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}

\frac{d}{d x} (8 8 x^{7} - 10)

\frac{d y}{d x} = arctan (- 6 x)

y = (\frac{2 x + 1}{3 x - 1})^{4}

\int e^{- 6 x} d x

n = 1 \sum \infty (x - 3)^{n}

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