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

The solution for y^{''}+y^'-2y=xe^x+e^x is y=c_{1}e^x+c_{2}e^{-2x}+(3e^xx^2+4e^xx)/(18)

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{3 e ^{x} x ^{2} + 4 e ^{x} x}{18}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{3 e ^{x} x ^{2} + 4 e ^{x} x}{18}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{3 e ^{x} x ^{2} + 4 e ^{x} x}{18}

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{2 x} + 1

y^{^{''}} - y^{^{'}} - 2 y = 5 e^{4 x} + 6 x

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

y^{^{''}} + 4 y - 5 = 0

What is the solution for y^{''}+y^'-2y=xe^x+e^x ?
The solution for y^{''}+y^'-2y=xe^x+e^x is y=c_{1}e^x+c_{2}e^{-2x}+(3e^xx^2+4e^xx)/(18)