What is the solution for y^{''}+16y=e^{-t} ?

The solution for y^{''}+16y=e^{-t} is y=c_{1}cos(4t)+c_{2}sin(4t)+1/17 e^{-t}

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y^{''}+16y=e^{-t}

y^{^{''}} + 16 y = e^{- t}

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{17} e^{- t}

Solution steps

y^{''} (t) + 16 y = e^{- t}

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{17} e^{- t}

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{17} e^{- t}

\frac{\partial}{\partial x} (e^{- x} cos (x))

9 y ln (x) - x y^{^{'}} = 0

36 y^{^{''}} + 84 y^{^{'}} + 49 y = 0

(x sin (3 x))^{^{'}}

\frac{d}{d x} (\frac{tan ( x )}{cos ( x ) - 4 x})

What is the solution for y^{''}+16y=e^{-t} ?
The solution for y^{''}+16y=e^{-t} is y=c_{1}cos(4t)+c_{2}sin(4t)+1/17 e^{-t}