Calculus y '= Xe ^ 2x-y, when x = 1 / 2, y = 0, the ball is the special solution of the differential equation

Calculus y '= Xe ^ 2x-y, when x = 1 / 2, y = 0, the ball is the special solution of the differential equation

y'-y=xe^(2x)
e^(-x)(y'-y)=xe^x
(e^(-x)y)'=xe^x
Two side integral: e ^ (- x) y = ∫ Xe ^ xdx = ∫ XD (e ^ x) = Xe ^ X - ∫ e ^ xdx = Xe ^ x-e ^ x + C
y=(x-1)e^(2x)+C