P42. 1. 求下列方程的解：
（1）$\frac{dy}{dx}=2xy$，并求满足初始条件$x=0,y=1$的特解
解：$\frac{dy}{y}=2xdx$
$\int \frac{1}{y}dy=\int 2xdx+lnC$
$ln|y|=x^{2}+lnC$
$y=e^{x^{2}+lnC}$
$y=Ce^{x^{2}}$
带入$x=0,y=1$ 得$1=e^{C}$，即$C=1$此时特解为$y=e^{x^{2}}$
（3）$\frac{dy}{dx}=\frac{1+y^{2}}{xy+x^{3}y}$
解：$xydy=\frac{1+y^{2}}{1+x^{2}}dx$
$\frac{y}{1+y^{2}}dy=\frac{1}{x(1+x^{2})}dx$
$\frac{1}{2} \int\frac{1}{1+y^{2}}d(1+y^{2})=\int \frac{1}{x}-\frac{x}{1+x^{2}}dx+ln|C|$
$\frac{1}{2}ln|1+y^{2}|=ln|x|-\frac{1}{2}ln|1+x^{2}|+ln|C|$
$\frac{1}{2}(ln|1+y^{2}|+ln|1+x^{2}|)=ln|x|+ln|C|$
$(1+y^{2})(1+x^{2})=Cx$
（5）$(y+x)dy+(x-y)dx=0$
解：$\frac{dy}{dx}=\frac{y-x}{y+x}$
令$u=\frac{y}{x}$则$y=ux,\frac{dy}{dx}=u+x\frac{du}{dx}$
得：$u+x\frac{du}{dx}=\frac{ux-x}{ux+x}\rightarrow u+x\frac{du}{dx}=\frac{u-1}{u+1}$
$x\frac{du}{dx}=\frac{u-1}{u+1}-u\rightarrow x\frac{du}{dx}=\frac{-2}{u+1}$
$-\frac{u+1}{2}du=\frac{1}{x}dx$
$-\int \frac{u+1}{2}du=\int \frac{1}{x}dx+lnC\rightarrow -(\frac{1}{4}u^{2}+\frac{1}{2}u)=ln|Cx|$
$-(\frac{1}{4}(\frac{y}{x})^{2}+\frac{y}{2x})=ln|Cx|$
（7）$tanydx-cotxdy=0$
解：$\frac{dy}{dx}=\frac{tany}{cotx}$
$\frac{1}{tany}dy=\frac{1}{cotx}dx\rightarrow cotydy=tanxdx$
$\int cotydy=\int tanxdx+ln|C|\rightarrow ln|siny|=-ln|cosx|+ln|C|$
$siny=-Ccosx$
（9）$x(lnx-lny)dy-ydx=0$
解：$-(lny-lnx)dy-\frac{y}{x}dx=0\rightarrow -ln\frac{y}{x}dy=\frac{y}{x}dx$
$\frac{dy}{dx}=\frac{y}{-xln\frac{y}{x}}$
令$u=\frac{y}{x}$则$y=ux,\frac{dy}{dx}=u+x\frac{du}{dx}$
得：$u+x\frac{du}{dx}=\frac{-u}{lnu}\rightarrow -\frac{lnu}{u(1-lnu)}du=\frac{1}{x}dx$
$-\int \frac{lnu}{u(1-lnu)}du=\int \frac{1}{x}dx+ln|C|\rightarrow \int \frac{lnu}{1-lnu}d(1-lnu)=ln|Cx|$
$\int \frac{1-(1-lnu)}{1-lnu}d(1-lnu)=ln|Cx|\rightarrow ln|1-lnu|-(1-lnu)=ln|Cx|$
$ln|1-ln\frac{y}{x}|+ln\frac{y}{x}=ln|Cx|$
2. 做适当的变量变换求解下列方程
（1）$\frac{dy}{dx}=(x+y)^{2}$
解：令$u=x+y$，则$y=u-x,\frac{dy}{dx}=\frac{du}{dx}-1$
得：$\frac{du}{dx}=u^{2}+1\rightarrow \frac{1}{u^{2}+1}du=dx$
$\int \frac{1}{u^{2}+1}du=\int dx\rightarrow arctanu=x$
$arctan(x+y)=x$
（3）$\frac{dy}{dx}=\frac{2x-y+1}{x-2y+1}$
解：其方程组$2x-y=1,x-2y=-1$的解为$x=1,y=1$
令$x=X+1,y=Y+1$消去常数得$\frac{dY}{dX}=\frac{2X-Y}{x-2Y}$
令$u=\frac{Y}{X}$则$Y=uX,\frac{dY}{dX}=u+X\frac{du}{dX}$
得：$u+X\frac{du}{dX}=\frac{2-u}{1-2u}\rightarrow \frac{1-2u}{1+u}du=\frac{1}{X}dX$
