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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{4} - 2 d^{3} + d^{2}) y = x^{3}

d^{2} (d^{2} + 4) y = 96 x^{2} + sin^{2} (x) - k

(y^{^{''}} - 4 y^{^{'}} + 3 y) = 2 x + 5

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

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3} x}{x}

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