Question

高等数学求通解

看图片大神们求过程

Answer 1

Answer 2

let
u= y/x
du/dx = (1/x).dy/dx - (1/x^2)y
= (1/x).dy/dx - (u/x)
x.du/dx = dy/dx - u
dy/dx = u+x.du/dx
y' = (1/x)y + tan(y/x)
u+x.du/dx = u + tanu
x.du/dx = tanu
∫cotu.du =∫dx/x
ln|sinu| = ln|x| + C'
sinu = Cx
sin(y/x) = Cx
y = x.arcsin(Cx)

Answer 3

方法如下图所示，请作参考，祝学习愉快：