∵令t=lny,則y=e^t,dy=e^tdt
∴代入原方程得
dy/dx=y/(x+y^4)
==>ydx/dy=(x+y^4)
==>e^tdx/(e^tdt)=(x+e^(4t))
==>dx/dt=x+e^(4t).(1)
∵很容易求出齊次方程dx/dt=x的通解是
x=Ce^t (C是積分常數(shù))
∴根據(jù)常數(shù)變易法,設(shè)方程(1)的解為x=C(t)e^t (C(t)是關(guān)于t的函數(shù))
∵代入方程(1)得
C'(t)e^t+C(t)e^t=C(t)e^t+e^(4t)
==>C'(t)e^t=e^(4t)
==>C'(t)=e^(3t)
==>C(t)=e^(3t)/3+C (C是積分常數(shù))
∴方程(1)的通解是x=(e^(3t)/3+C)e^t=e^(4t)/3+Ce^t
故原方程的通解是x=y^4/3+Cy (C是積分常數(shù)).