What is the solution for y^{''}+25y=e^{-t} ?

The solution for y^{''}+25y=e^{-t} is y=c_{1}cos(5t)+c_{2}sin(5t)+1/26 e^{-t}

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y^{''}+25y=e^{-t}

y^{^{''}} + 25 y = e^{- t}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{26} e^{- t}

Solution steps

y^{''} (t) + 25 y = e^{- t}

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{26} e^{- t}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{26} e^{- t}

k = 1 \sum \infty \frac{1}{k ( k + 1 )}

\int 0.02 e^{- 0.02 x} d x

\frac{8 x ^{2} + 2 x + 4}{x}

\frac{1}{4} q^{^{''}} + 10 q^{^{'}} + 1000 q = 0

x \to 0 + lim ((1 + x)^{\frac{7}{x}})

What is the solution for y^{''}+25y=e^{-t} ?
The solution for y^{''}+25y=e^{-t} is y=c_{1}cos(5t)+c_{2}sin(5t)+1/26 e^{-t}