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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 9 y = sin (3 x) + e^{- x}

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 x)

(D^{2} - 2 D + 1) y = x^{2} e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 x + 3

(d^{2} - 1) y = x sin (x) + (1 + x^{2}) e^{x}

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