微分方程y'=1/(x+y)^2的通解是什么?

Question

微分方程y'=1/(x+y)^2的通解是什么?

Answer 1

令x+y=t,则y'=t'-1
代入原方程,得t'-1=1/t²
==>t'=(1+t²)/t²
==>t²dt/(1+t²)=dx
==>(1-1/(1+t²))dt=dx
==>t-arctant=x-C (C是任意常数)
==>x+y-arctan(x+y)=x-C
==>arctan(x+y)=y+C
==>x+y=tan(y+C)
==>x=tan(y+C)-y
故原方程的通解是x=tan(y+C)-y.