What is the solution for y^{''}+y^'-2y=3xe^x ?

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

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y^{''}+y^'-2y=3xe^x

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

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

Solution steps

y^{''} (x) + y^{'} (x) - 2 y = 3 x e^{x}

Solve linear ODE:

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

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

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