$\int \frac{1-2u}{1+u}du=\int \frac{1}{X}dX+ln|C|\rightarrow -\int \frac{2(1+u)-3}{1+u}d(1+u)=ln|CX|$
$-2(1+u)+3ln|1+u|=ln|CX|\rightarrow 3ln|1+u|-2u=ln|CX|$
$3ln|1+\frac{y-1}{x-1}|-2\frac{y-1}{x-1}=ln|C(x-1)|$
（5）$\frac{dy}{dx}=(x+1)^{2}+(4y+1)^{2}+8xy+1$
解：$\frac{dy}{dx}=x^{2}+1+2x+16y^{2}+1+8y+8xy+1=(x+4y+1)^{2}+2$
令$u=x+4y+1$,则$y=\frac{u-x-1}{4},\frac{dy}{dx}=\frac{1}{4}(\frac{du}{dx}-1)$
得：$\frac{1}{4}(\frac{du}{dx}-1)=u^{2}+2\rightarrow \frac{1}{4u^{2}+9}du=dx$
$\frac{1}{2}\int \frac{1}{4u^{2}+9}d(2u)=\int dx+C\rightarrow \frac{1}{6}arctan\frac{2u}{3}=x+C$
$arctan\frac{2(x+4y+1)}{3}=6x+C$
（7）$\frac{dy}{dx}=\frac{2x^{3}+3xy^{2}+x}{3x^{2}y+2y^{3}-y}$
解：$\frac{dy}{dx}=\frac{x(2x^{2}+3y^{2}+1)}{y(3x^{2}+2y^{2}-1)}\rightarrow \frac{dy^{2}}{dx^{2}}=\frac{2x^{2}+3y^{2}+1}{3x^{2}+2y^{2}-1}$
令$u=y^{2},v=x^{2}$
得：$\frac{du}{dv}=\frac{2v+3u+1}{3v+2u-1}$
解方程组$2v+3u+1=0,3v+2u-1=0$得$u=-1,v=1$
令$u=U-1,v=V+1$消去常数得:$\frac{dU}{dV}=\frac{2V+3U}{3V+2U}$
令$t=\frac{U}{V}$，则$U=tV,\frac{dU}{dV}=t+V\frac{dt}{dV}$
$t+V\frac{dt}{dV}=\frac{2+3t}{3+2t}\rightarrow \frac{2(t-1)+5}{t-1}d(t-1)=\frac{1}{V}dV$
$\int \frac{2(t-1)+5}{t-1}d(t-1)=\int \frac{1}{V}dV+ln|C|\rightarrow 2(t-1)+5ln|t-1|=ln|CV|$
$2t+5ln|t-1|=ln|CV|\rightarrow 2\frac{y^{2}+1}{x^{2}-1}+5ln|\frac{y^{2}+1}{x^{2}-1}-1|=ln|C(x^{2}-1)|$
5.求具有性质$x(t+s)=\frac{x(t)+x(s)}{1-x(t)x(s)}$的函数$x(t)$，已知$x'(0)$存在
解：令$t=s=0$，得$x(0)=\frac{2x(0)}{1-x^{2}(0)}$
若$x(0)\neq 0$即$x^{2}(0)= -1$，所以$x(0)=0$
$x'(t)=lim_{\Delta t\rightarrow 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}$
又$x(t+\Delta t)=\frac{x(t)+x(\Delta t)}{1-x(t)x(\Delta t)}$带入$x'(t)$
得$lim_{\Delta t\rightarrow 0}\frac{x(t)+x(\Delta t)-x(t)[1-x(t)x(\Delta t)]}{\Delta t[1-x(t)x(\Delta t)]}$
$lim_{\Delta t\rightarrow 0}\frac{x(\Delta t)(1+x^{2}(t))}{\Delta t(1-x(t)x(\Delta t))}=x'(0)(1+x^{2}(t))$
$x'(t)=\frac{dx(t)}{dt}=x'(0)(1+x^{2}(t))\rightarrow \frac{dx(t)}{1+x^{2}(t)}=x'(0)dt$
$\int \frac{dx(t)}{1+x^{2}(t)}=\int x'(0)dt+C\rightarrow arctanx(t)=x'(0)t+C$
得$x(t)=tan(x'(0)+C)$，当$t=0$时，$x(0)=0$即得$C=0$
即函数$x(t)=tanx'(0)$
9.证明满足习题1.2第8题(7)所给条件的曲线是抛物线族
解：建立曲线的微分方程得$\frac{dy}{dx}=kx$
求解该方程得$dy=kxdx\rightarrow \int dy=\int kxdx+C\rightarrow y=\frac{k}{2}x^{2}+C$
由$C$为任意常数可知$y=\frac{k}{2}x^{2}+C$为抛物线族
第二章第二次作业
P49.
