习题答案

原书第 316 页

习题答案与提示

第七章

习题7-1

(1) $ \left| (x,y) \mid y^{2}-2x+1\gt 0 \right| $;

(2) $ \left|\left(x,y\right)\right|\left|x+y\gt 0,x-y\gt 0\right| $;

(3) $ \left| (x,y) \mid x\gt 0, y\geqslant0 \right| $, $ x^{2}\geqslant y $;

(4) $ \{(x,y) \mid y \gt x, x \geq 0, x^2 + y^2 \lt 1\} $;

(5) $ \left| (x,y) \mid y \geqslant 0, y \geqslant -2x \text{ 或 } y \leqslant 0, y \leqslant -2x \text{ 且 } x \neq 0, y \neq 0 \right| $;

(6) $ \left| (x,y) \mid 1 \lt x^{2} + y^{2} + z^{2} \leqslant 2^{2} \right| $

(1) 1; (2) +\infty; (3) 0; (4) 0; (5) 0; (6) $ \frac{1}{6} $.

(1) 不存在；(2) 不存在；(3) 提示：取路径 $ y = kx^{2} - x (k \neq 0) $，可证明极限不存在；(4) 提示：取路径 y = 0 与 y = x，可证明极限不存在.

5.（1）在直线 $ x+y=0 $ 上的点不连续；

（2）在（0，0）处不连续；（3）在全平面xOy上连续；

（4）在全平面 xOy 上连续.

习题7-2

0, 6.

$ \frac{1}{2} $

$ \frac{\pi}{4} $

(1) $ \frac{\partial z}{\partial x} = y + \frac{1}{y} $, $ \frac{\partial z}{\partial y} = x - \frac{x}{y^{2}} $;

(2) $ \frac{\partial z}{\partial x} = \sin(x + y) + x\cos(x + y) $, $ \frac{\partial z}{\partial y} = x\cos(x + y) $;

(3) $ \frac{\partial z}{\partial x} = \frac{e^y}{y^2} $, $ \frac{\partial z}{\partial y} = \frac{x e^y}{y^2} \left( 1 - \frac{2}{y} \right) $;

原书第 317 页

(4)

$$ \frac{\partial z}{\partial x}=\frac{2}{y\sin\frac{2x}{y}},\quad\frac{\partial z}{\partial y}=-\frac{2x}{y^{2}\sin\frac{2x}{y}}, $$

(5) $ \frac{\partial z}{\partial x} = y^2 (1 + xy)^{y-1} $, $ \frac{\partial z}{\partial y} = (1 + xy)^y \left[ \ln(1 + xy) + \frac{xy}{1 + xy} \right] $;

(6) $ \frac{\partial z}{\partial x} = y \cos xy e^{\sin xy} $, $ \frac{\partial z}{\partial y} = x \cos xy e^{\sin xy} $;

(7)

$$ \frac{\partial u}{\partial x}=\frac{y^{2}}{|y|\left(x^{2}+y^{2}\right)}; $$

(8) $ \frac{\partial u}{\partial x} = \frac{y^2}{x} x^y^2 $, $ \frac{\partial u}{\partial y} = \frac{zy' \ln x}{y} x^y^2 $, $ \frac{\partial u}{\partial z} = y^2 x^2 \ln x \ln y $.

$ \frac{\partial^{2}u}{\partial x^{2}} = (2+x)yze^{x+y+z}, \frac{\partial^{2}u}{\partial y^{2}} = (2+y)xze^{x+y+z}, \frac{\partial^{2}u}{\partial z^{2}} = (2+z)xye^{x+y+z}. $

$ \frac{\partial^{2}z}{\partial x\partial y}=\frac{\partial^{2}z}{\partial y\partial z}=\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}}. $

习题7-3

$ \Delta z = -0.20404 $, $ dz = -0.2 $.

$ \mathrm{du}\bigg|_{(1,1,1)}=\frac{1}{2}\mathrm{dx}-\frac{1}{2}\mathrm{dy}+\mathrm{dz}. $

(1) $ \frac{ydx-xdy}{|y|\sqrt{y^{2}-x^{2}}}; $

(2) $ \mathrm{e}^{x} \left[ \sin(x+y) + \cos(x+y) \right] \mathrm{d}x + \mathrm{e}^{x} \cos(x+y) \mathrm{d}y $;

(3) $ -\frac{x\mathrm{d}x+y\mathrm{d}y+z\mathrm{d}z}{\left(x^{2}+y^{2}+z^{2}\right)^{3/2}} $;

(4)

$$ \mathrm{d}x+\left(\frac{1}{2}\cos\frac{y}{2}+ze^{y}\right)\mathrm{d}y+ye^{y}\mathrm{d}z; $$

$$ z\left(\frac{x}{y}\right)^{s-1}\frac{1}{y}\mathrm{d}x-\frac{xz}{y^{2}}\left(\frac{x}{y}\right)^{s-1}\mathrm{d}y+\left(\frac{x}{y}\right)^{s}\ln\frac{x}{y}\mathrm{d}z; $$

(6) $ \frac{zx}{1+x^{2}+y^{2}}dx+\frac{2y}{1+x^{2}+y^{2}}dy. $

4.（1）2.95；（2）0.97；（3）0.5023.

$ -200\pi \, cm^{3} $

习题7-4

$$ \frac{\partial z}{\partial x}=-\frac{2y}{x}\left[\frac{y}{x^{2}}\ln\left(3y-2x\right)+\frac{y}{x\left(3y-2x\right)}\right] $$

$$ \frac{\partial z}{\partial y}=\frac{y}{x}\left[\frac{2}{x}\ln(3y-2x)+\frac{3y}{x(3y-2x)}\right] $$

原书第 318 页

$ \frac{\partial z}{\partial u}=3u^{2}\sin v\cos v(\cos v-\sin v) $,

$$ \frac{\partial z}{\partial v}=-2u^{3}\sin v\cos v\left(\sin v+\cos v\right)+u^{3}\left(\sin^{3}v+\cos^{3}v\right). $$

$$ \frac{\partial u}{\partial x}=\left(x-y\right)^{2+y^{2}}\left[\frac{x^{2}+y^{2}}{x-y}+2x\ln\left(x-y\right)\right] $$

$$ \frac{\partial u}{\partial y}=\left(x-y\right)^{2+y^{2}}\left[2y\ln\left(x-y\right)-\frac{x^{2}+y^{2}}{x-y}\right]. $$

$$ \frac{\partial z}{\partial x}=2xf^{\prime}_{1}+ye^{xy}f^{\prime}_{2},\quad\frac{\partial z}{\partial y}=-2yf^{\prime}_{1}+xe^{xy}f^{\prime}_{2}. $$

$ \frac{\partial z}{\partial x} = \frac{(x-2)e^x}{x^3} \cos \frac{e^x}{x^2} $.

$$ \frac{\partial z}{\partial t}=-\left(\mathbf{e}^{t}+\mathbf{e}^{-t}\right). $$

$ \frac{\partial z}{\partial x}=2x+\frac{1}{2}\frac{\cos x}{\sqrt{\sin x}} $.

$$ \frac{\partial u}{\partial x}=\frac{1}{y}f_{1}^{\prime},\quad\frac{\partial u}{\partial y}=-\frac{x}{y^{2}}f_{1}^{\prime}+\frac{1}{z}f_{2}^{\prime},\quad\frac{\partial u}{\partial z}=-\frac{y}{z^{2}}f_{2}^{\prime}; $$

(2)

$$ \frac{\partial u}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\cdot\frac{\partial\varphi}{\partial t}\cdot\frac{\partial\psi}{\partial t},\quad\frac{\partial u}{\partial y}=\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\left(\frac{\partial\varphi}{\partial y}+\frac{\partial\varphi}{\partial t}\cdot\frac{\partial\varphi}{\partial y}\right). $$

11.

$$ \frac{\partial^{2}z}{\partial x^{2}}=2f^{\prime}+4x^{2}f^{\prime \prime},\quad\frac{\partial^{2}z}{\partial x\partial y}=4xyf^{\prime \prime},\quad\frac{\partial^{2}z}{\partial y^{2}}=2f^{\prime}+4y^{2}f^{\prime \prime}; $$

(2)

$$ \frac{\partial^{2}z}{\partial x\partial y}=4f^{\prime\prime}{}_{21}-\frac{2y^{3}}{x^{2}}f^{\prime\prime}{}_{22}, $$

$$ \frac{\partial^{2}z}{\partial x^{2}}=4f_{1}^{\prime}+4xf_{11}^{\prime\prime}-4\frac{y^{2}}{x}f_{12}^{\prime\prime}+\frac{y^{4}}{x^{3}}f_{22}^{\prime\prime}, $$

$$ \frac{\partial^{2}z}{\partial y^{2}}=2f_{2}^{\prime}+\frac{4y^{2}}{x}f^{\prime\prime}_{22}. $$

12.

$$ \frac{\partial^{2}z}{\partial x\partial y}=x\mathrm{e}^{2y}\frac{\partial^{2}f}{\partial u^{2}}+\mathrm{e}^{y}\frac{\partial^{2}f}{\partial u\partial y}+x\mathrm{e}^{y}\frac{\partial^{2}f}{\partial x\partial u}+\frac{\partial^{2}f}{\partial x\partial y}+\mathrm{e}^{y}\frac{\partial f}{\partial u}. $$

14.

$$ \mathrm{d}u=\left(y+z\right)f^{\prime}\mathrm{d}x+\left(x+z\right)f^{\prime}\mathrm{d}y+\left(x+y\right)f^{\prime}\mathrm{d}z, $$

$$ \frac{\partial u}{\partial x}=\left(y+z\right)f^{\prime},\frac{\partial u}{\partial y}=\left(x+z\right)f^{\prime},\frac{\partial u}{\partial z}=\left(x+y\right)f^{\prime}. $$

$ \mathrm{d}z = (2xf'_1 + yf'_2)dx + (xf'_2 - 2yf'_1)dy $,

$$ \frac{\partial z}{\partial x}=2xf_{1}^{\prime}+yf_{2}^{\prime}+\frac{\partial z}{\partial y}=xf_{2}^{\prime}-2yf_{1}^{\prime}. $$

a=3.

