What is the solution for y^{''}+4y^'+8y=4e^{-2x} ?

The solution for y^{''}+4y^'+8y=4e^{-2x} is y=e^{-2x}(c_{1}cos(2x)+c_{2}sin(2x))+e^{-2x}

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y^{''}+4y^'+8y=4e^{-2x}

y^{^{''}} + 4 y^{^{'}} + 8 y = 4 e^{- 2 x}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 8 y = 4 e^{- 2 x}

Solve linear ODE:

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

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

y^{^{''}} - 4 y = 10 e^{- 2 x}

y^{^{''}} + 2 y^{^{'}} + 10 y = x, y (0) = 0, y^{^{'}} (0) = - 1

y^{^{''}} + 17 y = e^{6 x}

y^{^{''}} - y = sin (t), y (0) = 1, y (0) = - 1

y^{^{''}} + y = 6 e^{- 2 x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'+8y=4e^{-2x} ?
The solution for y^{''}+4y^'+8y=4e^{-2x} is y=e^{-2x}(c_{1}cos(2x)+c_{2}sin(2x))+e^{-2x}