What is the solution for y^{''}+4y+5y=e^{-t} ?

The solution for y^{''}+4y+5y=e^{-t} is y=c_{1}cos(3t)+c_{2}sin(3t)+1/10 e^{-t}

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y^{''}+4y+5y=e^{-t}

y^{^{''}} + 4 y + 5 y = e^{- t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{10} e^{- t}

Solution steps

y^{''} (t) + 4 y + 5 y = e^{- t}

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{10} e^{- t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{10} e^{- t}

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} \cdot e^{3 t}

y^{^{''}} - 36 y = 9 e^{6 x}

\frac{d ^{2} y}{d x ^{2}} + 4 y = 8

y^{^{''}} - 4 y^{^{'}} + 3 y = 4 cos (x) - 3 sin (x)

y^{^{'''}} - 3 y^{^{''}} - y^{^{'}} + 3 y = 3

What is the solution for y^{''}+4y+5y=e^{-t} ?
The solution for y^{''}+4y+5y=e^{-t} is y=c_{1}cos(3t)+c_{2}sin(3t)+1/10 e^{-t}