求dy/dx=x+xy^/y+yx^满足初始条件y｜(下面是x=0) =2的特解．

题目

求dy/dx=x+xy^/y+yx^满足初始条件y｜(下面是x=0) =2的特解．
^是2,2次方

答案

dy/dx = (x+xy²)/(y+yx²)(y+yx²)dy = (x+xy²)dxydy + yx²dy - xdx - xy²dx = 0ydy - xdx + 1/2 * (x²dy² - y²dx²) = 0dy² - dx² + x²dy² - ...