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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 36 x = 10 sin (4 t), x (0) = 0

y^{^{''}} - 3 y^{^{'}} = 3 e^{3 t}

y^{^{''}} + 16 y = 10 cos (4 x) + 15 sin (4 x)

y^{^{''}} - 10 y^{^{'}} + 24 y = 48 x^{2} + 10

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

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