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

The solution for y^{''}+y^'-2y=xe^x is y=c_{1}e^x+c_{2}e^{-2x}+(e^xx^2)/6-(e^xx)/9

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{x} x ^{2}}{6} - \frac{e ^{x} x}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{x} x ^{2}}{6} - \frac{e ^{x} x}{9}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{x} x ^{2}}{6} - \frac{e ^{x} x}{9}

y^{^{''}} + 11 y^{^{'}} + 28 y = - 2 e^{2 x} + 1

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

4 y^{^{''}} + 9 y = 15

y^{^{''}} - 3 y^{^{'}} + 2 y = sin (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + y = sec^{3} (x)

What is the solution for y^{''}+y^'-2y=xe^x ?
The solution for y^{''}+y^'-2y=xe^x is y=c_{1}e^x+c_{2}e^{-2x}+(e^xx^2)/6-(e^xx)/9