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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 1.44 y = 24 cos (1.2 x)

y^{^{''}} + y = 5 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

6 y^{^{''}} - y^{^{'}} - 2 y = 3 sin (10 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = 5 x - 8

y^{^{''}} + y = 10 sin (2 t) + 7 t cos (2 t)

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