1.求解下列方程的解
（1）$\frac{dy}{dx}=y+sinx$
解：先计算齐次方程$\frac{dy}{dx}=y$得
$ln|y|=x+C\rightarrow y=e^{x}e^{C}$
令$C=c(x)$即
$\frac{dy}{dx}=e^{x}e^{c(x)}+\frac{dc(x)}{dx}e^{x}e^{c(x)}$带入$\frac{dy}{dx}=y+sinx$得
$e^{c(x)}dc(x)=\frac{sinx}{e^{x}}dx$同时求积分得
$e^{c(x)}=\frac{1}{2}e^{-x}(-sinx-cosx)+C$带入$y=e^{x}e^{c(x)}$得
$y=Ce^{x}-\frac{1}{2}(sinx+cosx)$
（3）$\frac{ds}{dt}=-scost+\frac{1}{2}sin2t$
解：先计算齐次方程$\frac{ds}{dt}=-scost$得
$ln|s|=-sint+C\rightarrow s=e^{-sint}e^{C}$
令$C=c(t)$即
$\frac{ds}{dt}=-coste^{-sint}e^{c(t)}+\frac{dc(t)}{dt}e^{-sint}e^{c(t)}$带入$\frac{ds}{dt}=-scost+\frac{1}{2}sin2t$得
$e^{c(t)}dc(t)=\frac{1}{2}e^{sint}sin2tdt\rightarrow e^{c(t)}dc(t)=e^{sint}sintd(sint)$同时求积分得
$e^{c(t)}=\frac{1}{2}e^{sint}(sint-cost)+C$带入$s=e^{-sint}e^{c(x)}$得
$s=\frac{1}{2}sint-\frac{1}{2}cost+Ce^{-sint}$
（5）$\frac{dy}{dx}+\frac{1-2x}{x^{2}}y-1=0$
解：$y=e^{\int \frac{2x-1}{x^{2}}dx}(\int e^{\int \frac{1-2x}{x^{2}}dx}dx+c)$
$y=e^{lnx^{2}+\frac{1}{x}}(\int e^{-lnx^{2}-\frac{1}{x}}dx+c)=x^{2}e^{\frac{1}{x}}(\int x^{-2}e^{-\frac{1}{x}}dx+c)$
$y=x^{2}e^{\frac{1}{x}}(e^{\frac{-1}{x}}+c)=x^{2}(1+ce^{\frac{1}{x}})$
（7）$\frac{dy}{dx}-\frac{2y}{x+1}=(x+1)^{3}$
解：$\frac{dy}{dx}=\frac{2}{x+1}y+(x+1)^{3}$
即$P(x)=\frac{2}{x+1},Q(x)=(x+1)^{3}$
$e^{\int P(x)dx}=e^{\int \frac{2}{x+1}dx}=e^{ln(x+1)^2}=(x+1)^{2}$
$y=e^{\int P(x)dx}(\int Q(x)e^{-\int P(x)dx}dx+c)$
$y=(x+1)^{2}(\int \frac{1}{(x+1)^{2}}(x+1)^{3}dx+c)=(x+1)^{2}(\int x+1dx+c)$
$y=(x+1)^{2}(\frac{1}{2}x^{2}+x+c)$
（9）$\frac{dy}{dx}=\frac{ay}{x}+\frac{x+1}{x}$（$a$为常数）
解：$P(x)=\frac{a}{x},Q(x)=\frac{x+1}{x}$
$e^{\int P(x)dx}=e^{\int \frac{a}{x}dx}=e^{lnx^{a}}=x^{a}$
$y=e^{\int P(x)dx}(\int Q(x)e^{-\int P(x)dx}dx+c)$
$y=x^{a}(\int \frac{1}{x^{a}}\frac{x+1}{x}dx+c)$
当$x=0$时$y=(\int \frac{x+1}{x}dx+c)=x+ln|x|+c$
当$x=1$时$y=x(\int \frac{x+1}{x^{2}}dx+c)=xln|x|+cx-1$
当$x\neq 0,1$时
$y=cx^{a}+\frac{x}{1-a}-\frac{1}{a}$
（11）$\frac{dy}{dx}+xy=x^{3}y^{3}$
解：$\frac{dy}{dx}=-xy+x^{3}y^{3}\rightarrow \frac{dy}{y^{3}dx}=-xy^{-2}+x^{3}$
即$\frac{dy^{-2}}{dx}=-2(-xy^{-2}+x^{3})$
令$t=y^{-2}$得
$\frac{dt}{dx}=2xt-2x^{3}$
$P(x)=2x,Q(x)=-2x^{3}$
$e^{\int P(x)dx}=e^{\int 2xdx}=e^{x^{2}}$
$t=e^{\int P(x)dx}(\int e^{-\int P(x)dx}Q(x)dx+c)$
$t=e^{x^{2}}(\int e^{-x^{2}}(-2x^{3})dx+c)=e^{x^{2}}(-\int e^{-x^{2}}(-x^{2})d(-x^{2})+c)$
