隐函数e^y*y'-sin(x^2+y)*(2x+y')=0如何化成y'=的形式啊..能写出过程么,

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隐函数e^y*y'-sin(x^2+y)*(2x+y')=0如何化成y'=的形式啊..能写出过程么,

解出来就行了啊
e^y*y'-sin(x^2+y)*(2x+y')=0
e^y*y'-sin(x^2+y)*2x-sin(x^2+y)*y'=0
e^y*y'-sin(x^2+y)*y'=sin(x^2+y)*2x
y'=sin(x^2+y)*2x/[e^y-sin(x^2+y)]

e^y*y'-sin(x^2+y)*(2x+y')=0