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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = 2 x^{3} + 3 x^{2} - 4 x, y = x^{2}

y^{^{''}} + 4 y^{^{'}} + 34 y = 0

f (x) = \frac{2}{4 x + 3}

\int_{1}^{e^{2}} \frac{ln ^{6} ( x ^{2} )}{x} d x

f (x) = \frac{x - 4}{x + 2}

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