$t=e^{x^{2}}(-(e^{-x^{2}}(-x^{2}-1))+c)=x^{2}+1+ce^{x^{2}}$
即$y^{2}(x^{2}+1+ce^{x^{2}})=1$
（13）$2xydy=(2y^{2}-x)dx$
解：这是一个$n=-1$的伯努利方程
$\frac{dy}{dx}=\frac{1}{x}y-\frac{1}{2y}\rightarrow y\frac{dy}{dx}=\frac{1}{x}y^{2}-\frac{1}{2}$
令$z=y^{2}$得$\frac{dz}{dx}=2y\frac{dy}{dx}$带入$\frac{dy}{dx}=\frac{1}{x}y^{2}-\frac{1}{2}$得
$\frac{dz}{dx}=\frac{2}{x}z-1$
$P(x)=\frac{2}{x},Q(x)=-1$
$e^{\int P(x)dx}=e^{\int \frac{2}{x}dx}=e^{lnx^{2}}=x^{2}$
$z=e^{\int P(x)dx}(\int e^{-\int P(x)dx}Q(x)dx+c)$
$z=x^{2}(-\int x^{-2}dx+c)=x^{2}(x^{-3}+c)=\frac{1}{x}+cx^{2}$
即$y^{2}=\frac{1}{x}+cx^{2}$
（15）$\frac{dy}{dx}=\frac{1}{xy+x^{3}y^{3}}$
解：这是一个$n=3$的伯努利方程
$\frac{dx}{dy}=xy+x^{3}y^{3}\rightarrow \frac{d(x^{-2})}{dy}=-2(x^{-2}y+y^{3})$
令$z=x^{-2}$得$\frac{dz}{dy}=-2zy-2y^{3}$
$P(y)=-2y,Q(y)=-2y^{3}$
$e^{\int P(y)dy}=e^{\int -2ydy}=e^{-y^{2}}$
$z=e^{\int P(y)dy}(\int e^{-\int P(y)dy}Q(y)dy+c)$
$z=e^{-y^{2}}(\int e^{y^{2}}(-2y^{3})dy+c)=e^{-y^{2}}(-\int e^{y^{2}}y^{2}dy^{2}+c)$
$z=e^{-y^{2}}(-e^{y^{2}}(y^{2}-1)+c)=-y^{2}+1+ce^{-y^{2}}$
即$x^{-2}=-y^{2}+1+ce^{-y^{2}}\rightarrow x^{2}(1-y^{2}+ce^{-y^{2}})=1$
5.试证：
（2.28）：$\frac{dy}{dx}=P(x)y+Q(x)$ （2.3）：$\frac{dy}{dx}=P(x)y$
（1）一阶非齐次线性微分方程（2.28）的任两解之差必为相应的齐次线性微分方程（2.3）之解
解：设$y_{1},y_{2}$是（2.28）的任意两个解，则有
$\frac{dy_{1}}{dx}=P(x)y_{1}+Q(x)$-----（1）
$\frac{dy_{2}}{dx}=P(x)y_{2}+Q(x)$-----（2）
（2）-（1）得$\frac{dy_{2}-dy_{1}}{dx}=P(x)(y_{2}-y_{1})=\frac{d(y_{2}-y_{1})}{dx}$
令$y=y_{2}-y_{1}$满足方程（2.3），即命题得证
（2）若$y=y(x)$是（2.3）的非零解，而$y=\tilde{y}(x)$是（2.28）的解，则方程（2.28）的通解可表为$y=cy(x)+\tilde{y}(x)$，其中$c$为任意常数
解：由题意可得$\frac{dy(x)}{dx}=P(x)y$------（3）
$\frac{d\tilde{y}(x)}{dx}=P(x)\tilde{y}(x)+Q(x)$-------（4）
$c\times (3)+(4)$得
$\frac{cdy}{dx}+\frac{d\tilde{y}}{dx}=cP(x)y+P(x)\tilde{y}+Q(x)$
即$\frac{d(cy+\tilde{y})}{dx}=P(x)(cy+\tilde{y})+Q(x)$
所以$y=cy+\tilde{y}$是（2.28）的一个解
设$y_{1}$是（2.28）的一个解，则有$\frac{dy_{1}}{dx}=P(x)y_{1}+Q(x)$------（5）
由（5）-（4）可得$\frac{d(y_{1}-\tilde{y})}{dx}=P(x)(y_{1}-\tilde{y})$
即$y_{1}-\tilde{y}=ce^{\int P(x)dx}=cy\rightarrow y_{1}=\tilde{y}+cy$
命题得证
（3）方程（2.3）任一解的常数倍或任两解之和（或差）任是方程（2.3）的解
解：设$y_{3},y_{4}$是（2.3）的任意两解，则有
$\frac{dy_{3}}{dx}=P(x)y_{3}$--------（6）
$\frac{dy_{4}}{dx}=P(x)y_{4}$--------（7）
$c\times (6)$得$\frac{cdy_{3}}{dx}=cP(x)y_{3}\rightarrow \frac{d(cy_{3})}{dx}=P(x)(cy_{3})$其中$c$为任意常数
