解:∵(x-2)*dy/dx=y+2*(x-2)³ ==>(x-2)dy=[y+2*(x-2)³]dx ==>(x-2)dy-ydx=2*(x-2)³dx ==>[(x-2)dy-ydx]/(x-2)²=2*(x-2)dx ==>d[y/(x-2)]=d[(x-2)²] ==>y/(x-2)=(x-2)²+C (C是积分常数) ==>y=(x-2)³+C(x-2) ∴原方程的通解是y=(x-2)³+C(x-2) (C是积分常数)。不明白这里怎么从上面的式子得到下面的?