Q.Solve : $(D^2 - 4D + 1)y = x^2$

Solve the auxiliary equation for the complementary function, then find a polynomial particular integral by the inverse-operator expansion method.

1. Given ODE: $(D^2-4D+1)y = x^2$.
2. Complementary Function (C.F.): Auxiliary equation $m^2-4m+1=0$. $m = \dfrac{4\pm\sqrt{16-4}}{2} = \dfrac{4\pm\sqrt{12}}{2} = 2\pm\sqrt3$. $\text{C.F.} = C_1 e^{(2+\sqrt3)x} + C_2 e^{(2-\sqrt3)x}$.
3. Particular Integral (P.I.): $\text{P.I.} = \dfrac{1}{D^2-4D+1}x^2 = \dfrac{1}{1-(4D-D^2)}x^2$.
4. Expand as a binomial series in the operator $u=4D-D^2$ (only terms up to $D^2$ matter, since $D^3x^2=0$): $\dfrac{1}{1-u} = 1+u+u^2+\cdots$
5. Compute $u(x^2) = 4D(x^2)-D^2(x^2) = 4(2x)-2 = 8x-2$.
6. Compute $u^2(x^2)$: $u^2 = (4D-D^2)^2 = 16D^2-8D^3+D^4$; applied to $x^2$: $16D^2x^2-8D^3x^2+D^4x^2 = 16(2)-0+0=32$. …