What is the solution for y^{''}+5y^'+6y=e^{-x} ?

The solution for y^{''}+5y^'+6y=e^{-x} is y=c_{1}e^{-2x}+c_{2}e^{-3x}+1/2 e^{-x}

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y^{''}+5y^'+6y=e^{-x}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- x}

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{2} e^{- x}

Solution steps

y^{''} (x) + 5 y^{'} (x) + 6 y = e^{- x}

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{2} e^{- x}

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{2} e^{- x}

(4 d^{2} + 12 d + 9) y = 2022

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

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

y^{^{''}} - 2 y^{^{'}} + 10 = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 5, y^{^{'}} (0) = 3

What is the solution for y^{''}+5y^'+6y=e^{-x} ?
The solution for y^{''}+5y^'+6y=e^{-x} is y=c_{1}e^{-2x}+c_{2}e^{-3x}+1/2 e^{-x}