第十题，微分方程求解。

Question

如题所述

Answer 1

用全微分法：
y' = (x-y+1)/(x+y²+3)
(-x+y-1)dx+(x+y²+3)dy=0
d/dy(-x+y-1)=1=d/dx(x+y²+3)
f(x,y)=∫(-x+y-1)dx=-x²/2+xy-x+g(y)
f'y=x+g'(y)=x+y²+3
g'(y)=y²+3
g(y)=y³/3+3y
f(x,y)=-x²/2+xy-x+y³/3+3y
所以通解为
x²/2+xy-x+y³/3+3y=c1

Answer 2

（10）解：∵y'=(x-y+1)/(x+y^2+3)
==>dy/dx=(x-y+1)/(x+y^2+3)
==>(x+y^2+3)dy-(x-y+1)dx=0
==>(xdy+ydx)+(y^2+3)dy-(x+1)dx=0
==>∫(xdy+ydx)+∫(y^2+3)dy-∫(x+1)dx=0
==>xy+y^3/3+3y-x^2/2-x=C (C是积分常数)
∴原方程的通解是xy+y^3/3+3y-x^2/2-x=C。本回答被提问者采纳