Question

求微分方程dy/+dx+1/+x×+y=x^4的通解

Answer 1

问：求微分方程dy/+dx+1/+x×+y=x^4的通解

答：∵令t=lny,则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)∵很容易求出齐次方程dx/dt=x的通解是x=Ce^t (C是积分常数)∴根据常数变易法,设方程(1)的解为x=C(t)e^t (C(t)是关于t的函数)∵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是积分常数)∴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是积分常数).