What is the solution for y^{''}+3y^'+2y=e^{-2x} ?

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

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y^{''}+3y^'+2y=e^{-2x}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 2 x}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 1 = 0

y^{^{''}} - 4 y^{^{'}} + 13 y = 4 + 3 e^{t}

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

y^{^{''}} + 2 y = sec (x)

y^{^{''}} + 3 y^{^{'}} - 54 y = - 174 t + 108 t^{2}

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