Solve the Differential equation by variation of parameters

Question

y'' - 4y' + 3y = xe^-x

Answer 1

Characteristic equation: r² - 4r + 3 = 0; r = 1 or r = 3;

y_GH = Ae^x _{+ Be}^3x; W(e^x_, e^3x) = - 2e^4x_.

y_P = (- ∫e^3xxe^-x/(-2e^4x) dx)e^x + (∫e^xxe^-x/(-2e^4x)dx)e^3x = (1/2∫xe^-2xdx)e^x - (1/2∫xe^-4x)e^3x = (- 1/4x - 1/8)e^-x +

(1/8x + 1/32)e^-x = (- 1/8 - 3/32)e^-x.

y_G = Ae^x + Be^3x - (1/8x + 3/32)e^-x