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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 12 y = - sin (2 x) + 3 cos (2 x)

y^{^{''}} + 2 y^{^{'}} + y = e^{- t}, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 e^{- 2 t}

x^{^{''}} + 25 x = 2 cos (5 t)

x^{^{''}} + 9 x = sin (3 t)

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