What is the solution for y^{'''}+9y^'=27x^2 ?

The solution for y^{'''}+9y^'=27x^2 is y=c_{1}+c_{2}cos(3x)+c_{3}sin(3x)+x^3-(2x)/3

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y^{'''}+9y^'=27x^2

y^{^{'''}} + 9 y^{^{'}} = 27 x^{2}

y = c_{1} + c_{2} cos (3 x) + c_{3} sin (3 x) + x^{3} - \frac{2 x}{3}

Solution steps

y^{'''} (x) + 9 y^{'} (x) = 27 x^{2}

Solve linear ODE:

y = c_{1} + c_{2} cos (3 x) + c_{3} sin (3 x) + x^{3} - \frac{2 x}{3}

y = c_{1} + c_{2} cos (3 x) + c_{3} sin (3 x) + x^{3} - \frac{2 x}{3}

4 y^{^{''}} - y = e^{\frac{x}{2}}

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 5 e^{- 3 x}

(D^{2} - 5 D + 6) y = 3 e^{3 x} + e^{2 x}

y^{^{''}} - y^{^{'}} - 2 y = 3 cos (2 x)

6 y^{^{'''}} + 4 y^{^{'}} = e^{x} cos (2 x)

What is the solution for y^{'''}+9y^'=27x^2 ?
The solution for y^{'''}+9y^'=27x^2 is y=c_{1}+c_{2}cos(3x)+c_{3}sin(3x)+x^3-(2x)/3