Question

求微分方程通解，如图😀

Answer 1

y`=(x-y)/(x+y)=(1+y/x)/(1-y/x)
设u=y/x，则y=ux，y`=u`x+u，代入原方程得
u`x+u=(1+u)/(1-u)，整理得
u`x=(1+u²)/(1-u)，分离变量得
(1-u)du/(1+u²)=dx/x
∫(1-u)du/(1+u²)
=-(1/2)ln(1+u²)+arctanx+C1
故对(1-u)du/(1+u²)=dx/x两边积分得
-(1/2)ln(1+u²)+arctanx=ln|x|-(1/2)lnC
即ln(1+u²)-2arctanx=-2ln|x|+lnC
ln(1+u²)+2ln|x|-lnC=2arctanx，ln[(1+y²/x²)x²/C]=2arctanx
ln[( x²+y²)/C]=2arctanx，e^(2arctanx)=( x²+y²)/C
x²+y²=Ce^(2arctanx)

Answer 2

令u=y/x，则dy=xdu+udx
(xdu+udx)(1-u)=(1+u)dx
x(1-u)du=(1+u²)dx
(1-u)/(1+u²)du=1/xdx
arctanu-1/2ln(u²+1)=ln|x|+C
通解x=Ceˆ(arctanu-1/2ln(u²+1))=Ceˆ(arctan(y/x)-1/2ln(y²/x²+1))