What is the solution for y^{''}+6y^'+8y=cos(3t) ?

The solution for y^{''}+6y^'+8y=cos(3t) is y=c_{1}e^{-2t}+c_{2}e^{-4t}+18/325 sin(3t)-1/325 cos(3t)

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y^{''}+6y^'+8y=cos(3t)

y^{^{''}} + 6 y^{^{'}} + 8 y = cos (3 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{18}{325} sin (3 t) - \frac{1}{325} cos (3 t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 8 y = cos (3 t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{18}{325} sin (3 t) - \frac{1}{325} cos (3 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{18}{325} sin (3 t) - \frac{1}{325} cos (3 t)

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What is the solution for y^{''}+6y^'+8y=cos(3t) ?
The solution for y^{''}+6y^'+8y=cos(3t) is y=c_{1}e^{-2t}+c_{2}e^{-4t}+18/325 sin(3t)-1/325 cos(3t)