What is the solution for (d^2x)/(dt^2)+9x=-sin(3t) ?

The solution for (d^2x)/(dt^2)+9x=-sin(3t) is x=c_{1}cos(3t)+c_{2}sin(3t)+(tcos(3t))/6

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(d^2x)/(dt^2)+9x=-sin(3t)

\frac{d ^{2} x}{d t ^{2}} + 9 x = - sin (3 t)

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

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 9 x = - sin (3 t)

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (ln (\frac{2 x y}{z}))

\int \frac{5 x}{3 x ^{\frac{1}{3}}} d x

y = 1.1 x + e^{x}, (0, 1)

y = 0, y = x^{2}, y = 4 x - x^{2}, [0, 2]

What is the solution for (d^2x)/(dt^2)+9x=-sin(3t) ?
The solution for (d^2x)/(dt^2)+9x=-sin(3t) is x=c_{1}cos(3t)+c_{2}sin(3t)+(tcos(3t))/6