Question

求解微分方程(x^2-1)y'+2xy-cosx=0,万分感谢!

Answer 1

原方程=>dy/dx+2x/(x^2-1)*y=cosx/(x^2-1)

对应的齐次方程：dy/dx+2x/(x^2-1)*y=0

分离变量得 1/y*dy=-2x/(x^2-1)*dx

两边积分得 ln|y|=-ln|x^2-1|+lnC1

y=C/(x^2-1)

下面用待定系数法求解非齐次方程通解

令y=C(x)/(x^2-1)

则C'(x)/(x^2-1)=cosx/(x^2-1)

C'(x)=cosx

C(x)=∫cosxdx=sinx+C

非齐次方程通解:y=(sinx+C)/(x^2-1)