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

The solution for x^{''}+9x=sin(t) is x=c_{1}cos(3t)+c_{2}sin(3t)+1/8 sin(t)

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x^{''}+9x=sin(t)

x^{^{''}} + 9 x = sin (t)

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} sin (t)

Solution steps

x^{''} (t) + 9 x = sin (t)

Solve linear ODE:

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} sin (t)

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} sin (t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{(2 t)}

y^{^{''}} + 2 y^{^{'}} + 10 y = 10

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{x}

y^{^{''}} + 25 y^{^{'}} = - 24 e^{- t}

y^{^{''}} + 4 y^{^{'}} + 4 y = sin (x), y (0) = 1, y^{^{'}} (0) = 0

What is the solution for x^{''}+9x=sin(t) ?
The solution for x^{''}+9x=sin(t) is x=c_{1}cos(3t)+c_{2}sin(3t)+1/8 sin(t)