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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 5 y = 2 cos (3 x)

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 8 x e^{x}

y^{^{''}} + 2 y^{^{'}} + y = 4 cos (2 x)

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