习题7-5

$$ \frac{\partial y}{\partial x}=\frac{\partial-x^{2}}{y^{2}-x} $$

原书第 319 页

$$ \frac{\partial^{2}z}{\partial x^{2}}=\frac{\left(2+z\right)^{2}-x^{2}}{\left(2+z\right)^{3}}. $$

$$ 3.\ \frac{\partial z}{\partial x}=\frac{yz-\sqrt{xyz}}{\sqrt{xyz}-xy},\quad\frac{\partial z}{\partial y}=\frac{xz*2\sqrt{xyz}}{\sqrt{xyz}-xy},\quad\frac{\partial x}{\partial y}=\frac{xz-2\sqrt{xyz}}{\sqrt{xyz}-yz}. $$

$$ \begin{aligned}\frac{\partial z}{\partial x}&=\frac{1}{2ay}\left[1+\cos\left(2az\right)\right],\frac{\partial z}{\partial x}=-\frac{1}{2ay}\sin\left(2az\right)\\\frac{\partial^{2}z}{\partial x^{2}}&=-\frac{1}{2ay^{2}}\left[1+\cos\left(2az\right)\right]\sin\left(2az\right),\\\frac{\partial^{2}z}{\partial y^{2}}&=\frac{1}{2ay^{2}}\left[1+\cos\left(2az\right)\right]\sin\left(2az\right),\\\frac{\partial^{2}z}{\partial x\partial y}&=-\frac{1}{2ay^{2}}\left[1+\cos\left(2az\right)\right]\cos\left(2az\right).\end{aligned} $$

$$ \frac{\partial z}{\partial x}=\frac{z\ln z}{x(\ln z-1)},\frac{\partial z}{\partial y}=\frac{z^{2}}{xy(1-\ln y)},\mathrm{d}z=\frac{yz\ln\mathrm{d}z-z^{2}\mathrm{d}y}{xy(\ln z-1)}; $$

(2)

$$ \begin{aligned}&\frac{\partial z}{\partial x}=-\frac{\sin2x}{\sin2z},\quad\frac{\partial z}{\partial y}=-\frac{\sin2y}{\sin2z},\\&dz=-\frac{\sin2x\mathrm{d}x+\sin2y\mathrm{d}y}{\sin2z};\\ \end{aligned} $$

(3)

$$ \frac{\partial z}{\partial x}=\frac{ye^{-xy}}{e^{x}-2},\quad\frac{\partial z}{\partial y}=\frac{xe^{-xy}}{e^{x}-2},\quad\mathrm{d}z=\frac{e^{-xy}(y\mathrm{d}x+x\mathrm{d}y)}{e^{x}-2}; $$

(4)

$$ \frac{\partial z}{\partial x}=\frac{zf_{1}^{\prime}}{1-xf_{1}^{\prime}-f_{2}^{\prime}},\quad\frac{\partial z}{\partial y}=\frac{-f_{2}^{\prime}}{1-xf_{1}^{\prime}-f_{2}^{\prime}},\quad\mathrm{d}z=\frac{zf_{1}^{\prime}\mathrm{d}x-f_{2}^{\prime}\mathrm{d}y}{1-xf_{1}^{\prime}-f_{2}^{\prime}}. $$

$$ \begin{align*}\oint_{.}\frac{\partial z}{\partial x}&=-\frac{y F_{2}^{\prime}}{F_{1}^{\prime}+y F_{2}^{\prime}}\ ,\ \frac{\partial z}{\partial y}=-\frac{F_{1}^{\prime}+\left(x+z\right)F_{2}^{\prime}}{F_{1}^{\prime}+y F_{2}^{\prime}}\ ,\\\frac{\partial^{2}z}{\partial x^{2}}&=-\frac{y^{2}}{\left(F_{1}^{\prime}+y F_{2}^{\prime}\right)^{3}}\big(F_{2}^{\prime2}\cdot F_{11}^{\prime \prime}-2F_{1}^{\prime}\cdot F_{2}^{\prime}\cdot F_{12}^{\prime \prime}+F_{1}^{\prime2}\cdot F_{22}^{\prime \prime}\big)\ .\end{align*} $$

$$ \frac{\partial y}{\partial x}=\frac{x}{3z+1},\quad\frac{\mathrm{d}z}{\mathrm{d}x}=-\frac{x(6z+1)}{(3z+1)}. $$

10.

$$ \frac{\partial u}{\partial x}=-\frac{u}{u^{2}+v^{2}},\quad\frac{\partial u}{\partial y}=\frac{v}{u^{2}+v^{2}},\quad\frac{\partial v}{\partial x}=\frac{v}{u^{2}+v^{2}},\quad\frac{\partial v}{\partial y}=\frac{u}{u^{2}+v^{2}}. $$

11.

$$ \frac{\mathrm{d}z}{\mathrm{d}x}=\frac{\left(f+xf^{\prime}\right)F y-xf^{\prime}F x}{F y+xf^{\prime}F z}\qquad\left(F y+xf^{\prime}F z\neq0\right). $$

12.

$$ \frac{\mathrm{d}^{2}z}{\mathrm{d}x^{2}}=\frac{\partial^{2}f}{\partial x^{2}}+\frac{2\varphi^{\prime}(t)}{\left(1+\cos t\right)^{2}}\frac{\partial^{2}f}{\partial x\partial y}+\left[\frac{\varphi^{\prime}(t)}{1+\cos t}\right]^{2}\frac{\partial^{2}f}{\partial y^{2}}+\frac{\left(1+\cos t\right)\varphi^{\prime\prime}(t)+\sin t\varphi^{\prime}(t)}{\left(1+\cos t\right)^{3}}\frac{\partial f}{\partial y} $$

(1) -2; (2) -1.

习题7-6

$$ \frac{\sqrt{2}}{2} $$

$$ \frac{81}{\sqrt{5}} $$

原书第 320 页

$ \frac{\partial z}{\partial l} $; (2) $ \frac{2}{3} $; (3) $ \frac{9}{1183} $; (4) $ \frac{43}{15} $; (5) $ \frac{-3\sqrt{2}}{2} $.

$ \frac{\partial z}{\partial l}\big|_{(-1,1)}=-\frac{3}{\sqrt{5}} $, $ \mathrm{grad}f(-1,1)=|-3,3| $. 减小最快的方向是 $ \{\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\} $; 变化率为零的方向是 $ \{\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\} $或 $ \{-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\} $.

(1) {16, 18}; (2) $ \{\frac{1}{2}, 0, \frac{1}{2}\}; (3) \{-\frac{9\sqrt{3}}{2}, -\frac{3\sqrt{3}}{2}\} $;

(4) $ \{-\frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4}, 0\} $; (5) \{1, 1, 1\}.

沿 $ \left|-1,2,-1\right| $ 的方向导数值取最大值 $ 2\sqrt{6} $

$ -\frac{q}{4\pi\varepsilon r^3} $ | $ x $, $ y $, $ z $|.

$ \frac{\partial f}{\partial l} = \cos \alpha \sin \alpha $.

$ 4[-1+x_{0}(x_{0}-2y_{0})+x_{0}^{2}y_{0}^{2}]\{y_{0}-x_{0}-x_{0}y_{0}^{2},x_{0}-x_{0}^{2}y_{0}\}. $

球面 $ (x-a)^{2}+(y-b)^{2}+(z-c)^{2}=1 $上所有点.

习题7-7

(1) (1, 1, 2); (2) $ 2x + y - 4 = 0 $; (3) $ \{0, \sqrt{\frac{2}{5}}, \sqrt{\frac{3}{5}}\} $;

(4) $ \frac{x-1}{2} = \frac{y+2}{-4} = \frac{z-2}{6} $.

切线 $ \{\begin{array}{l}x-\frac{1}{2}=-(z-\frac{1}{2})\\y=\frac{1}{2};\end{array}. $ 法平面 x-z=0.

