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}

csc (x + 5)

\frac{d y}{d x} - (\frac{1 + x}{x}) y = \frac{y ^{2}}{x}

\int_{1}^{3} 4 d x

f (x) = sin (2 x^{3})

\int_{1}^{2} x x^{2} - 1 d x

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}