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

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

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

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

y = c_{1} + c_{2} e^{- 5 t} - \frac{1}{4} e^{- t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 5 t} - \frac{1}{4} e^{- t}

y = c_{1} + c_{2} e^{- 5 t} - \frac{1}{4} e^{- t}

y^{^{'}} + y cos (x) = 2 cos (x)

y = \frac{10}{x ^{2} - 2 x - 24}, x = - 2, x = 2

h^{^{'}} = 6 - 5 cos (\frac{π}{15} t)

\int \frac{1 - sin ( x )}{x + cos ( x )} d x

x \to + (- 1) lim (x)

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