$ (-1,\ 1,\ -1) $, $ \left(-\frac{1}{3},\ \frac{1}{9},\ -\frac{1}{27}\right) $.

$$ \frac{x-\frac{1}{\sqrt{2}}}{1}=\frac{y-\frac{1}{\sqrt{2}}}{-1}=\frac{z-\frac{1}{\sqrt{z}}}{-1},\quad x-y-z+\frac{1}{\sqrt{2}}=0. $$

切线： $ \frac{x-1}{16}=\frac{y-1}{9}=\frac{z-1}{-1} $，

法平面： $ 16x+9y-z-24=0 $.

$ ax_{0}x + by_{0}y + cz_{0}z = 1 $; $ \frac{x - x_{0}}{ax_{0}} = \frac{y - y_{0}}{by_{0}} = \frac{z - z_{0}}{cz_{0}} $.

$$ (x-3)+5(y+1)+2(z-2)=0,\frac{x-3}{1}=\frac{y+1}{5}=\frac{z-2}{2}. $$

$ z-4=2(x-1)+6(y-1) $; $ \frac{x-1}{2}=\frac{y-1}{6}=\frac{z-4}{-1} $.

原书第 321 页

$ \pm\frac{5\sqrt{10}}{99} $, $ \{\frac{1}{\sqrt{10}}, -\frac{1}{27\sqrt{10}}, \frac{\sqrt{10}}{27}\} $.

$ \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} $, $ x + 2y + 3z - 8 = 0 $.

$ \frac{1}{ab}\sqrt{2(a^{2}+b^{2})} $.

$ \left(\frac{1}{3}, \frac{4}{3}, \frac{4}{3}\right) $ 及 $ \left(-\frac{1}{3}, -\frac{4}{3}\right) $.

提示：设点 $ M_{0} $ 为所求，曲面在点 $ M_{0} $ 处的法线的三个方向数应相等，即 $ F_{x}(M_{0}) = F_{y}(M_{0}) = Fz(M_{0}) $.

$ \frac{x-a}{bc}=\frac{y-b}{ca}=\frac{z-c}{ab} $, $ bcx+acy+abz-3=0 $.

a = -5, b = -2.

习题7-8

极大值 2. 极大值 $ f(-1,-1)=-2 $，极小值 $ f(1,1)=-2 $.

极大值 $ f(-3,2)=31 $

极小值 $ f(1,0) = -1 $，极小值 $ f(-1,0) = -1 $.

极小值 $ f\left(\frac{1}{2}, -1\right) = -\frac{e}{2} $.

极小值 $ z(-2,0)=1 $，极大值 $ z\left(\frac{16}{7},0\right)=-\frac{8}{7} $.

极大值 $ f\left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right) = 4\ln\frac{5}{2\sqrt{3}} $

$ \left(\frac{8}{5},\frac{3}{5}\right) $

长，宽，高分别为 $ \frac{2R}{\sqrt{3}} $， $ \frac{2R}{\sqrt{3}} $， $ \frac{R}{\sqrt{3}} $.

在点 $ \left(\frac{\pi}{4}, \frac{\pi}{4}\right) $处取得最大值 $ \frac{1}{2} $.

$ \left(\frac{a}{\sqrt{3}}, \frac{b}{\sqrt{3}}, \frac{c}{\sqrt{3}}\right) $

最大值 $ z\left(\frac{5}{2}, \frac{5}{4}\right) = \frac{625}{64} $，最小值 $ z(0, 0) = 0 $.

最大值 $ f(\pm2, 0)=4 $，最小值 $ f(0, \pm2)=-4 $.

提示：设 $ a = e^x $， $ \beta = y $， $ \gamma = |z| $，把问题转化为求三个正数的积 $ u = \alpha \cdot \beta \cdot \gamma $ 在约束条件 $ \alpha + \beta + \gamma = 3 $ 下的最大值，在 $ x = 0 $， $ y = 1 $， $ |z| = 1 $ 时取得最大值 1.

提示：在约束条件 $ 2x^{2}+2y^{2}+z^{2}=1 $ 下求目标函数 $ \frac{\partial f}{\partial l}=\sqrt{2}(x-y) $ 的最大值，点

原书第 322 页

$ \left(\frac{1}{2}, -\frac{1}{2}, 0\right) $ 为所示.

提示：将问题转化为求目标函数 $ u = x^2 + y^2 + (z-1)^2 $ 在约束条件 $ x^2 + y^2 - R^2 = 0 $ 及 $ x + y + z - 1 = 0 $ 下的最大值及最小值，长半轴 $ a = \sqrt{3}R $，短半轴 $ b = R $.

y=0.387 2x+3.622 4.

习题7-9

$ \ln(1+x+y)=x+y-\frac{1}{2}(x+y)^2+\frac{1}{3}(x+y)^3+R_3 $,

其中 $ R_{3} = -\frac{1}{4} \frac{(x+y)^{4}}{(1+\theta x+\theta y)^{4}} $ （ $ 0 \lt \theta \lt 1 $）.

$$ \mathrm{e}^{x}\ln\left(1+y\right)=y+\frac{1}{2!}\left(2xy-y^{2}\right)+\frac{1}{3!}\left(3x^{2}y-3xy^{2}+2y^{3}\right)+R_{3}, $$

其中

$$ R_{3}=\frac{e^{4s}}{24}\left[x^{4}\ln\left(1+\theta y\right)+\frac{4x^{3}y}{1+\theta y}-\frac{6x^{2}y^{2}}{\left(1+\theta y\right)^{2}}+\frac{8xy^{3}}{\left(1+\theta y\right)^{3}}-\frac{6y^{4}}{\left(1+\theta y\right)^{4}}\right]. $$

$ \sin(x^{2}+y^{2})=x^{2}+y^{2}+o(\rho^{2}) $，其中 $ \rho=\sqrt{x^{2}+y^{2}} $.

总练习题七

(1) e; (2) 1; (3) $ \frac{dx+dy}{4} $; (4) $ \frac{1}{2} $; (5) $ \{\frac{2}{a}, \frac{4}{9}, -\frac{4}{9}\} $; (6) 6.

(1) D; (2) D; (3) B; (4) C; (5) B; (6) D.

$ \left| (x,y) \right| \leq 0 \lt x^{2} \lt y^{2} \lt 1 $, $ y^{2} \leq 4x $, $ \frac{\sqrt{2}}{\ln \frac{3}{4}} $.

提示：(1) $ 0 \leqslant \frac{|xy|}{\sqrt{x^2 + y^2}} \leqslant \frac{1}{2} \sqrt{x^2 + y^2} $，应用夹逼准则可得

$$ \liminf_{x\to0\atop x\to0}(x,\ y)=0=f(0,\ 0); $$

(2) $ f_{x}(x, y)=\{\begin{aligned}&\frac{y^{3}}{(x^{2}+y^{2})^{3/2}},&(x, y)\neq(0,0),\\&0,&(x, y)=(0,0),\end{aligned}. $ 在点 $ (0,0) $

的任一邻域内有界，其中，在（0，0）去心邻域

$$ \left|f_{x}(x,y)\right|=\left|\frac{\rho^{3}\sin^{3}\theta}{\rho^{3}}\right|=\left|\sin^{3}\theta\right|\leqslant1. $$

同理可证： $ f_{y}(0,0) $ 在点（0，0）任一邻域内也有界.

$$ \lim_{\substack{\Delta x\to0\\\Delta y\to\Delta x}}\frac{\Delta f-\left[f_{x}(0,0)\Delta x+f_{y}(0,0)\Delta y\right]}{\rho}=\lim_{\substack{\Delta x\to0\\\Delta y\to\Delta x}}\frac{\Delta x\cdot\Delta y}{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}}=\frac{1}{2}\neq0; $$

（4）设 $ l=\left|\cos\alpha,\sin\alpha\right|\quad(0\leqslant\alpha\leqslant2\pi) $

原书第 323 页

$$ \frac{\partial f}{\partial l}=\lim_{\substack{x\to0\\ y\to0}}\frac{f(x,y)-f(0,0)}{\sqrt{x^{2}+y^{2}}}=\lim_{\rho\to0}\frac{\rho^{\prime}\cos\alpha\sin\alpha}{\rho}=\cos\alpha\sin\alpha. $$

$ \frac{\partial z}{\partial x} = e^x \sin y \cdot f_1 + 2x \cdot f_2 - \frac{1}{x^2} \cdot f_3 $,

$$\frac{\partial^{2}\partial z}{\partial x\partial y}=e^{x}\cos y\cdot f_{1}+\frac{1}{2}e^{2x}\sin2y\cdot f_{11}+2e^{x}(y\sin y+x\cos y)\cdot f_{12}+4xye^{x}\cos y\cdot f_{22}-\frac{1}{x^{2}}e^{x}\cos y\cdot f_{31}-\frac{2y}{x^{2}}f_{32}.$$

6.（1）-32；（2）-4.

$ f(u) = c_1 e^u + c_2 e^{-u} $. 提示：解二阶常系数齐次微分方程 $ f''(u) - f(u) = 0 $.