即$y=cy_{3}$满足方程（2.3）
$(6)\pm (7)$可得$\frac{dy_{3}}{dx}\pm \frac{dy_{4}}{dx}=P(x)y_{3}\pm P(x)y_{4}$
即$\frac{d(y_{3}\pm y_{4})}{dx}=P(x)(y_{3}\pm y_{4})$
即$y=y_{3}\pm y_{4}$满足方程（2.3）命题得证
7.求解下列方程：
（1）$(x^{2}-1)y'-xy+1=0$
解：$y'=\frac{1-xy}{x^{2}-1}\rightarrow y'=-\frac{x}{x^{2}-1}y+\frac{1}{x^{2}-1}$
$P(x)=-\frac{x}{x^{2}-1},Q(x)=\frac{1}{x^{2}-1}$
$e^{\int P(x)dx}=e^{\int -\frac{x}{x^{2}-1}dx}=e^{-\frac{1}{2}\int \frac{1}{x^{2}-1}d(x^{2}-1)}=(x^{2}-1)^{-\frac{1}{2}}$
$y=e^{\int P(x)dx}(\int e^{-\int P(x)dx}Q(x)dx+c)$
$y=(x^{2}-1)^{-\frac{1}{2}}(\int (x^{2}-1)^{\frac{1}{2}}\frac{1}{x^{2}-1}dx+c)$
$y=(x^{2}-1)^{-\frac{1}{2}}(\int \frac{1}{\sqrt{x^{2}-1}}dx+c)=(x^{2}-1)^{-\frac{1}{2}}(ln|x+\sqrt{x^{2}-1}|+c)$
$y=\frac{ln|x+\sqrt{x^{2}-1}|+c}{\sqrt{x^{2}-1}}$
P60
1.验证下列方程是恰当微分方程，并求出方程的解：
（1）$(x^{2}+y)dx+(x-2y)dy=0$
解：$M(x,y)=(x^{2}+y),N(x,y)=(x-2y)$
$\frac{\partial M}{\partial y}=1,\frac{\partial N}{\partial x}=1$
即$\frac{\partial M}{\partial y}=1,\frac{\partial N}{\partial x}=1$
所以此方程为恰当方程
现求u使得它同时满足
$\frac{\partial u}{\partial x}=M(x,y)=(x^{2}+y)$----(1
$\frac{\partial u}{\partial y}=N(x,y)=(x-2y)$-----(2
对（1求积分得$u=\frac{1}{3}x^{3}+xy+\varphi (y)$带入(2得
$x+\frac{d\varphi (y)}{dy}=x-2y\Rightarrow \varphi(y)=\int -2ydy=-y^{2}$
即$\frac{1}{3}x^{3}+xy-y^{2}=c$
（3）$[\frac{y^{2}}{(x-y)^{2}}-\frac{1}{x}]dx+[\frac{1}{y}-\frac{x^{2}}{(x-y)^{2}}]dy=0$
解：$M(x,y)=\frac{y^{2}}{(x-y)^{2}}-\frac{1}{x},N(x,y)=\frac{1}{y}-\frac{x^{2}}{(x-y)^{2}}$
$\frac{\partial M}{\partial y}=\frac{2y(x-y)^{2}+2y^{2}(x-y)}{(x-y)^{4}}=\frac{2xy}{(x-y)^{3}}$
$\frac{\partial N}{\partial x}=\frac{2x(x-y)^{2}+2x^{2}(x-y)}{(x-y)^{4}}=\frac{2xy}{(x-y)^{3}}$
得$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$即此方程为恰当方程
即有$u$使得
$\frac{\partial u}{\partial x}=M=\frac{y^{2}}{(x-y)^{2}}-\frac{1}{x}$----(3
$\frac{\partial u}{\partial y}=N=\frac{1}{y}-\frac{x^{2}}{(x-y)^{2}}$----(4
对(3求积分得$u=\int \frac{y^{2}}{(x-y)^{2}}dx-\int \frac{1}{x}dx+\varphi (y)=-\frac{y^{2}}{x-y}-ln|x|+\varphi ((y))$带入(4得
$\frac{y^{2}+(x-y)2y}{(x-y)^{2}}+\frac{d\varphi (y)}{dy}=\frac{-2xy+y^{2}}{(x-y)^{2}}+\frac{d\varphi (y)}{dy}=\frac{1}{y}-\frac{x^{2}}{(x-y)^{2}}$
即$\frac{d\varphi (y)}{dy}=\frac{1}{y}-\frac{x^{2}}{(x-y)^{2}}-\frac{-2xy+y^{2}}{(x-y)^{2}}=\frac{1}{y}-1$
