What is the solution for (d^4+d^2+16)y=16x^2+256 ?

The solution for (d^4+d^2+16)y=16x^2+256 is

(d^4+d^2+16)y=16x^2+256

(d^{4} + d^{2} + 16) y = 16 x^{2} + 256

y = e^{\frac{7 x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + e^{- \frac{7 x}{2}} (c_{3} cos (\frac{3 x}{2}) + c_{4} sin (\frac{3 x}{2})) + x^{2} + \frac{127}{8}

Solution steps

(d^{4} + d^{2} + 16) y = 16 x^{2} + 256

Solve linear ODE:

y = e^{\frac{7 x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + e^{- \frac{7 x}{2}} (c_{3} cos (\frac{3 x}{2}) + c_{4} sin (\frac{3 x}{2})) + x^{2} + \frac{127}{8}

y = e^{\frac{7 x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + e^{- \frac{7 x}{2}} (c_{3} cos (\frac{3 x}{2}) + c_{4} sin (\frac{3 x}{2})) + x^{2} + \frac{127}{8}

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{- 5} x

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{- 2} x

(d^{2} + 3 d + 2) y = e^{e^{x}}

(d^{2} - 6 d + 9) y = 2 e^{3} x

What is the solution for (d^4+d^2+16)y=16x^2+256 ?
The solution for (d^4+d^2+16)y=16x^2+256 is