$ \frac{x-1}{8}=\frac{y+1}{10}=\frac{z-2}{7} $, $ 8x+10y+7z-12=0 $.

提示：两直线 $ \frac{x-x_{1}}{m_{1}}=\frac{y-y_{1}}{n_{1}}=\frac{z-z_{1}}{p_{1}} $与 $ \frac{x-x_{2}}{m_{2}}=\frac{y-y_{2}}{n_{2}}=\frac{z-z_{2}}{p_{2}} $共面的充要条件是

$$ \left|\begin{array}{c c c}{x_{2}-x_{1}}&{y_{2}-y_{1}}&{z_{2}-z_{1}}\\ {}&{}&{}\\ {m_{1}}&{n_{1}}&{p_{1}}\\ {}&{}&{}\\ {m_{2}}&{n_{2}}&{p_{2}}\\ \end{array}\right|=0. $$

提示：若令 $ \frac{a+b+c}{5}=S $，则问题转化为在 $ \frac{a+b+c}{5}=S $的条件下，证明函数 $ abc^3 $的最大值为 $ 27S^5 $，也可转化为在条件 $ abc^3=A $下，证明 $ a+b+c $有最小值 $ 5\left(\frac{A}{27}\right)^{\frac{1}{5}} $。

极大值 $ f(2,1)=4 $ ，最大值 $ f(2,1)=4 $ ，最小值 $ f(4,2)=-64 $

第八章

习题8-1

(1)(a) $ \iint\limits_{D}\left(x+y\right)^{2}\mathrm{d}\sigma\geqslant\iint\limits_{D}\left(x+y\right)^{3}\mathrm{d}\sigma; $

(2) (a) $ \iint_{D} e^{xy} d\sigma \leqslant \iint_{D} e^{2xy} d\sigma $;

$$ \iint\limits_{D}\mathbf{e}^{s y}\mathrm{d}\sigma\geqslant\iint\limits_{D}\mathbf{e}^{2s y}\mathrm{d}\sigma. $$

(1) $ 0 \leq I \leq 2 $; (2) $ \frac{\pi}{2} \leq I \leq \frac{\sqrt{2}\pi}{2} $.

$ f(x_{0}, y_{0}) $.

提示：用反证法.

提示：根据连续函数在闭区间上具有最值以及积分的性质.

习题8-2

(1) $ \int_{1}^{3} dx \int_{\frac{1}{x}}^{x} f(x, y) dy $ 或 $ \int_{\frac{1}{3}}^{1} dy \int_{\frac{1}{y}}^{3} f(x, y) dx + \int_{1}^{3} dy \int_{y}^{3} f(x, y) dx $;

原书第 324 页

(2) $ \int_{-1}^{1} \mathrm{d}x \int_{\sqrt{1-x^{2}}}^{\sqrt{4-x^{2}}} f(x, y) \, \mathrm{d}y + \int_{-1}^{1} \mathrm{d}x \int_{-\sqrt{4-x^{2}}}^{-\sqrt{1-x^{2}}} f(x, y) \, \mathrm{d}y + \int_{-2}^{-1} \mathrm{d}x \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} f(x, y) \, \mathrm{d}y + \int_{1}^{2} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} f(x, y) \, \mathrm{d}y $，或 $ \int_{1}^{2} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} f(x, y) \, \mathrm{d}y + \int_{-2}^{-1} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} f(x, y) \, \mathrm{d}y $， $ \int_{-1}^{1} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{-\sqrt{1-y^{2}}} f(x, y) \, \mathrm{d}y + \int_{-1}^{1} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} f(x, y) \, \mathrm{d}y $。

(1) $ \int_{0}^{4} dx \int_{\frac{x}{2}}^{\sqrt{x}} f(x, y) dy $;

(2) $ \int_{0}^{1} \mathrm{d}y \int_{2-y}^{1 + \sqrt{1 - y^{2}}} f(x, y) \, \mathrm{d}x $;

(3) $ \int_{0}^{2} \mathrm{d}x \int_{\frac{x}{2}}^{3-x} f(x, y) \, \mathrm{d}y $.

(1) $ \frac{6}{55} $;

(2) $ e-e^{-1} $;

(3) $ \pi^{2}-\frac{40}{9} $;

(4) $ 1 - \sin 1 $.

(1) $ \frac{a^{2}}{8}(15-16\ln2) $;

(2) $ \frac{2}{3}(p+q)\sqrt{pq} $.

(1) $ \frac{5}{6} $;

(2) $ \frac{88}{105} $

(1) $ \int_{0}^{\frac{\pi}{4}}\mathrm{d}\theta\int_{0}^{\sec\theta}f(r\cos\theta, r\sin\theta)r\mathrm{d}r+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\mathrm{d}\theta\int_{0}^{\cos\theta}f(r\cos\theta, r\sin\theta)r\mathrm{d}r $;

(2) $ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{d}\theta \int_{0}^{\sec\theta} f(r^{2}) r\mathrm{d}r $;

(3) $ \int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\int_{2\cos\theta}^{2}f(r\cos\theta, r\sin\theta)r\mathrm{d}r. $

(1) $ \frac{1}{3}(b^{3}-a^{3})\left(\frac{1}{\sqrt{1+\alpha^{2}}}-\frac{1}{\sqrt{1+\beta^{2}}}\right) $;

(2) $ \frac{a^{2}}{2}; $

(3) $ 7\pi $;

(4) $ \frac{\pi}{2} $

(1) $ 14a^{4} $;

(2) $ \frac{3}{64}\pi^{2} $;

(3) $ -\frac{3\pi}{2}; $

(4) $ \frac{1}{3}R^{3}\left(\pi-\frac{4}{3}\right) $

$ \frac{4}{3} $.

(1) $ \frac{\pi}{4}R^{2}a - \frac{2}{3}R^{3} $;

(2) $ \frac{5}{2}\pi $

(1) $ \pi $;

(2) $ \frac{2}{3}\pi ab. $

原书第 325 页

习题8-3

(1) $ \int_{-2}^{2} dx \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} dy \int_{0}^{x+y+10} f(x, y, z) dz $;

(2) $ \int_{-\frac{1}{2}}^{\frac{1}{2}} dx \int_{-\sqrt{1-4x^{2}}}^{\sqrt{1-4x^{2}}} dy \int_{3x^{2}+y^{2}}^{1-x^{2}} f(x, y, z) dz $;

(3) $ \int_{-1}^{1} \mathrm{d}x \int_{x^{2}}^{1} \mathrm{d}y \int_{0}^{x^{2}+y^{2}} f(x, y, z) \, \mathrm{d}z $;

(4) $ \int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}dx\int_{-\sqrt{\frac{1}{2}-x^{2}}}^{\sqrt{\frac{1}{2}-x^{2}}}dy\int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{1-x^{2}-y^{2}}}f(x,y,z)dz. $

(1) $ \frac{1}{2}\left(\ln 2 - \frac{5}{8}\right) $;

(2) $ \frac{1}{48} $;

(3) 0;

(4) $ \frac{\pi}{4}h^{2}R^{2} $;

(5) $ \frac{59\pi}{480}R^{5} $

$ \frac{4}{3} $

(1) $ \pi\left(1-\frac{1}{e}\right) $;

(2) $ \frac{7\pi}{12} $;

(3) $ \frac{16\pi}{3} $;

(4) 0.

(1) $ \pi R^{4} $;

(2) $ \frac{7\pi}{6}R^{4} $;

(3) $ \frac{\pi^{2}}{16}(2-\sqrt{2}) $

(1) $ \frac{1}{8} $;

(2) $ \frac{\pi}{10} $

(3) $ 8\pi $;

(4) $ \frac{\pi}{8}a^{4} $

$ k\pi R^{4} $ (k 为比例常数).

(1) $ 4\pi $;

习题8-4

(2) $ 4\pi abc $ (e-2).

(1) $ \bar{x} = -\frac{5a}{6} $, $ \bar{y} = 0 $;

(2) $ \overline{x} = \frac{b^2 + ab + a^2}{2(a+b)} $, $ \overline{y} = 0 $;

(3) $ \bar{x} = -\frac{1}{2} $, $ \bar{y} = \frac{8}{5} $;

(4) $ \bar{x} = \bar{y} = 0 $, $ \bar{z} = \frac{3}{8}R $.

原书第 326 页

$ (0, 0, \frac{5}{4}R) $.

(1) $ I_x = \frac{1}{3} \mu ab^3 $, $ I_y = \frac{1}{3} \mu ba^3 $;

(2) $ I_{x}=\frac{9}{4}\mu a^{4} $, $ I_{y}=\frac{9}{8}\mu a^{4} $;

(3) $ I_{v} = \frac{5}{4} \mu \pi R^{4} $;

(4) $ I_{y}=\frac{1}{4}\mu\pi a^{3}b $

$ \frac{4}{9}MR^{2} $

$$ F_{x}=F_{y}=0,\quad F_{z}=2\pi G a\mu\left[\frac{1}{\sqrt{R^{2}+a^{2}}}-\frac{1}{\sqrt{r^{2}+a^{2}}}\right]. $$

$$ F_{x}=F_{y}=0,\quad F_{z}=2\pi G\mu\left[\sqrt{\left(h-a\right)^{2}+R^{2}}-\sqrt{R^{2}+a^{2}}+h\right]. $$

总练习题八

(1) A; (2) B; (3) C; (4) D.

