Question

求该微分方程的特解，希望大神教

Answer 1

令y=xu
则y'=u+xu', 代入原方程：
(x+2xu)(u+xu')=xu-2x
即(1+2u)(u+xu')=u-2
u+xu'=(u-2)/(2u+1)
xu'=(u-2-2u^2-u)/(2u+1)
xdu/dx=-2(u^2+1)/(2u+1)
du(2u+1)/(u^2+1)=-2dx/x
d(u^2)/(u^2+1)+du/(u^2+1)=-2dx/x
积分：ln(u^2+1)+arctanu=-2ln|x|+C1
即(u^2+1)e^(arctanu)=C/x^2
代入u得：(y^2+x^2)e^(arctany/x)=C
代入y(1)=1,得：2e^arctan1=C, 即C=2e^(π/4)
所以特解为(y^2+x^2)e^arctan(y/x)=2e^(π/4)