y'(2y-y')=y^2*(sinx)^2求解,
问题描述：
y'(2y-y')=y^2*(sinx)^2求解,
答
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±cosxlny=x±sinx+Cy=e^(x±sinx+C)