可得$\varphi (y)=\int (\frac{1}{y}-1)dy=ln|y|-y$
所以可得通解$u=-\frac{y^{2}}{x-y}-ln|x|+ln|y|-y=ln|\frac{y}{x}|-\frac{xy}{x-y}=c$
（5）$(\frac{1}{y}sin\frac{x}{y}-\frac{y}{x^{2}}cos\frac{y}{x}+1)dx+(\frac{1}{x}cos\frac{y}{x}-\frac{x}{y^{2}}sin\frac{x}{y}+\frac{1}{y^{2}})dy=0$
解：$M(x,y)=(\frac{1}{y}sin\frac{x}{y}-\frac{y}{x^{2}}cos\frac{y}{x}+1)$
$N(x,y)=(\frac{1}{x}cos\frac{y}{x}-\frac{x}{y^{2}}sin\frac{x}{y}+\frac{1}{y^{2}})$
$\frac{\partial M}{\partial y}=-\frac{1}{y^{2}}sin\frac{x}{y}-\frac{x}{y^{3}}cos\frac{x}{y}-\frac{1}{x^{2}}cos\frac{y}{x}+\frac{y}{x^{3}}sin\frac{y}{x}$
$\frac{\partial N}{\partial x}=-\frac{1}{y^{2}}sin\frac{x}{y}-\frac{x}{y^{3}}cos\frac{x}{y}-\frac{1}{x^{2}}cos\frac{y}{x}+\frac{y}{x^{3}}sin\frac{y}{x}$
即$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$所以此方程是恰当方程
则有$u$满足$\frac{\partial u}{\partial x}=M,\frac{\partial u}{\partial y}=N$
对$M$求积分得$u=sin\frac{y}{x}-cos\frac{x}{y}+x+\varphi (y)$，对其求$y$的偏导得
$\frac{1}{x}cos\frac{y}{x}-\frac{x}{y^{2}}sin\frac{x}{y}+\frac{d\varphi (y)}{dy}=\frac{1}{x}cos\frac{y}{x}-\frac{x}{y^{2}}sin\frac{x}{y}+\frac{1}{y^{2}}$
得$\frac{d\varphi (y)}{dy}=\frac{1}{y^{2}}\Rightarrow \varphi (y)=-\frac{1}{y}$
故方程的通解为$sin\frac{y}{x}-cos\frac{x}{y}+x-\frac{1}{y}=c$
第三次作业
P60
2.求下列方程的解
（1）$2x(ye^{x^{2}}-1)dx+e^{x^{2}}dy=0$
解：$M=2x(ye^{x^{2}}-1),N=e^{x^{2}}$
$\frac{\partial M}{\partial y}=2xe^{x^{2}},\frac{\partial N}{\partial x}=2xe^{x^{2}}$即$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
所以此方程是恰当方程
$\frac{\partial u}{\partial x}=M=2x(ye^{x^{2}}-1)\Rightarrow u=y\int e^{x^{2}}dx^{2}-\int 2xdx+\varphi (y)$
得$u=ye^{x^{2}}-x^{2}+\varphi (y)\rightarrow \frac{du}{dy}=e^{x^{2}}+\frac{d\varphi (y)}{dy}=N=e^{x^{2}}$
即$\varphi (y)=\int 0dy=c\rightarrow u=ye^{x^{2}}-x^{2}=c$
（3）$2xydx+(x^{2}+1)dy=0$
解：$M=2xy,N=x^{2}+1$
$\frac{\partial M}{\partial y}=2x,\frac{\partial N}{\partial x}=2x$即$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
所以此方程是恰当微分方程
$\frac{\partial u}{\partial x}=M=2xy\rightarrow u=\int 2xydx+\varphi (y)=yx^{2}+\varphi (y)$
$\frac{\partial u}{\partial y}=x^{2}+\frac{d\varphi (y)}{dy}=N=x^{2}+1\rightarrow \varphi (y)=y$
$u=yx^{2}+y=c$
（5）$ydx-(x+y^{3})dy=0$
解：$\frac{dy}{dx}=\frac{y}{x+y^{3}}\rightarrow \frac{dx}{dy}=\frac{1}{y}x+y^{2}$
