What is the solution for y^{''}+2y^'+y=8e^{-t} ?

The solution for y^{''}+2y^'+y=8e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+4t^2e^{-t}

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y^{''}+2y^'+y=8e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 8 e^{- t}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 8 e^{- t}

Solve linear ODE:

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

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

\frac{d y}{d x} = - \frac{2 x y + y ^{2} + 1}{x ^{2} + 2 x y}

k = 1 \sum \infty \frac{2 ^{k}}{k !}

\frac{d}{d x} (x^{2} e^{y})

\int s 4^{s} d s

What is the solution for y^{''}+2y^'+y=8e^{-t} ?
The solution for y^{''}+2y^'+y=8e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+4t^2e^{-t}