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

The solution for y^{''}+9y=tsin(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)-(t^2cos(3t))/(12)+(tsin(3t))/(36)

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

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

y = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{t ^{2} cos ( 3 t )}{12} + \frac{t sin ( 3 t )}{36}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{t ^{2} cos ( 3 t )}{12} + \frac{t sin ( 3 t )}{36}

y = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{t ^{2} cos ( 3 t )}{12} + \frac{t sin ( 3 t )}{36}

\frac{d}{d x} (sec^{11} (x))

n = 0 \sum \infty 6 (0.9)^{n - 1}

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\int_{- 3}^{3} ∣ x ∣ d x

\int \frac{4 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 32} d x

What is the solution for y^{''}+9y=tsin(3t) ?
The solution for y^{''}+9y=tsin(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)-(t^2cos(3t))/(12)+(tsin(3t))/(36)