What is the solution for y^{''}+2y^'+y=4e^{-1} ?

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

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y^{''}+2y^'+y=4e^{-1}

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- 1}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 4 e^{- 1}

Solve linear ODE:

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

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

\frac{d}{d t} (tan (t + \frac{π}{4}))

\frac{d y}{d x} = y - y^{5}

\frac{d ^{2} y}{d x ^{2}} - 2 \cdot \frac{d y}{d x} + a y = 0, a > 1

\frac{x ^{5} + 1}{x}

\int_{0}^{1} \frac{8 y}{e ^{2 y}} d y

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