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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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