Assume you mean y''+3y'+2y=4x².
First solve homogeneous DE:
y''+3y'+2y=0.
The characteristic equation is found by factoring:
r²+3r+2=(r+2)(r+1), so r=-2, -1 giving the solution:
y₁=Ae⁻²ˣ+Be⁻ˣ.
Let y₂=ax²+bx+c,
y₂'=2ax+b, y₂''=2a, so:
y₂''+3y₂'+2y₂=
2a+6ax+3b+2ax²+2bx+2c≡4x².
Equating coefficients:
2a=4 (x²), a=2,
6a+2b=0 (x), 2b=-6a=-12, b=-6
2a+3b+2c=4-18+2c=0, c=7.
Particular solution y=y₁+y₂:
y=Ae⁻²ˣ+Be⁻ˣ+2x²-6b+7.