即$P(y)=\frac{1}{y},Q(y)=y^{2}$
$e^{\int P(y)dy}=e^{\int \frac{1}{y}dy}=y$
$x=e^{\int P(y)dy}(\int e^{-\int P(y)dy}Q(y)dy+c)=y(\int y^{-1}y^{2}dy+c)=\frac{1}{2}y^{3}+cy$
（7）$(y-x^{2})dx-xdy=0$
解：$\frac{dy}{dx}=\frac{1}{x}y-x$
即$P(x)=\frac{1}{x},Q(x)=x$
$e^{\int P(x)dx}=e^{\int \frac{1}{x}dx}=x$
$y=e^{\int P(x)dx}(\int e^{-\int P(x)dx}Q(x)dx+c)=x(\int x^{-1}xdx+c)=x^{2}+cx$
（9）$[xcos(x+y)+sin(x+y)]dx+xcos(x+y)dy=0$
解：$M=xcos(x+y)+sin(x+y),N=xcos(x+y)$
$\frac{\partial M}{\partial y}=-xsin(x+y)+cos(x+y),\frac{\partial N}{\partial x}=cos(x+y)-xsin(x+y)$
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$即此方程为恰当微分方程
$\frac{\partial u}{\partial x}=M\rightarrow u=\int Mdx=xsin(x+y)+\varphi (y)$
$\frac{\partial u}{\partial y}=xcos(x+y)+\frac{d\varphi (y)}{dy}=N=xcos(x+y)\rightarrow \varphi (y)=c$
$u=xsin(x+y)=c$
（11）$x(4ydx+2xdy)+y^{3}(3ydx+5xdy)=0$
解：两边同乘以$x^{2}y$得
$(4x^{3}y^{2}dx+2x^{4}ydy)+(3x^{2}y^{5}+5x^{3}ydy)=0\rightarrow d(x^{4}y^{2})+d(x^{3}y^{5})=0$
得$x^{4}y^{2}+x^{3}y^{5}=c$
3.试导出方程$M(x,y)dx+N(x,y)dy=0$分别具有形为$\mu (x+y)$和$\mu (xy)$的积分因子的充要条件
解：若方程具有积分因子$\mu (x+y)$则有
$\frac{\partial (\mu M)}{\partial y}=\frac{\partial (\mu N)}{\partial x}\rightarrow M\frac{\partial \mu}{\partial y}-N\frac{\partial \mu}{\partial x}=\mu (\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})$
令$z=x+y$即有$\frac{\partial \mu}{\partial x}=\frac{d\mu}{dz}\frac{\partial z}{\partial x}=\frac{d\mu}{dz},\frac{\partial \mu}{\partial y}=\frac{d\mu}{dy}$
得$(M-N)\frac{d\mu}{dz}=\mu(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\rightarrow \frac{d\mu}{\mu}=\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M-N}dz$
即当$\mu (x+y)$为积分因子的充要条件为$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M-N}$是关于$x+y$的函数
令$z=xy$即有$\frac{\partial \mu}{\partial x}=\frac{d\mu}{dz}\frac{\partial z}{\partial x}=y\frac{d\mu}{dz},\frac{\partial \mu}{\partial y}=x\frac{d\mu}{dy}$
得$(Mx-Ny)\frac{d\mu}{dz}=\mu(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\rightarrow \frac{d\mu}{\mu}=\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{Mx-Ny}dz$
即当$\mu (xy)$为积分因子的充要条件为$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{Mx-Ny}$是关于$xy$的函数
5.试证齐次微分方程$M(x,y)dx+N(x,y)dy=0$当$xM+yN\neq 0$时有积分因子$\mu =\frac{1}{xM+yN}$
