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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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