Question

求微分方程dy/dx+y/x=cosx的通解？

Answer 1

dy/dx + y/x = cosx
积分因子 = e^∫ 1/x dx = e^ln|x| = x,乘以方程两边
x · dy/dx + y = xcosx
d(xy)/dx = xcosx
xy = ∫ xcosx dx
xy = ∫ x d(sinx) = xsinx - ∫ sinx dx = xsinx + cosx + C
y = sinx + (cosx)/x + C/x,8,符合dy/dx+P(x)y=Q(x),则通解为
y=[e^(-∫p(x)dx)]*[∫Q(x)e^(p(x)dx)+C]
则P(x)=1/x,Q(x)=cosx,
则通解为y=[e^(-∫1/xdx)]*[∫cosxe^(1/xdx)+C]
=(1/x)*(∫x*cosxdx+C)
=(1/x)*(∫xdsinx+C)
=(1/x)*(xsinx-∫sinx+C )
=sinx+(cosx+C)/x,1,