(1) $ \frac{\pi}{2}-1 $;

(2) $ \frac{35}{12}\pi a^{4} $;

(3) $ \frac{\pi}{2}\ln2-\frac{\pi}{4} $

(1) $ \frac{1}{36} $;

(2) $ \frac{\pi}{4}-\frac{1}{2} $;

(3) $ \frac{\pi}{4e^{4}}\left(2e^{3}-5\right) $

(4) $ \frac{1}{6}a^{2}b^{2}c^{2} $

$ \left(\frac{2}{3}-\frac{b}{4a}\right)\pi b^{3} $

$ \frac{\pi}{2} $

$ \left(\frac{2\pi}{3}-\frac{7\sqrt{3}}{8}\right)R^{4} $

$ \frac{368}{105}\rho $

$ \frac{1}{4}k\rho\pi R $，k为引力常数， $ \rho $为锥体密度.

第九章

习题9-1

(1) $ 1 + \sqrt{2} $;

(2) $ \frac{1}{12}(5\sqrt{5}+6\sqrt{2}-1) $;

原书第 327 页

(3) $ a^{\frac{7}{3}} $;

(4) $ 2a^{2} $;

(5) $ \frac{256}{15}a' $;

(6) $ \frac{3}{2}\sqrt{14}+18 $;

(7) $ \frac{8\sqrt{2}}{15}-\frac{\sqrt{3}}{5} $; (8) $ \pi $.

$ \frac{a}{8}\left(3\sqrt{3}-1+\frac{3}{2}\ln\frac{3+2\sqrt{3}}{3}\right) $.

质心在扇形的对称轴上且与圆心距离为 $ \frac{2\sin\frac{\alpha}{2}}{\alpha} $

(1) $ \sqrt{a^{2}+b^{2}}\left(2\pi a^{2}+\frac{8}{3}\pi^{3}b^{2}\right) $;

(2) $ \overline{x} = \frac{6ab^2}{3a^2 + 4\pi^2b^2} $, $ \overline{y} = \frac{-6\pi ab^2}{3a^2 + 4\pi^2b^2} $, $ \overline{z} = \frac{3b(\pi a^2 + 2\pi^3b^2)}{3a^2 + 4\pi^2b^2} $;

(3) $ \frac{2}{3}\pi a^{2}\sqrt{a^{2}+b^{2}}(3a^{2}+4\pi^{2}b^{2}) $.

习题9-2

(1) $ -\frac{1}{2} $;

(2) $ -\frac{\pi}{2}a^{3} $;

(3) $ -2\pi a^{2} $;

(4) 0;

(5)-1;

(6) $ \sin 1 + e - 1 $;

（7）13；

(8) $ -2\pi $;

(9) -4;

(10) $ \frac{1}{4} $

$ -\frac{8}{15} $

(1) $ \frac{a^{2}-b^{2}}{2} $; (2) 0.

(1) $ \int_{L}\frac{P(x, y) + Q(x, y)}{\sqrt{2}}ds $;

(2) $ \int_{L}\frac{P(x, y) + 2xQ(x, y)}{\sqrt{1 + 4x^{2}}}ds $;

(3) $ \int_{L}\left[\sqrt{2x-x^{2}}P(x,y)+(1-x)Q(x,y)\right]ds. $

$ \int_{F}\frac{P+2xQ+3yR}{\sqrt{1+4x^{2}+9y^{2}}}ds. $

习题9-3

(1) $ \frac{3}{8}\pi a^{2} $;

(2) $ 12\pi $.

原书第 328 页

(1) $ \frac{1}{30} $;

(2) 8.

(1) $ -2\pi ab $;

(2) $ \frac{1}{4}\pi^{2} $;

(3) $ \frac{1}{8}m\pi a^{2} $;

（4）0；

(5) $ \frac{\sin 2}{4}-\frac{7}{6} $.

4.（1）0；

(2) $ 2\pi $.

(1) $ e^{\alpha}\cos b-1 $;

(2) 5;

(3) $ -m(\pi+2) $;

(4) 236.

$ e^{2}-\frac{7}{2} $

(1) $ (x+y-1)e^x - (x+1)e^y + 2 $;

(2) $ -\cos 2x \cdot \sin 3y $;

(3) $ x^3y + 4x^2y^2 - 12e^y + 12ye^y $;

(4) $ y^{2}\sin x + x^{2}\cos y $

习题9-4

略.

(1) $ 4\sqrt{61} $;

(2) $ \frac{1+\sqrt{2}}{2}\pi $

(3) $ \frac{64}{15}\sqrt{2}a^{4} $;

(4) $ 8\pi a^{4} $;

(5) $ \pi a(a^{2}-h^{2}) $.

$ \left(\sqrt{2}+\frac{3}{2}\right)\pi $

$ \frac{2\pi}{15}(6\sqrt{3}+1) $.

$ (0, 0, \frac{h}{3}) $.

习题9-5

略.

(1) $ \frac{1}{12} $;

(2) $ \frac{3}{2}\pi $;

(3) $ \frac{1}{8} $;

(4) $ 2\pi e^{2} $;

(5) $ 8\pi $.

原书第 329 页

$ -\frac{\pi}{2}a^{3} $

$ \pi $.

(1) $ \iint_{\Sigma}\left(\frac{3}{5}P + \frac{2}{5}Q + \frac{2\sqrt{3}}{5}R\right)\,\mathrm{d}S $;

(2) $ \iint_{\Sigma}\frac{2xP + 2yQ + R}{\sqrt{1 + 4x^{2} + 4y^{2}}}\,\mathrm{d}S $.

习题9-6

(1) $ \frac{1}{4} $;

(2) $ 12\pi $;

(3) $ -\frac{9}{2}\pi $;

(4) $ 2\pi $;

(5) $ 24\pi $

(1) $ \frac{12\pi a^{3}}{5} $;

(2) $ \frac{6}{5}\pi R^{5} + \frac{\pi}{2}R^{4} $

$ 34\pi $

(1) $ \sqrt{3}\pi a^{2} $;

(2) $ -2\pi a(a+b) $;

(3) $ -20\pi $

(4) 2.

(2) $ \frac{x^{3}}{3}+\frac{y^{3}}{3}+\frac{z^{3}}{3}-xyz+C $ (C为任意常数).

习题9-7

(1) 6;

(2) $ \frac{\pi}{2} $;

(3) $ \frac{5}{12}\pi; $

(4) $ 108\pi $.

(1) 8;

(2) 0.

(1) $ 2\pi $;

(2) $ 12\pi $.

(1) $ 2i+j+k $;

(2) -2yz_{i}-2xz_{j}-2xy_{k}.

(1) xyz

(2) $ x^{2} + xy + 2yz - 3z^{2} $

总练习题九

(1) $ \oint_{F}(P\cos\alpha+Q\cos\beta+R\cos\gamma)\mathrm{d}s $，切向量；

(2) $ \oint_{\Sigma} (P \cos \alpha + Q \cos \beta + R \cos \gamma) \, \mathrm{d}S $，法向量；

(3) $ \mathrm{e}^{x}(C-x) $; (4) 0;

(5) 0, \{-1, 0, -e\}.

(1) $2a^{2}$;

(2) $ \frac{8\sqrt{2}}{15}-\frac{\sqrt{3}}{5} $;

原书第 330 页

(3) $ \frac{2}{3}\pi a^{3} $;

(4) $ -2\pi a^{2} $;

(5) $ -\frac{87}{4}; $

(6) $ \frac{1}{35} $;

(7) $ \pi a^{2} $;

(8) 0.

(1) 0;

(2) $ 2\pi\arctan\frac{h}{R} $;

(3) $ -\frac{\pi}{4}h^{4} $;

(4) $ 2\pi R^{3} $;

(5) $ \frac{2}{15} $;

(6) $ 9\pi $

$ \frac{x^{2}}{2}e^{x}, \frac{x^{2}}{2}e^{x}y. $

$ \sqrt{2} $.

$ \frac{3}{2} $

$ \frac{1}{2}\ln|x^2+y^2| $.

当 a=1 时， $ I(a) $ 达到最小值.

$ \left(0,\ 0,\ \frac{a}{2}\right) $

第 1 十章

习题10-1

(1) 一阶；(2) 三阶；(3) 一阶；(4) 一阶；(5) 四阶.

2.（1）（a）；（2）（c）；（3）（a）；（4）（b）及（c）.

(1) C=9, $ x^{3}+y^{3}=9 $;

(2) C=2, $ y=\sqrt{2x^{2}+1} $;

(3) $ C_{1}=1 $, $ C_{2}=\frac{\pi}{2} $; $ y=-\cos x $.

