What is the solution for y^{''}+9y=t^2+cos(3t) ?

The solution for y^{''}+9y=t^2+cos(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(t^2)/9-2/81+(tsin(3t))/6

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y^{''}+9y=t^2+cos(3t)

y^{^{''}} + 9 y = t^{2} + cos (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t ^{2}}{9} - \frac{2}{81} + \frac{t sin ( 3 t )}{6}

Solution steps

y^{''} (t) + 9 y = t^{2} + cos (3 t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t ^{2}}{9} - \frac{2}{81} + \frac{t sin ( 3 t )}{6}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t ^{2}}{9} - \frac{2}{81} + \frac{t sin ( 3 t )}{6}

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f (x) = x^{3} + 4 e^{- 0.5 x}

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What is the solution for y^{''}+9y=t^2+cos(3t) ?
The solution for y^{''}+9y=t^2+cos(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(t^2)/9-2/81+(tsin(3t))/6