y'(2y-y')=y^2*(sinx)^2求解,

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y'(2y-y')/y^2=sin^2(x)
2y'/y-(y'/y)^2=sin^2(x)
(y'/y)^2-2y'/y+1=1-sin^2(x)=cos^2(x)
(y'/y-1)^2=cos^2(x)
y'/y-1=±cosx
(lny)'=1±cosx
lny=x±sinx+C
y=e^(x±sinx+C)