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

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

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

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

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + e^{- t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + e^{- t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + e^{- t}

3 y^{^{'}} + 2 y^{^{''}} = - 10 cos (3.14 \cdot \frac{t}{0.2})

y^{^{''}} + 36 y = sin (4 x)

y^{^{''}} + 3 y = - 48 x^{2} \cdot e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 14 y = 7 + 4 x

y^{^{''}} - 3 y + 2 y = sin (e^{- x})

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