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

The solution for y^{''}+2y^'-3y=sin(2t) is y=c_{1}e^t+c_{2}e^{-3t}-7/65 sin(2t)-4/65 cos(2t)

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

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

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{7}{65} sin (2 t) - \frac{4}{65} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{7}{65} sin (2 t) - \frac{4}{65} cos (2 t)

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{7}{65} sin (2 t) - \frac{4}{65} cos (2 t)

What is the solution for y^{''}+2y^'-3y=sin(2t) ?
The solution for y^{''}+2y^'-3y=sin(2t) is y=c_{1}e^t+c_{2}e^{-3t}-7/65 sin(2t)-4/65 cos(2t)