Question

一个微分方程求解的题，请给出详细步骤，谢谢！

Answer 1

dy/dx=(2y²-xy)/(y²-xy+x²)=(2(y/x)²-y/x)/((y/x)²-y/x+1)
令u=y/x,则y=ux, dy/dx=u+x·du/dx=(2u²-u)/(u²-u+1)
得x·du/dx=-u(u-1)(u-2)/(u²-u+1)
-(u²-u+1)/(u(u-1)(u-2)) du=dx/x=ln|x|+C1
(-1/(u-2)-1/(u(u-1)(u-2))) du=(-1/(u-2)-1/2(1/u+1/(u-2)-2/(u-1))) du
=-3/(2(u-2)) d(u-2)-1/2u du+1/(u-1) d(u-1)=-3/2 ln|u-2|-1/2 ln|u|+ln|u-1|+C2=ln|x|+C1
整理得 ln|x²(u-2)³u/(u-1)²|=C3
将u=y/x代入上式得 (y-2x)³y/(y-x)²=C;