What is the solution for x^{''}+16x^'+80x=e^{-8t} ?

The solution for x^{''}+16x^'+80x=e^{-8t} is x=e^{-8t}(c_{1}cos(4t)+c_{2}sin(4t))+1/16 e^{-8t}

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x^{''}+16x^'+80x=e^{-8t}

x^{^{''}} + 16 x^{^{'}} + 80 x = e^{- 8 t}

x = e^{- 8 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{16} e^{- 8 t}

Solution steps

x^{''} (t) + 16 x^{'} (t) + 80 x = e^{- 8 t}

Solve linear ODE:

x = e^{- 8 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{16} e^{- 8 t}

x = e^{- 8 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{16} e^{- 8 t}

y^{^{''}} + 2 y^{^{'}} - y = t^{- 1} \cdot e^{t}

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 x + e^{- x}

x^{^{''}} - x^{^{'}} - 12 x = e^{4 t}, x (0) = 1, x^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} - 40 y = - 2 e^{- 3 x} + cos (2 x)

y^{^{''}} - 2 y + y = x

What is the solution for x^{''}+16x^'+80x=e^{-8t} ?
The solution for x^{''}+16x^'+80x=e^{-8t} is x=e^{-8t}(c_{1}cos(4t)+c_{2}sin(4t))+1/16 e^{-8t}