求微分方程xy✀-y=y^2满足初始条件y(1)=1的解

2025-01-21 09:39:33
推荐回答(2个)
回答1:

解:∵xy'-y=y^2
==>(xy'-y)/y^2=1
==>1+(y-xy')/y^2=0
==>1+d(x/y)/dx=0
==>dx+d(x/y)=0
==>∫dx+∫d(x/y)=0
==>x+x/y=C (C是积分常数)
==>x(y+1)=Cy
∴此方程的通解是x(y+1)=Cy
∵y(1)=1
∴代入通解,得C=2
故所求特解是x(y+1)=2y。

回答2:

解:∵xy'-y=y^2
==>(xy'-y)/y^2=1
==>1+(y-xy')/y^2=0
==>1+d(x/y)/dx=0
==>dx+d(x/y)=0
==>∫dx+∫d(x/y)=0
==>x+x/y=C
(C是积分常数)
==>x(y+1)=Cy
∴此方程的通解是x(y+1)=Cy
∵y(1)=1
∴代入通解,得C=2
故所求特解是x(y+1)=2y。