What is the solution for y^{''}+9y=6sin(3t) ?

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

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y^{''}+9y=6sin(3t)

y^{^{''}} + 9 y = 6 sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) - t cos (3 t)

Solution steps

y^{''} (t) + 9 y = 6 sin (3 t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) - t cos (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) - t cos (3 t)

2 sec (x)

\frac{d}{d x} (4^{2 - 2 x})

\int_{0}^{1} 2 x (4 - x^{2}) d x

\frac{\partial}{\partial x} (10 x y e^{- (x^{2} + y^{2})})

\int_{- 1}^{1} \frac{x}{( x ^{2} + 1 ) ^{2}} d x

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