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

The solution for y^{''}+y^'-2y=(e^x+e^{-x})/2 is y=c_{1}e^x+c_{2}e^{-2x}+(e^xx)/6-1/4 e^{-x}

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{x} x}{6} - \frac{1}{4} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{x} x}{6} - \frac{1}{4} e^{- x}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{x} x}{6} - \frac{1}{4} e^{- x}

y^{^{''}} + y = 4 sin (3 x)

y^{^{''}} - 5 y^{^{'}} = 25 (H (t - 2) - H (t - 4))

x^{^{''}} - 3 x^{^{'}} - 15.25 x = e^{t}

(D^{2} + 3 D + 2) y = sin (e^{x})

y^{^{'''}} - 2 y^{^{'}} + 4 y = e^{x} cos (x) + x^{2}

What is the solution for y^{''}+y^'-2y=(e^x+e^{-x})/2 ?
The solution for y^{''}+y^'-2y=(e^x+e^{-x})/2 is y=c_{1}e^x+c_{2}e^{-2x}+(e^xx)/6-1/4 e^{-x}