解：设$\mu$为其积分因子，则有
$\frac{\partial \mu M}{\partial y}=\frac{\partial \frac{M}{xM+yN}}{\partial y}=\frac{\frac{\partial M}{\partial y}(xM+yN)-M(x\frac{\partial M}{\partial y}+N+y\frac{\partial N}{\partial y})}{(xM+yN)^{2}}$
$=\frac{yN\frac{\partial M}{\partial y}-MN-yM\frac{\partial N}{\partial y}}{(xM+yN)^{2}}=\frac{\partial \mu N}{\partial x}$
即，$\mu =\frac{1}{xM+yN}$为该方程的积分因子
10.设$\mu_{1}(x,y),\mu_{2}(x,y)$是方程（2.42）的两个积分因子，且$\frac{\mu_{1}}{\mu_{2}}\not\equiv $常数，求证$\frac{\mu_{1}}{\mu_{2}}=c$（任意常数）是方程（2.42）的通解
方程2.42：$M(x,y)dx+N(x,y)dy=0$
解：$\mu_{i}Mdx+\mu_{i}Ndy=0(i=1,2)\Rightarrow dy=-\frac{M}{N}dx$
又$N\frac{\partial \mu_{i}}{\partial x}-M\frac{\partial \mu_i}{\partial y}=\mu_{i}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})$
可得$d(\frac{\mu_{1}}{\mu_{2}})=\frac{\mu_{2}(\frac{\partial \mu_{1}}{\partial x}dx+\frac{\partial \mu_{1}}{\partial y}dy)-\mu_{1}(\frac{\partial \mu_{2}}{\partial x}dx+\frac{\partial \mu_{2}}{\partial y}dy)}{\mu_{2}^{2}}$
$=\frac{\mu_{2}(\frac{\partial \mu_{1}}{\partial x}dx-\frac{\partial \mu_{1}}{\partial y}\frac{M}{N}dx)-\mu_{1}(\frac{\partial \mu_{2}}{\partial x}dx-\frac{\partial \mu_{2}}{\partial y}\frac{M}{N}dx)}{\mu_{2}^{2}}$
$=\frac{[\mu_{2}(N\frac{\partial \mu_{1}}{\partial x}-M\frac{\partial \mu_{1}}{\partial y})-\mu_{1}(N\frac{\partial \mu_{2}}{\partial x}-M\frac{\partial \mu_{2}}{\partial y})]dx}{N\mu_{2}^{2}}$
$=\frac{[\mu_{2}\mu_{1}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})-\mu_{1}\mu_{2}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})]dx}{N\mu_{2}^{2}}=0$
即当$\frac{\mu_{1}}{\mu_{2}}\not\equiv $常数时，$\frac{\mu_{1}}{\mu_{2}}=c$时方程的解
P69.
1.求解下列方程：
（1）$xy'^{3}=1+y'$
解：令$y'=p=\frac{1}{t}$则有$x=\frac{1+\frac{1}{t}}{\frac{1}{t^{3}}}=t^{3}+t^{2}$
则有$y=\int pdx+c=\int \frac{1}{t}d(t^{3}+t^{2})+c=\int 3t+2dt+c=\frac{3}{2}t^{2}+2t+c$
于是求得方程参数形式的通解为$x=t^{3}+t^{2},y=\frac{3}{2}t^{2}+2t+c$
（3）$y=y'^{2}e^{y'}$
解：令$y'=p$则有$y=p^2e^p$
即有$x=\int \frac{1}{p}d(p^2e^p)+c=\int \frac{1}{p}(2pe^p+p^2e^p)dp+c=e^p(1+p)+c$
于是求得方程参数形式的通解为$x=e^p(1+p)+c,y=p^2e^p$
（5）$x^{2}+y'^{2}=1$
解：令$y'=p=cost$则有$x=\sqrt{1-cos^2t}=sint$
即有$y=\int costd(sint)+c=\int cos^2tdt+c=\int \frac{1+cos2t}{2}dt+c=\frac{1}{2}t+\frac{1}{4}sin2t+c$
于是求得方程参数形式的通解为$x=sint,y=\frac{1}{2}t+\frac{1}{4}sin2t+c$