(1) $ yy'+2x=0 $;

(2) $ \frac{dp}{dT}=k\frac{p}{T^{2}} $

(i) (4); (ii) (3); (iii) (1); (iv) (2).

(1) $ (2y')^{3}-27ay=0 $;

(2) $ xy'' + 2y' - xy = 0 $.

原书第 331 页

习题10-2

(1) $ 10^{-7} + 10^{8} = C $;

(2) $ \frac{y}{1-ay}=C(a+x) $;

(3) $ (x-1)(y+1)=Ce^{y-x} $;

(4) $ 3x^4 + 4(y+1)^3 = C $;

(5) $ \arcsin x + \arcsin y = C $;

(6) $ y = C \cos x $;

(7) ten xtan y=C;

(8) $ \ln y = C \tan \frac{x}{2} $;

(9) $ 3\sqrt{y}=x^{\frac{3}{2}}+C(y\gt 0) $ 或 $ 3\sqrt{-y}=(-x)^{\frac{3}{2}}+C(y\lt 0) $;

(10) $ \sqrt{1+e^{2x}}(e^{y}-1)=C $.

(1) $ \cos x - \sqrt{2} \cos y = 0 $;

(2) $ x=\frac{(t+2)^{3}-8}{24} $;

(3) $ 1 + e^x + 2\cos y = 0 $;

(4) $ 3e^{-y^2} - 2e^{3x} = 3 - 2e $.

$ f(x) = f(0) e^{-x} $

xy = 6.

7.5 m/s.

$ R = R_0 e^{-0.000433t} $

3.9 kg.

习题10-3

(1) $ y^{2}=2x^{2}\ln CX $;

(2) $ y = xe^{ax+1} $;

(3) $ x^{3}-2y^{3}=CX $;

(4) $ S = t \left( \ln \frac{S}{t} \right)^{2} $;

(5) $ \arctan \frac{y}{x} - \frac{1}{2} \ln(x^2 + y^2) = C $.

(1) $ 2y - \sqrt{2} \arctan \frac{x + 2y}{\sqrt{2}} = C $ （令 $ u = x + 2y $）;

(2) $ \ln\left[(y+3)^2+(x+2)^2\right]+2\arctan\left(\frac{y+3}{x+2}\right)=C $ 令 $ u=\frac{y+3}{x+2} $;

(3) $ y = \frac{1}{4} \left[ (2x - 3) + \frac{C}{2x - 3} + 5 \right] $ （令 X = 2x - 3，Y = 4y - 5）.

原书第 332 页

(4) $ y = -x + \tan(x + C) $ （令 $ x + y = u $）;

(5) $ y^{4}+2xy^{2}-x^{2}=C $ （令 $ u=\frac{y^{2}}{x} $）；

(6) $ \tan y = (x^{2} + C)e^{-x^{2}} $ (令 $ \tan y = u $).

$ x^{2}+(y-C)^{2}=C^{2} $

习题10-4

(1) $ y = \frac{1}{x} (C - \cos x) $;

(2) $ 3\rho = 2 + Ce^{-3\theta} $;

(3) $ x = Ce^t - \frac{1}{2} (\cos t + \sin t) $;

(4) $ y = Cx^3 - x^2 $;

(5) $ y = x + \frac{C - x}{\ln x} $;

(6) $ x = Ce^{2y} + \frac{1}{4}(2y^2 + 2y + 1) $;

(7) $ y = \frac{1}{3} x^{2} + \frac{3}{2} x + 2 + \frac{C}{x} $;

(8) $ 2x\ln y = \ln^2 y + C $, $ x = \frac{1}{2}\ln y + \frac{C}{\ln y} $;

(9) $ y = (x - 2)^3 + C(x - 2) $;

(10) $ x = Cy^3 + \frac{y^2}{2} $.

(1) $ y = 2e^{2x} - e^x + \frac{x}{2} + \frac{1}{4} $;

(2) $ y = \frac{x}{\cos x} $;

(3) $ y \sin x + 5 e^{\cos x} = 1 $;

(4) $ y = \sin x - 1 + 2e^{-\sin x} $;

(5) $ xy = -1 + \ln y $.

$ y=\{\begin{aligned}&x(1-4\ln x),& 当 0\lt x\leq1,\\ &0,& 当 x=0.\end{aligned}. $

$ f(x)=\frac{2}{3}x+\frac{1}{3\sqrt{x}} $

(1) $ \frac{1}{y} = -\sin x + Ce^x $;

(2) $ \left(1+\frac{3}{y}\right)e^{\frac{3}{2}x^{2}}=C $;

(3) $ (Ce^{x}-2x-1)y^{3}=1 $.

原书第 333 页

习题10-5

(1) 是， $ x^{3}+3x^{2}y^{2}+\frac{4}{3}y^{3}=C $;

(2) 是， $ a^{2}x - x^{2}y - xy^{2} - \frac{1}{3}y^{3} = C $;

(3) 是， $ xe^{y}-y^{2}=C $;

（4）是， $ y\cos x+x\sin y=C $;

(5) 是， $ \frac{1}{3}x^{3}-xy=C $;

（6）非：

(7) $ \rho(e^{2\theta}+1)=C $;

（8）非.

(1) 积分因子是 $ \frac{1}{x+y} $，通解为 $ x-y-\ln(x+y)=C $;

(2) 积分因子是 $ \frac{1}{y^{2}} $，通解为 $ \frac{x}{y} + \frac{x^{2}}{2} = C $；

(3) 积分因子是 $ \frac{1}{y^{2}} $，通解为 $ \frac{x^{2}}{2} - \frac{1}{y} - 3xy = C $；

（4）积分因子是 $ \frac{1}{2(x^{2}+y^{2})} $，通解为 $ x^{2}+y^{2}=Ce^{2x} $

习题10-6

(1) $ y = \frac{1}{6} x^{3} - \sin x + C_{1} x + C_{2} $;

(2) $ y=(x-3)e^{x}+C_{1}x^{2}+C_{2}x+C_{3} $;

(3) $ y = C_1 \ln |x| + C_2 $;

(4) $ y = C_1(x - e^{-x}) + C_2 $;

(5) $ y = 1 - (C_1 x + C_2)^{-1} $;

(6) $ C_{1}y^{2}-1=(C_{1}x+C_{2})^{2} $;

(7) $ y = \arcsin(C_2 e^x) + C_1 $;

(8) $ y = -\cos(\pm x + C_1) + C_2 x + C_3 $

或 $ y = -\sin(C_{1} \pm x) + C_{2} x + C_{3} $

(1) $ y = \sqrt{2x - x^{2}} $ (0 < x < 2);

(2) $ y = 2 \arctan e^x $;

(3) $ y = 1 - \sqrt{1 - 2x} $;

(4) $ y = \tan\left(x + \frac{\pi}{4}\right) $.

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

原书第 334 页

习题10-7

$ y = C_{1}\cos\omega x + C_{2}\sin\omega x $

$ y = C_1(y_1 - y_2) + C_2(y_1 - y_3) $.

$ y = C_{1} e^{x} + C_{2} (2x + 1) $.

$ y = C_{1}x + \frac{C_{2}}{x} $

提示：观察出 $ y_{1}=\frac{1}{x} $ 为原方程的一个特解。令 $ y_{2}=\frac{1}{x}u(x) $，利用公式（6）求出 $ y_{2} $。

习题10-8

(1) $ y = C_1 e^x + C_2 e^{-2x} $;

(2) $ y = (C_1 x + C_2) e^{3/2 x} $;

(3) $ S = C_{1} + C_{2} e^{-t} $;

(4) $ y = C_{1} \cos x + C_{2} \sin x $;

(5) $ y = e^{-3x} (C_1 \cos 2x + C_2 \sin 2x) $;

(6) $ y = C_{1} + (C_{2}x^{2} + C_{3}x + C_{4})e^{2x} $;

(7) $ y = C_{1} + (C_{2}x + C_{3}) \cos x + (C_{4}x + C_{5}) \sin x $;

(8) $ y = C_1 \cos \sqrt{2} x + C_2 \sin \sqrt{2} x + C_3 e^{\sqrt{2} x} + C_4 e^{-\sqrt{2} x} $.

(1) $ y = (2 + x) e^{-\frac{x}{2}} $;

(2) $ y = 3e^{-2x} \sin 5x $;

(3) $ y = e^{\frac{b}{x}} - \cos x $;

(4) $ y = \cos ax $.

$ \varphi(x) = \mathrm{ecos} x $.

$ y = \cos 3x + \frac{1}{3} \sin 3x $

$ x = \frac{2v_0 e^{-\frac{k_2}{2}t}}{\sqrt{k_2^2 + 4k_1}} sh \frac{\sqrt{k_2^2 + 4k_1}}{2} t $.

