What is the solution for y^{''}+8y^'+16y=2e^{-4t} ?

The solution for y^{''}+8y^'+16y=2e^{-4t} is y=c_{1}e^{-4t}+c_{2}te^{-4t}+e^{-4t}t^2

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y^{''}+8y^'+16y=2e^{-4t}

y^{^{''}} + 8 y^{^{'}} + 16 y = 2 e^{- 4 t}

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 2 e^{- 4 t}

Solve linear ODE:

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

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

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\int_{1}^{4} (5 x^{3} - 6) d x

\frac{d y}{d x} = 5 y e^{x + 8}

\int sec^{3} (7 x) d x

What is the solution for y^{''}+8y^'+16y=2e^{-4t} ?
The solution for y^{''}+8y^'+16y=2e^{-4t} is y=c_{1}e^{-4t}+c_{2}te^{-4t}+e^{-4t}t^2