令v0=-e/h,u0=[(be/h)-c]/a则原方程化为du/dv+b/a=f(hv+e)/g(a(u-u0)+b(v-v0))令x=a(u-u0)+b(v-v0)则(1/a)dx/dv=d(u-u0)/dv + b/a=du/dv + b/a=f(hv+e)/g(x)所以g(x)dx=adv/(hv+e)即为可分离变量方程
