Question

求微分方程的通解如图

Answer 1

求微分方程 dy/dx+[2x/(x²-1)]y=(cosx)/(x²-1)的通解
解：先求齐次方程dy/dx+[2x/(x²-1)]y=0的通解。
分离变量得 dy/y=-[2x/(x²-1)]dx
取积分 lny=-∫[2x/(x²-1)]dx=-∫d(x²-1)/(x²-1)
=-ln(x²-1)+lnc₁=ln[c₁/(x²-1)]
故y=c₁/(x²-1);
将c₁换成x的函数u，得y=u/(x²-1)...............①
对①取导数:
dy/dx=[(x²-1)(du/dx)-2ux]/(x²-1)²=(du/dx)/(x²-1)-2ux/(x²-1)²...............②
将①②代入原式得：
(du/dx)/(x²-1)-2ux/(x²-1)²+[2x/(x²-1)][u/(x²-1)]=(cosx)/(x²-1)
化简得 (du/dx)/(x²-1)=(cosx)/(x²-1)
分离变量得 du=cosxdx
积分之得u=sinx+c.............③
将③代入①式，即得通解为： y=(sinx+c)/(x²-1)