习题10-9

$$ y=C_{1}(x-1)+C_{2}x^{2}+1. $$

(1) $ y = C_{1} e^{-x} + C_{2} e^{-4x} - \frac{1}{2} x + \frac{11}{8} $;

(2) $ y = C_1 + C_2 e^{3x} + x^2 $;

(3) $ y = C_1 e^{\frac{x}{2}} + C_2 e^{-x} + e^x $;

(4) $ y = C_1 \cos ax + C_2 \sin ax + \frac{e^x}{1 + a^2} $;

原书第 335 页

(5) $ y = C_{1} e^{-x} + C_{2} e^{-2x} + \left( \frac{3}{2} x^{2} - 3x \right) e^{-x} $;

(6) $ y = (C_1 + C_2 x) e^{3x} + \frac{x^2}{2} \left( \frac{x}{3} + 1 \right) e^{3x} $;

(7) $ y = e^{x} $ ( $ C_{1}\cos 2x + C_{2}\sin 2x $) $ -\frac{1}{4}xe^{x}\cos 2x $;

(8) $ y = C_{1}\cos 2x + C_{2}\sin 2x + \frac{1}{3}x\cos x + \frac{2}{9}\sin x $;

(9) $ y = C_{1}\cos x + C_{2}\sin x + \frac{e^{x}}{2} + \frac{x}{2}\sin x $;

(10) $ y = C_{1} e^{x} + C_{2} e^{-x} + \frac{1}{10} \cos 2x - \frac{1}{2} $.

(1) $ y = -5e^x + \frac{7}{2}e^{2x} + \frac{5}{2} $;

(2) $ y = -\cos x - \frac{1}{3} \sin x + \frac{1}{3} \sin 2x $

（1）当 $ a \neq 0 $ 时， $ y^{*} = \frac{b^{2}}{a^{2}} $；当 a = 0 时， $ y^{*} = \frac{b^{2}}{2}x^{2} $；

(2) 当 $ a \neq b $时， $ y^* = \frac{8}{a^2 - b^2} \cos bx $;

$ a = b \neq 0 $ 时， $ y^* = \frac{4}{a} x \sin ax $;

a=b=0 时， $ y^{*}=4x^{2} $

$ y = \frac{1}{2} \sin x + \frac{x}{2} e^{x} $.

$ f(x)=\frac{5}{2}\sin x-\frac{3}{2}\cos x+\frac{3}{2}e^{-x} $

$ f(x) = 2\cos x + \sin x + x^{2} - 2 $.

总练习题十

(1) 3;

(2) $ y' = -\sqrt{1 - y^2} $;

(3) $ y'' - 2y' + y = 0 $;

(4) $ y = e^{-\frac{x^2}{2}} $;

(5) $ y = C_{1} e^{x} + C_{2} x^{2} + 3 $.

(1) C; (2) D; (3) C; (4) D.

$ \alpha = -3 $, $ \beta = 2 $, r = -1; $ y = C_1 e^{2x} + C_2 e^x + xe^x $.

(1) $ x\sin\left(\frac{y^{2}}{x}\right)=C $ (令 $ y^{2}=xu $);

(2) $ \ln |xy| + \arctan \frac{y}{x} = C $;

原书第 336 页

(3) $ x = -2(e^{-cos y} + cos y - 1) $ （方程变形为 $ \frac{dx}{dy} - x\sin y = \sin 2y $）;

(4) $ 4[C_1(2+y)-1]=C_1^2(x+C_2)^2 $;

(5) $ x = C_1 e^{-2y} + C_2 e^{2y} + \frac{1}{4} y e^{2y} $. 提示： $ \frac{d^2 x}{dy^2} = -\frac{y^n}{(y^{m+1})^3} $.

(1) $ y(x) = e^{-xx} \cdot \int_{0}^{x} f(t) e^{st} dt $.

$ y - x = -x^{3}y $ （或 $ y = \frac{x}{1 + x^{3}} $）.

$ -y^{n}=1+(y^{\prime})^{2} $

$$ y=\ln\left|\cos\left(\frac{\pi}{4}-x\right)\right|+1+\frac{1}{2}\ln2, $$

无极小值；当 $ x=\frac{\pi}{4} $ 时，y 有极大值 $ y=1+\frac{1}{2}\ln 2 $.

$ f(x) = Cx^{-\frac{n-1}{n}} + \frac{1}{n-1} $

$ f(x)=\frac{1}{2}\sin x+\frac{x}{2}\cos x $.

第十一章

习题11-1

(1) $ \frac{1}{2n-1} $; (2) $ (-1)^{n-1}\frac{n+1}{n} $; (3) $ \frac{\sin n}{2^n} $;

(4) $ \frac{\cos n!}{n(n+1)}; $ (5) $ \frac{x^{\frac{n}{2}}}{(2n)!!}; $ (6) $ (-1)^{n-1}\frac{a^{n+1}}{2n+1} $

(1) $ -\sin\frac{\pi}{4}+\sin\frac{\pi}{2}-\sin\frac{3\pi}{4}+\sin\pi-\cdots $

(2) $ \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\cdots $

(3) $ -\frac{1}{1\times2}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{4\times5}-\cdots; $

(4) $ \frac{1}{2} + \frac{1 \cdot 3}{2 \cdot 4} - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8} - \cdots $.

3.（1）收敛；（2）发散；（3）收敛；（4）发散.

4.（1）发散；（2）收敛；（3）发散；（4）发散；（5）收敛.

5.（略）.

习题11-2

1.（1）收敛；（2）收敛；（3）收敛；（4） $ 0\lt \alpha\lt 1 $ 时发散， $ \alpha\gt 1 $ 时收敛.

原书第 337 页

2.（1）收敛；（2）发散；（3）收敛；（4）收敛.

3.（1）收敛；（2）发散；（3）收敛；（4）收敛。

4.（1）发散；（2）发散；（3）收敛；（4）收敛.

(1) 提示：考察级数 $ \sum_{k=1}^{\infty}\frac{k^{k}}{(2k)!} $ 的收敛性.

(2) 提示：考察级数 $ \sum_{k=1}^{\infty}\frac{k^{k}}{(k!)^{2}} $ 的收敛性.

6.（1）条件收敛；（2）绝对收敛；（3）发散；（4）绝对收敛；（5）条件收敛；

(6) $ |x| \lt 2 $ 时，绝对收敛；x = -2 时，条件收敛； $ |x| \gt 2 $ 或 x = 2 时，发散.

提示：用 p-判别法.

证明略. $ \sum_{k=1}^{\infty}\frac{1}{1+a_{n}} $ 发散，因为 $ \lim_{n\to\infty}\frac{1}{1+a_{n}}=1\neq0 $.

习题11-3

1.（1）R=1，收敛域为 $ [-1,1] $; （2）R=3，收敛域为 $ (-3,3) $;

(3) R=a，收敛域为 $ [-a, a) $; (4) R=2，收敛域为 $ [-2, 2] $;

(5) $ R=\infty $，收敛域为 $ (-\infty, +\infty) $; (6) R=1，收敛域为 $ [4, 6) $;

(1) 收敛域为 $ [-1, 1) $，和函数为 $ \{\begin{aligned}-\frac{\ln(1-x)}{x},&0\lt |x|\lt 1\text{ 或 }x=-1,\\1,&x=0;\end{aligned}. $

(2) 收敛域为 $ (-1,1) $，和函数为 $ \frac{x}{(1-x)^{2}} $;

（3）收敛域为 $ (-∞,+∞) $，和函数为 $ \frac{1}{2}(e^{x}+e^{-x}) $；

(4) 收敛域为 $ (-1, 1) $，和函数为 $ \{\begin{aligned}\frac{1}{1-x}+\frac{\ln(1-x)}{x},&0\lt |x|\lt 1,\\0,&x=0;\end{aligned}. $

（5）收敛域为 $ (-1,1) $，和函数为 $ \frac{1}{2}\ln\frac{1+x}{1-x} $， $ \sum_{n=1}^{\infty}\frac{1}{(2n-1)2^{n}} $的和为 $ \frac{\sqrt{2}}{2}\ln(1+\sqrt{2}) $。

习题11-4

(I) $ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n}}{n!},\quad(-\infty,+\infty) $;

(2) $ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+1}}{n!\cdot2^{n}},(-\infty,+\infty) $;

(3) $ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!\cdot2^{2n+1}} $, $ (-\infty,+\infty) $;

(4) $ 1 + \sum_{n=1}^{\infty}(-1)^{n}\frac{2^{2n-1}}{(2n)!}x^{2n}, (-\infty, +\infty) $;

原书第 338 页

(5) $ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{2^{2n-1}}{(2^n)!}x^{2n},(-\infty,+\infty) $

(6) $ \sum_{n=0}^{\infty}(-1)^{n}\frac{t^{2n+1}}{n!\cdot(2n+1)},\quad(-\infty,+\infty) $;

(7) $ \sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n)!\cdot(4n+1)}t^{4n+1} $, $ (-\infty,+\infty) $;

(8) $ \sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1},\quad[-1,1] $;

(9) $ \frac{1}{3}\sum_{n=0}^{\infty}\left(1-\frac{(-1)^{n}}{2^{n}}\right)x^{n}, (-1,1) $;

(10) $ \sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n-1)!},\quad(-\infty,+\infty) $.

