题目

求微分方程dy={(y+x)/(x+1)²}dx通解,

提问时间：2020-10-29

答案

求微分方程dy={(y+x)/(x+1)²}dx的通解
dy/dx=(y+x)/(x+1)²=y/(x+1)²+x/(x+1)²
dy/dx-y/(x+1)²=x/(x+1)².(1)
先求齐次方程 dy/dx-y/(x+1)²=0.(2) 的通解；
分离变量得 dy/y=dx/(x+1)²,积分之得：lny=-1/(x+1)+lnC₁
即(2)的通解为：y=C₁e^[-1/(x+1)]
再用参数变易法,将通解中的C₁换成x的函数u而令y=ue^[-1/(x+1)].(3)
将(3)两边对x取导数得：
dy/dx=(du/dx)e^[-1/(x+1)]+{ue^[-1/(x+1)]}/(x+1)².(4)
将(3)(4)代入(1)以求u：
(du/dx)e^[-1/(x+1)]+{ue^[-1/(x+1)]}/(x+1)²-{ue^[-1/(x+1)]}/(x+1)²= x/(x+1)²
化简得 (du/dx)e^[-1/(x+1)]=x/(x+1)²
故得du/dx=[x/(x+1)²]e^[1/(x+1)]
于是得u=∫[x/(x+1)²]e^[1/(x+1)]dx=-∫xde^[1/(x+1)]=-xe^[1/(x+1)]+∫e^[1/(x+1)]dx .(5)
将(5)代入(3)式即得原方程的通解为：
y={-xe^[1/(x+1)]+∫e^[1/(x+1)]dx }e^[-1/(x+1)]=-x+e^[-1/(x+1)]∫e^[1/(x+1)]dx
其中积分∫e^[1/(x+1)]dx 不好求,你自己找本比较全的积分表查一查吧!