Find GS: $(D-1)^{3}(D-2)^{1}(D^{2}-6D+10)y(x)=0$ where $D=\frac{d}{dx}$

Question

Find GS: $(D-1)^{3}(D-2)^{1}(D^{2}-6D+10)y(x)=0$ where $D=\frac{d}{dx}$

Is this one even solvable? Should I multiply everything out?

Answer 1

$$(D-1)^{3}(D-2)^{1}(D^{2}-6D+10)y(x)=0$$

1) $(D-1)^3$

$$D=1 \text { multiplicity 3 } \implies y_1(x)=c_1e^x+c_2xe^x+c_3x^2e^x$$

2)$(D-2)$

$$D=2 \implies y_2(x)=c_4e^{2x}$$

3)$ D^{2}-6D+10$ $$\Delta=-4 \implies D={3\pm 2i} \implies y_3(x)=c_5e^{3x}\cos(2x)+c_6e^{3x}\sin(2x)$$ Therefore the general solution of the equation is :

$$y(x)=y_1+y_2+y_3$$ $$\boxed{y(x)=c_1e^x+c_2xe^x+c_3x^2e^x+c_4e^{2x}+c_5e^{3x}\cos(2x)+c_6e^{3x}\sin(2x)}$$

Answer 2

HINT

Note the roots of the quadratic term are $$ \frac{6 \pm \sqrt{6^2-4\cdot 1 \cdot 10}}{2\cdot 1} = \frac{6 \pm \sqrt{-4}}{2} = 3 \pm 2i, $$ where $i = \sqrt{-1}$.

So the general solution for the first term has root of $1$ with multiplicity $3$, hence contributes $$e^x + xe^x + x^2e^x.$$

Can you finish the rest?