求解微分方程的y'-ycotx=2xsinx的通解!要过程,谢谢!
2019-05-07
求解微分方程的y'-ycotx=2xsinx的通解!要过程,谢谢!
优质解答
y'-ycotx=0
dy/dx=y*cotx
1/y*dy=cotx*dx
ln|y|=ln(sin(x))+c1
y=c2*sin(x)
令 y=c(x)*sin(x) 常数变易法
y'=c(x)cos(x)+c'(x)*sin(x)
y'-y=c(x)cos(x)+c'(x)*sin(x)-c(x)cos(x)=2x*sin(x)
=> c'(x)=2x => c(x)=x^2 + c3
即
y'-ycotx=2xsinx的通解为
y=(x^2+c3)*sin(x) c3为任意常数
y'-ycotx=0
dy/dx=y*cotx
1/y*dy=cotx*dx
ln|y|=ln(sin(x))+c1
y=c2*sin(x)
令 y=c(x)*sin(x) 常数变易法
y'=c(x)cos(x)+c'(x)*sin(x)
y'-y=c(x)cos(x)+c'(x)*sin(x)-c(x)cos(x)=2x*sin(x)
=> c'(x)=2x => c(x)=x^2 + c3
即
y'-ycotx=2xsinx的通解为
y=(x^2+c3)*sin(x) c3为任意常数