求微分方程y'-2xy=xy^2的通解

Question

求微分方程y'-2xy=xy^2的通解

Answer 1

y'-2xy=xy²,则y'=xy²+2xy=xy(y+2).所以dy/[y(y+2)]=xdx
又1/[y(y+2)]=(1/2){(1/y)-[1(y+2)]},
所以(1/2){(1/y)-[1(y+2)]}dy=xdx,两边积分,得
(1/2) [lny-ln(y+2)] = x²/2 +C ,(C为任意常数)
故(1/2) ln[y/(y+2)]=x²/2 +C ,(C为任意常数).
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(经验证,答案正确.)