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

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

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{5} e^{- t}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{5} e^{- t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{5} e^{- t}

7 (4 e^{x} - 27)^{5}

\int_{1}^{5} 1 + 2^{2} d x

x \to \infty lim (\frac{1}{9})

7^{^{'}}

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

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