$ \text{ The Auxiliary equation is } \\ $

$ D^2 - 1 = 0 \\ $

$D = \pm1 \\ $

$ \text{ C.F. is } y_c= c_1e^x + c_2e^{-x}\\ $

$ \text{P.I = }y_p = \frac{1}{D^2 - 1}x^2 sin3x\\ $

$=\text {Imaginary part of }\frac{1}{(D^2-1)}x^2e^{3ix} \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{(D+ 3i)^2 - 1}x^2 \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{(D^2 + 6iD - 9 -1) }x^2 \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{(D^2 + 6iD -10)}x^2 \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{ -10[1- \frac{D^2 + 6iD}{10}]}x^2 \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{-10} [1- (\frac{D^2 + 6iD}{10})]^{-1}x^2 \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{-10} [1+ (\frac{D^2 + 6iD}{10}) + \frac{(D^2 + 6iD)^2}{100}]x^2 \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{-10} [(x^2+\frac{2}{10} + 2x(\frac{6i}{10}) - (\frac{36(2)}{10})] \\ $

$=\text {Imaginary part of } e^{3ix} \frac{1}{-10} [(x^2+\frac{6}{5}xi - (\frac{13}{25})] \\ $

$=\text {Imaginary part of } (cos3x + isin3x) \frac{1}{-10} [(x^2+\frac{6}{5}xi - (\frac{13}{25})] \\ $

$ = \frac{1}{-10} [(\frac{6}{5}x cos3x + (x^2-\frac{13}{25}) sin3x)] \\ $

$\therefore \text{ The complete solution is y = C.F. + P.I. } \\ $

$ \therefore y = c_1e^x + c_2e^{-x}+ \frac{1}{-10} [(\frac{6}{5}x cos3x + (x^2-\frac{13}{25}) sin3x)] \\ $

How to computer coding