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 = \frac{2}{sin ( 2 x )}

x^{^{''}} + 3 x^{^{'}} + 2 x = 5 cos (t) + sin (2 t)

3 x^{^{''}} + 53 x = 5 cos (4 t)

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

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{a 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