What is the solution for y^{''}+9y=e^{-2t} ?

The solution for y^{''}+9y=e^{-2t} is y=c_{1}cos(3t)+c_{2}sin(3t)+1/13 e^{-2t}

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y^{''}+9y=e^{-2t}

y^{^{''}} + 9 y = e^{- 2 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{13} e^{- 2 t}

Solution steps

y^{''} (t) + 9 y = e^{- 2 t}

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{13} e^{- 2 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{13} e^{- 2 t}

y^{^{''}} + 6 y^{^{'}} + 73 y = 80 e^{x} cos (4 x)

y^{^{''}} + 2 y^{^{'}} + y = 15 e^{- x} x + 1

y^{^{''}} + 4 y = 13 sec^{2} (2 t)

y^{^{''}} + y^{^{'}} = 3 + e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = t, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+9y=e^{-2t} ?
The solution for y^{''}+9y=e^{-2t} is y=c_{1}cos(3t)+c_{2}sin(3t)+1/13 e^{-2t}