What is the solution for (d^4y)/(dx^4)+6(dy)/(dx)=-5x^2 ?

The solution for (d^4y)/(dx^4)+6(dy)/(dx)=-5x^2 is

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(d^4y)/(dx^4)+6(dy)/(dx)=-5x^2

\frac{d ^{4} y}{d x ^{4}} + 6 \frac{d y}{d x} = - 5 x^{2}

y = c_{1} + c_{2} e^{- 36 x} + e^{3 \frac{3}{4} x} (c_{3} cos (\frac{6 243 x}{4 ^{\frac{1}{3}}}) + c_{4} sin (\frac{6 243 x}{4 ^{\frac{1}{3}}})) - \frac{5 x ^{3}}{18}

Solution steps

\frac{d ^{4} y}{d x ^{4}} + 6 \frac{d y}{d x} = - 5 x^{2}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 36 x} + e^{3 \frac{3}{4} x} (c_{3} cos (\frac{6 243 x}{4 ^{\frac{1}{3}}}) + c_{4} sin (\frac{6 243 x}{4 ^{\frac{1}{3}}})) - \frac{5 x ^{3}}{18}

y = c_{1} + c_{2} e^{- 36 x} + e^{3 \frac{3}{4} x} (c_{3} cos (\frac{6 243 x}{4 ^{\frac{1}{3}}}) + c_{4} sin (\frac{6 243 x}{4 ^{\frac{1}{3}}})) - \frac{5 x ^{3}}{18}

y^{^{''}} + y^{^{'}} - 8 y = 9 e^{3 t}

y^{^{''}} + 4 y = 3 e^{3 t}

y^{^{''}} + 2 y^{^{'}} + y = 18 e^{2 x} - 4 sin (x) - x^{2}

3 y^{^{''}} - 2 y - y = 2 t^{2} - t - 3 e^{- 3 t}

y^{^{''}} - 2 y^{^{'}} - 15 y = (6 t + 3 tt + 1) e^{t}

What is the solution for (d^4y)/(dx^4)+6(dy)/(dx)=-5x^2 ?
The solution for (d^4y)/(dx^4)+6(dy)/(dx)=-5x^2 is