Differential equation question

I have a differential equation homework question that I need help with:

(a) Find the value of A so that the equation y ' - xy - 4x = 0 has a solution of the form
y(x) = A + Be^((x^2)/2) for any constant B.


(b) If y(0) = 1, find B.

I'm really stuck on how to do this, so any bit of help would be greatly appreciated. Thanks so much.

y(x) = A + Be^[(x^2/2)]

y ' = Be^[(x^2/2)]*x

y ' = Bx e^[x^2/2)]

substitute y ' and y values in DE

Bxe^[(x^2/2)] - x [A + Be^[(x^2/2)] - 4x = 0

=> Bxe^[(x^2/2)] - Bxe^[(x^2/2)] - Ax - 4x = 0

=> -x(A + 4) = 0

since x can't be zero, A + 4 = 0 ==> A = -4

y(x) = -4 + Be^[(x^2/2)]

b)

y(0) = 1 ==> -4 + B = 1 ==> B = 5

y' = xy + 4x
dy/dx = x (y + 4)
dy/(y + 4) = x dx
Integrate:
ln|y + 4| = x^2/2 + C
y + 4 = B e^(x^2/2)
y = Be^(x^2/2)-4
Let x = 0, y = 1
1 = B-4
B = 5