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

The solution for y^{''}+2y^'+5y=e^{-5x} is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+1/20 e^{-5x}

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

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

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{20} e^{- 5 x}

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{20} e^{- 5 x}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{20} e^{- 5 x}

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

y^{^{''}} - 6 y^{^{'}} + 13 y = 4.29 e^{3 t} \cdot cos (4.29 t)

5 y^{^{''}} - 10 y^{^{'}} + 10 y = e^{x} \cdot sec (x)

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

y^{^{''}} - 2 y = 18 (cot (t))^{3}

What is the solution for y^{''}+2y^'+5y=e^{-5x} ?
The solution for y^{''}+2y^'+5y=e^{-5x} is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+1/20 e^{-5x}