题目
6.求下列全微分的原函数:-|||-(1) ((x)^2+2xy-(y)^2)dx+((x)^2-2xy-(y)^2)dy-|||-(2) ^x[ e'(x-y+2)+y] dx+(e)^x[ (e)^y(x-y)+1] dy-|||-(3) (sqrt ({x)^2+(y)^2})xdx+f(sqrt ({x)^2+(y)^2})ydy
题目解答
答案
解析
(1) 步骤 1：验证全微分条件
设 $P(x,y) = {x}^{2}+2xy-{y}^{2}$, $Q(x,y) = {x}^{2}-2xy-{y}^{2}$, 则
$\frac {\partial P}{\partial y} = 2x-2y$, $\frac {\partial Q}{\partial x} = 2x-2y$,
所以 $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$, 满足全微分条件。
步骤 2：求原函数
原函数 $u(x,y)$ 满足 $\frac {\partial u}{\partial x} = P(x,y)$, $\frac {\partial u}{\partial y} = Q(x,y)$,
所以 $u(x,y) = \int P(x,y)dx + \int Q(x,y)dy - \int \frac {\partial u}{\partial y}dy$,
$u(x,y) = \int ({x}^{2}+2xy-{y}^{2})dx + \int ({x}^{2}-2xy-{y}^{2})dy - \int (2x-2y)dy$,
$u(x,y) = \frac {{x}^{3}}{3}+{x}^{2}y-x{y}^{2}+C$。
(2) 步骤 1：验证全微分条件
设 $P(x,y) = {e}^{x}[ {e}^{y}(x-y+2)+y]$, $Q(x,y) = {e}^{x}{[ e'(x-y)}+1]$,
则 $\frac {\partial P}{\partial y} = {e}^{x}[ {e}^{y}(x-y+1)]$, $\frac {\partial Q}{\partial x} = {e}^{x}[ {e}^{y}(x-y+1)]$,
所以 $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$, 满足全微分条件。
步骤 2：求原函数
原函数 $u(x,y)$ 满足 $\frac {\partial u}{\partial x} = P(x,y)$, $\frac {\partial u}{\partial y} = Q(x,y)$,
所以 $u(x,y) = \int P(x,y)dx + \int Q(x,y)dy - \int \frac {\partial u}{\partial y}dy$,
$u(x,y) = \int {e}^{x}[ {e}^{y}(x-y+2)+y] dx + \int {e}^{x}{[ e'(x-y)}+1] dy - \int {e}^{x}[ {e}^{y}(x-y+1)] dy$,
$u(x,y) = {e}^{x+y}(x-y-1)+{e}^{x}(x+y+1)$。
(3) 步骤 1：验证全微分条件
设 $P(x,y) = f(\sqrt {{x}^{2}+{y}^{2}})x$, $Q(x,y) = f(\sqrt {{x}^{2}+{y}^{2}})y$,
则 $\frac {\partial P}{\partial y} = \frac {xy}{\sqrt {{x}^{2}+{y}^{2}}}f'(\sqrt {{x}^{2}+{y}^{2}})$, $\frac {\partial Q}{\partial x} = \frac {xy}{\sqrt {{x}^{2}+{y}^{2}}}f'(\sqrt {{x}^{2}+{y}^{2}})$,
所以 $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$, 满足全微分条件。
步骤 2：求原函数
原函数 $u(x,y)$ 满足 $\frac {\partial u}{\partial x} = P(x,y)$, $\frac {\partial u}{\partial y} = Q(x,y)$,
所以 $u(x,y) = \int P(x,y)dx + \int Q(x,y)dy - \int \frac {\partial u}{\partial y}dy$,
$u(x,y) = \int f(\sqrt {{x}^{2}+{y}^{2}})xdx + \int f(\sqrt {{x}^{2}+{y}^{2}})ydy - \int \frac {xy}{\sqrt {{x}^{2}+{y}^{2}}}f'(\sqrt {{x}^{2}+{y}^{2}})dy$,
