What is the solution for x^{''}+3x^'+2x=sin(3t) ?

The solution for x^{''}+3x^'+2x=sin(3t) is x=c_{1}e^{-t}+c_{2}e^{-2t}-7/130 sin(3t)-9/130 cos(3t)

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x^{''}+3x^'+2x=sin(3t)

x^{^{''}} + 3 x^{^{'}} + 2 x = sin (3 t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} - \frac{7}{130} sin (3 t) - \frac{9}{130} cos (3 t)

Solution steps

x^{''} (t) + 3 x^{'} (t) + 2 x = sin (3 t)

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 2 t} - \frac{7}{130} sin (3 t) - \frac{9}{130} cos (3 t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} - \frac{7}{130} sin (3 t) - \frac{9}{130} cos (3 t)

What is the solution for x^{''}+3x^'+2x=sin(3t) ?
The solution for x^{''}+3x^'+2x=sin(3t) is x=c_{1}e^{-t}+c_{2}e^{-2t}-7/130 sin(3t)-9/130 cos(3t)