(1) $ \sum_{n=0}^{\infty}\frac{1}{2^{n+1}}(x-1)^n $, $ x\in(-1,3) $;

$$ \frac{\sqrt{2}}{2}\sum_{n=0}^{\infty}(-1)^{n}\left[\frac{x^{2n}}{(2n)!}+\frac{x^{2n+1}}{(2n+1)!}\right],x\in(-\infty,+\infty); $$

(3) $ \sum_{n=0}^{\infty}\left(\frac{1}{2^{n+1}}-\frac{1}{3^{n+1}}\right)^{n}(x+4)^{n}, x\in(-6,-2) $;

(4) $ \sum_{n=0}^{\infty}(-1)^{n}\frac{1}{3^{n+1}}(x-3)^{n}, x\in(0,6). $

(1) $ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)(2n+1)!}, x\in(-\infty,+\infty) $;

(2) $ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)^2}, x\in[-1,1]. $

$ \sum_{n=1}^{\infty}\frac{nx^{n-1}}{(n+1)!},\ x\in(-\infty,+\infty) $.

1.606.

1.098 6.

习题11-5

1.（略）；2.（略）.

$ S(x) = \begin{cases} -2, & -\pi \lt x \lt 0, \\ 2, & 0 \lt x \lt \pi, \\ 0, & x = 0, \pm \pi. \end{cases} $

(1) $ f(x) = \frac{e^{2\pi} - e^{-2\pi}}{\pi} \left[ \frac{1}{4} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2 + 4} (2\cos nx - n\sin nx) \right] $

$ (x \neq (2n+1)\pi, n=0, \pm1, \pm2, \cdots) $;

(2) $ f(x)=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos2nx}{4n^{2}-1},x\in(-\infty,+\infty) $;

原书第 339 页

(3)

$$ \begin{aligned}f(x)&=\frac{1}{2}+\frac{2(\pi+1)}{\pi}\sin x-\sin2x+\frac{2(\pi+1)}{3\pi}\sin3x-\frac{2}{4}\sin4x+\cdots\\&\left(x\in(-\infty,+\infty),x\neq n\pi,n=0,\pm1,\pm2,\cdots\right);\end{aligned} $$

(4)

$$ f(x)=\pi-\sum_{n=1}^{\infty}\frac{2}{n}\sin nx\quad(x\in(-\infty,+\infty),x\neq0,\pm2\pi,\pm4\pi,\cdots). $$

$$ 2\sin\frac{x}{3}=\frac{18\sqrt{3}}{\pi}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{9n^{2}-1}\sin nx,\quad x\in(-\pi,\pi); $$

(2)

$$ \begin{align*}2)\quad&f(x)=\frac{1+\pi-\mathrm{e}^{-\pi}}{2\pi}+\frac{1}{\pi}\sum_{n=1}^{\infty}\{\frac{1-(-1)^{n}\mathrm{e}^{-\pi}}{1+n^{2}}\cos\ n x+.\\&.[\frac{n((-1)^{n}\mathrm{e}^{-\pi}-1)}{1+n^{2}}+\frac{1-(-1)^{n}}{n}]\sin\ n x\},\quad(x\in(-\pi,\ \pi)).\end{align*} $$

$$ x^{2}=\frac{2}{\pi}\sum_{n=1}^{\infty}\left[-\frac{2}{n^{3}}+\left(-1\right)^{n}\left(\frac{2}{n^{3}}-\frac{\pi^{2}}{n}\right)\right]\sin n x,\quad x\in\left[0,\pi\right); $$

$$ x^{2}=\frac{\pi^{3}}{3}+4\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{2}}\cos nx,\quad x\in\left[0,\pi\right]; $$

$$ \begin{align*}(2)\ \mathrm{e}^{x}&=\{\frac{\mathrm{e}^{\pi}+1}{2}+\sum_{n=2}^{\infty}\frac{n}{n^{2}-1}[\mathrm{e}^{\pi}(-1)^{n}-1]\}\sin n x,\ x\in(0,\ \pi);\\\mathrm{e}^{x}&=\frac{\mathrm{e}^{\pi}-1}{\pi}+\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}[\mathrm{e}^{\pi}(-1)^{n}-1]\cos n x,\ x\in[0,\ \pi].\end{align*} $$

$ \cos x = \frac{2}{\pi} \sum_{n=2}^{\infty} \frac{n}{n^2 - 1} \left[ (-1)^{n+1} - 1 \right] \sin nx, x \in (-\pi, 0). $

习题11-6

$$ 1.f(x)=\frac{k}{2}+\frac{k}{\pi}\sum_{n=1}^{\infty}\frac{1-(-1)^{n}}{n}\sin\frac{n\pi x}{2}\quad(x\in(-\infty,+\infty),x\neq0,\pm2,\pm4,\cdots). $$

$$ 2.f(x)=-\frac{1}{2}+\frac{6}{\pi}\sum_{n=1}^{\infty}\left[\frac{1-(-1)^{n}}{n^{2}\pi}\cos\frac{n\pi x}{3}+\frac{(-1)^{n+1}}{n}\sin\frac{n\pi x}{3}\right] $$

$$ (x\in(-\infty,+\infty),x\neq3(2k+1),k=0,\pm1,\pm2,\cdots). $$

$$ 1-x^{2}=\frac{11}{12}+\frac{1}{\pi^{2}}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}\cos2n\pi x,\quad x\in\left[-\frac{1}{2},\frac{1}{2}\right] $$

(2)

$$ f(x)=\frac{1}{\pi}+\cos\frac{\pi x}{2}+\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{4n^{2}-1}\cos\frac{n\pi x}{2},x\in[-2,2]. $$

$$ \begin{align*}4.\ &\frac{4l}{\pi^{2}}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{\left(2n-1\right)^{2}}\sin\frac{\left(2n-1\right)\pi x}{l},\left[0,l\right);\\&\frac{l}{4}-\frac{2l}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^{2}}\cos\frac{2\left(2n-1\right)\pi x}{l},\left[0,l\right).\end{align*} $$

$$ 5.x(4-x)=\frac{128}{\pi^{3}}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{3}}\sin\frac{(2n-1)\pi x}{4},x\in\left[0,4\right]. $$

$$ u(t)=\frac{h\tau}{T}+\frac{h}{\pi}\sum_{n=-\infty}^{\infty}\frac{1}{n}\sin\frac{n\pi\tau}{T}e^{\frac{2n\pi i}{T}} $$

原书第 340 页

$$ \left(x\in(-\infty,+\infty),t\neq\pm\frac{\tau}{2},\pm\frac{T}{2},\cdots\right). $$

总练习题十一

(1) B; (2) C; (3) A; (4) B.

2.（1）发散；（2）收敛；（3）收敛；（4）收敛；（5）发散；（6）发散；（7）发散；

（8）收敛；（9）当 b<a 时，收敛；当 b>a 时，发散；当 b=a 时，不能确定；

(10) 收敛.

3.（1）条件收敛；（2）条件收敛；（3）绝对收敛；（4）条件收敛.

(1) 0;

(2) $ \sqrt[4]{8} $，提示：化成 $ 2^{\frac{1}{3} \times \frac{2}{3^2} \times \frac{3}{3^3} \times \cdots \times \frac{n}{3^n}} $。

5.（1）收敛域： $ \left(-\frac{4}{3}, -\frac{2}{3}\right) $;

(2) p=0 时，收敛域为 $ (-1,1) $; $ 0\lt p\leq1 $ 时，收敛域为 $ [-1,1) $; p>1 时，收敛域为 $ [-1,1] $.

(1) $ (-1, 1) $, $ \frac{2x}{(1-x)^{3}} $;

(2) $ (-1, 1) $, $ \frac{1}{4}\ln\frac{1+x}{1-x}+\frac{1}{2}\arctan x-x $.

(1) $ (1-x)\ln(1+x)=x+\sum_{n=2}^{\infty}(-1)^n\frac{2n-1}{n(n-1)}x^n,x\in[-1,1] $;

(2) $ \arcsin x = x + \sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!} \cdot \frac{x^{2n+1}}{2n+1} $, $ x \in [-1, 1] $;

$$ \frac{x}{\sqrt{1+x^{2}}}=x+\sum_{n=1}^{\infty}(-1)^{n}\frac{(2n)!}{2^{2n}\left(n!\right)^{2}}x^{2n+1},\quad x\in(-1,1]; $$

(4) $ \ln\frac{1}{1-x}=1+\sum_{n=1}^{\infty}\frac{x^{n}}{n},\ x\in[-1,\ 1) $.

(1) $ \frac{1}{x^2} = \sum_{n=1}^{\infty} n(x+1)^{n-1} $, $ x \in (-2, 0) $;

(2) $ \lg x = \frac{1}{\ln 10} \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{(x-1)^n}{n} $, $ x \in (0, 2] $.

$$ f(x)=\frac{2}{\pi}\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}\sin\frac{n\pi}{2}+(-1)^{n+1}\cdot\frac{\pi}{2n}\right]\sin nx(x\neq(2n+1)\pi,n=0,\pm1,\pm2,\cdots) $$