$u(x,y) = \frac {1}{2}\int f(\sqrt {{x}^{2}+{y}^{2}})d(\sqrt {{x}^{2}+{y}^{2}})$。
设 $P(x,y) = {x}^{2}+2xy-{y}^{2}$, $Q(x,y) = {x}^{2}-2xy-{y}^{2}$, 则
$\frac {\partial P}{\partial y} = 2x-2y$, $\frac {\partial Q}{\partial x} = 2x-2y$,
所以 $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$, 满足全微分条件。
步骤 2：求原函数
原函数 $u(x,y)$ 满足 $\frac {\partial u}{\partial x} = P(x,y)$, $\frac {\partial u}{\partial y} = Q(x,y)$,
所以 $u(x,y) = \int P(x,y)dx + \int Q(x,y)dy - \int \frac {\partial u}{\partial y}dy$,
$u(x,y) = \int ({x}^{2}+2xy-{y}^{2})dx + \int ({x}^{2}-2xy-{y}^{2})dy - \int (2x-2y)dy$,
$u(x,y) = \frac {{x}^{3}}{3}+{x}^{2}y-x{y}^{2}+C$。
(2) 步骤 1：验证全微分条件
设 $P(x,y) = {e}^{x}[ {e}^{y}(x-y+2)+y]$, $Q(x,y) = {e}^{x}{[ e'(x-y)}+1]$,
则 $\frac {\partial P}{\partial y} = {e}^{x}[ {e}^{y}(x-y+1)]$, $\frac {\partial Q}{\partial x} = {e}^{x}[ {e}^{y}(x-y+1)]$,
所以 $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$, 满足全微分条件。
步骤 2：求原函数
原函数 $u(x,y)$ 满足 $\frac {\partial u}{\partial x} = P(x,y)$, $\frac {\partial u}{\partial y} = Q(x,y)$,
所以 $u(x,y) = \int P(x,y)dx + \int Q(x,y)dy - \int \frac {\partial u}{\partial y}dy$,
$u(x,y) = \int {e}^{x}[ {e}^{y}(x-y+2)+y] dx + \int {e}^{x}{[ e'(x-y)}+1] dy - \int {e}^{x}[ {e}^{y}(x-y+1)] dy$,
$u(x,y) = {e}^{x+y}(x-y-1)+{e}^{x}(x+y+1)$。
(3) 步骤 1：验证全微分条件
设 $P(x,y) = f(\sqrt {{x}^{2}+{y}^{2}})x$, $Q(x,y) = f(\sqrt {{x}^{2}+{y}^{2}})y$,
则 $\frac {\partial P}{\partial y} = \frac {xy}{\sqrt {{x}^{2}+{y}^{2}}}f'(\sqrt {{x}^{2}+{y}^{2}})$, $\frac {\partial Q}{\partial x} = \frac {xy}{\sqrt {{x}^{2}+{y}^{2}}}f'(\sqrt {{x}^{2}+{y}^{2}})$,
所以 $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$, 满足全微分条件。
步骤 2：求原函数
原函数 $u(x,y)$ 满足 $\frac {\partial u}{\partial x} = P(x,y)$, $\frac {\partial u}{\partial y} = Q(x,y)$,
所以 $u(x,y) = \int P(x,y)dx + \int Q(x,y)dy - \int \frac {\partial u}{\partial y}dy$,
$u(x,y) = \int f(\sqrt {{x}^{2}+{y}^{2}})xdx + \int f(\sqrt {{x}^{2}+{y}^{2}})ydy - \int \frac {xy}{\sqrt {{x}^{2}+{y}^{2}}}f'(\sqrt {{x}^{2}+{y}^{2}})dy$,
$u(x,y) = \frac {1}{2}\int f(\sqrt {{x}^{2}+{y}^{2}})d(\sqrt {{x}^{2}+{y}^{2}})$。