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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 6 y = 12 e^{3 t} + 12 e^{- 2 t}

y^{^{''}} + 3 y^{^{'}} = 2 x^{4}

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

y^{^{''}} + 10 y^{^{'}} + 16 y = 2581 cos (5 t)

y^{^{''}} = k y - c

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