Find the general solution of xy' + ( 1 + x )y - e^(-x) cos2x = 0.

Answer 1

y' + (1/x + 1)y = e^-xcos2x/x; M = e^{∫(1/x + 1)dx} = xe^x. After multiplication we get: (yxe^x)' = cos2x;

yxe^x = 1/2sin2x + C; y = (1/2sin2x + C)/(xe^x)