What is the solution for y^{''}-4y^'-12y=sin(3t) ?

The solution for y^{''}-4y^'-12y=sin(3t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-7/195 sin(3t)+4/195 cos(3t)

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y^{''}-4y^'-12y=sin(3t)

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (3 t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{7}{195} sin (3 t) + \frac{4}{195} cos (3 t)

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = sin (3 t)

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{7}{195} sin (3 t) + \frac{4}{195} cos (3 t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{7}{195} sin (3 t) + \frac{4}{195} cos (3 t)

What is the solution for y^{''}-4y^'-12y=sin(3t) ?
The solution for y^{''}-4y^'-12y=sin(3t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-7/195 sin(3t)+4/195 cos(3t)