What is the solution for y^{''}+6y^'+18y=-9e^{-3t} ?

The solution for y^{''}+6y^'+18y=-9e^{-3t} is y=e^{-3t}(c_{1}cos(3t)+c_{2}sin(3t))-e^{-3t}

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y^{''}+6y^'+18y=-9e^{-3t}

y^{^{''}} + 6 y^{^{'}} + 18 y = - 9 e^{- 3 t}

y = e^{- 3 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) - e^{- 3 t}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 18 y = - 9 e^{- 3 t}

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) - e^{- 3 t}

y = e^{- 3 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) - e^{- 3 t}

What is the solution for y^{''}+6y^'+18y=-9e^{-3t} ?
The solution for y^{''}+6y^'+18y=-9e^{-3t} is y=e^{-3t}(c_{1}cos(3t)+c_{2}sin(3t))-e^{-3t}