What is the solution for y^{''}+4y^'+5y=e^{-2x} ?

The solution for y^{''}+4y^'+5y=e^{-2x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+e^{-2x}

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y^{''}+4y^'+5y=e^{-2x}

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- 2 x}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = e^{- 2 x}

Solve linear ODE:

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

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

y^{^{''}} + 4 y = sin (x), y^{^{'}} (π) = 0, y (0) = 0

y^{^{'''}} - 19 y^{^{'}} + 30 y = x^{2} e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{x} )}{1 + x ^{2}}

y^{^{''}} - y^{^{'}} + 144 y = 12 sin (12 t)

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

What is the solution for y^{''}+4y^'+5y=e^{-2x} ?
The solution for y^{''}+4y^'+5y=e^{-2x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+e^{-2x}