Question

求方程(x+xy²)dx-(x²y+y)dy=0的通解，求大神顺便教教我全微分方程怎么解

Answer 1

解：∵(3x²+2xy-y²)dx+(x²-xy)dy=90同除以x^2(3+2y/x-(y/x)^2)dx+(1-y/x)dy=0y/x=uy=uxy'=u'x+u(3+2u-u^2)+(1-u)(u'x+u)=0(3+2u-u^2)/(u-1)-u=u'x(3+2u-u^2-u^2+u)/(u-1)=u'x(-2u^2+3u+3)/(u-1)=u'x(u^2-1.5u-1.5)/(u-1)=-1/2u'x(u-1)/(u^2-1.5u-1.5)=(u-1)/[(u-0.75)^2-33/16]＝16(u-1)/[33[16/33(u-0.75)^2]-1]做到这里自然可设16/33(u-0.75)^2=sec^2t虽然可做出，但很麻繁，可能是题目错了。、、题目改成一下：解：∵(3x²+2xy-y²)dx+(x²-2xy)dy=0==>3x²dx+2xydx-y²dx+x²dy-2xydy=0==>d(x³)+yd(x²)-y²dx+x²dy-xd(y²)=0==>d(x³)+[yd(x²)+x²dy]-[y²dx+xd(y²)]=0==>d(x³)+d(x²y)-d(xy²)=0==>x³+x²y-xy²=C(C是积分常数)

追问

原题答案是y²=2x²+c不复杂啊就是不知道怎么求的

Answer 2

先进行分离变量 得( y/1+y²)dy=(x/1+x²)dx 再两边同时积分∫( y/1+y²)dy=∫(x/1+x²)dx 得到 1/2ln|1+y²|=1/2ln|1+x²|+ln|C| 解得 y²=x²+C