Doubt in method of Undetermined Coefficients $(D^3+3D^2+2D)y=x^2$.

Question

Solve $(D^3+3D^2+2D)y=x^2, D \equiv \dfrac{d}{dx}$

Choosing, trial solution as $y=a_0+a_1x+a_2x^2$

If i substitute this in given differential equation,

$0+3(2a_2)+2(a_1+2a_2x)=x^2 $

Comparing corresponding coefficients, i get,

$a_0=0=a_1 = a_2$ --> it seems i made wrong some where, Pls correct me.

Answer 1

Hint: Try with $y_p=A+B x+C x^2+D x^3$ then $$(D^3+3D^2+2D)y=(2 B + 6 C + 6 D)+ (4 C + 18 D)x+ 6 Dx^2\equiv x^2$$

Answer 2

$$(D^3+3D^2+2D)y=x^2,$$ If you integrate then you will see that you need a polynomial of at least degree 3 $$(D^2+3D+2)y=\frac {x^3}3+K_3$$ $$(D+1)(D+2)y=\frac {x^3}3+K_3$$ $$\implies y_h=K_1e^{-x}+K_2e^{-2x}$$ $$\implies y_p=ax^3+bx^2+cx$$