^{2} - 2ab\ \text{cos}\ C}

\end{array}$

Addition and Subtraction Formulas

$\begin{matrix}

{\text{sin}\ (x + y)} & = & {\text{sin}\ x\ \text{cos}\ y + \text{cos}\ x\ \text{sin}\ y} \\

{\text{sin}\mspace{2mu}(x - y)} & = & {\text{sin}\ x\ \text{cos}\ y - \text{cos}\ x\ \text{sin}\ y} \\

{\text{cos}\mspace{2mu}(x + y)} & = & {\text{cos}\ x\ \text{cos}\ y - \text{sin}\ x\ \text{sin}\ y} \\

{\text{cos}\mspace{2mu}(x - y)} & = & {\text{cos}\ x\ \text{cos}\ y + \text{sin}\ x\ \text{sin}\ y} \\

{\text{tan}\mspace{2mu}(x + y)} & = & \frac{\text{tan}\ x + \text{tan~}y}{1 - \text{tan}\ x\ \text{tan~}y} \\

{\text{tan}(x - y)} & = & \frac{\text{tan}\ x - \text{tan~}y}{1 + \text{tan}\ x\ \text{tan~}y}

\end{matrix}$

Double-Angle Formulas

$\begin{array}{rll}

{\text{sin}\ 2x} & = & {2\mspace{2mu}\text{sin}\ x\ \text{cos}\ x} \\

{\text{cos}\ 2x} & = & {\text{cos}^{2}x - \text{sin}^{2}x = 2\mspace{2mu}\text{cos}^{2}x - 1 = 1 - 2\mspace{2mu}\text{sin}^{2}x} \\

{\text{tan}\ 2x} & = & \frac{2\mspace{2mu}\text{tan}\ x}{1 - \text{tan}^{2}x}

\end{array}$

Half-Angle Formulas

$\begin{array}{rll}

{\text{sin}^{2}x} & = & \frac{1 - \text{cos}\ 2x}{2} \\

{\text{cos}^{2}x} & = & \frac{1 + \text{cos}\ 2x}{2}

\end{array}$

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Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction

Citation information

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/c-review-of-pre-calculus

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