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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 13 (t) = 0

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

y^{^{''}} + 9 y = 12 cos (3 t)

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

y^{^{'''}} + 3 y^{^{'}} = sin (x) + 1

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