线性微分方程x^2y''=(y')^2+2xy'
问题描述：
线性微分方程x^2y''=(y')^2+2xy'
答
x^2y''-2xy'=y'^2(x^2y''-2xy')/y'^2=1(x^2/y')'=1x^2/y'=pdp/dx=1p=x+Cx^2/y'=x+Cy'=x^2/(x+C)dy=x^2dx/(x+C) ∫x^2dx/(x+C)=x^2/2-∫Cdx+C^2∫dx/(x+c)=X^2/2-Cx+C^2ln(x+C)+Cy=(x^2/2)-Cx+C^2ln(x+C)+C