What is the solution for y^{''}+9y=2x^2-5 ?

The solution for y^{''}+9y=2x^2-5 is y=c_{1}cos(3x)+c_{2}sin(3x)+(2x^2)/9-49/81

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y^{''}+9y=2x^2-5

y^{^{''}} + 9 y = 2 x^{2} - 5

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x ^{2}}{9} - \frac{49}{81}

Solution steps

y^{''} (x) + 9 y = 2 x^{2} - 5

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x ^{2}}{9} - \frac{49}{81}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x ^{2}}{9} - \frac{49}{81}

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 4 e^{4 x} ln (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{1}{t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 9 x

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 x} + 50 sin (x)

What is the solution for y^{''}+9y=2x^2-5 ?
The solution for y^{''}+9y=2x^2-5 is y=c_{1}cos(3x)+c_{2}sin(3x)+(2x^2)/9-49/81