What is the solution for y^{''}+5y^'+4y=e^{-t} ?

The solution for y^{''}+5y^'+4y=e^{-t} is y=c_{1}e^{-t}+c_{2}e^{-4t}+(e^{-t}t)/3

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y^{''}+5y^'+4y=e^{-t}

y^{^{''}} + 5 y^{^{'}} + 4 y = e^{- t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{- t} t}{3}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{- t} t}{3}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{- t} t}{3}

y^{^{''}} + 49 y = 11 sec^{2} (7 t)

y^{^{''}} - 10 y^{^{'}} + 25 y = 3 x + 5

y^{^{''}} + y^{^{'}} = 2 x

y^{^{''}} + 2 y^{^{'}} + 5 y = 16 e^{- t} cos (2 t)

y^{^{''}} + y^{^{'}} = x^{3} - 1

What is the solution for y^{''}+5y^'+4y=e^{-t} ?
The solution for y^{''}+5y^'+4y=e^{-t} is y=c_{1}e^{-t}+c_{2}e^{-4t}+(e^{-t}t)/3