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。
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。