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

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

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

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

y = c_{1} e^{x} + c_{2} + e^{x} x^{2} - 2 (e^{x} x - e^{x})

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + e^{x} x^{2} - 2 (e^{x} x - e^{x})

y = c_{1} e^{x} + c_{2} + e^{x} x^{2} - 2 (e^{x} x - e^{x})

y^{^{'''}} - 6 y^{^{''}} = 2 - cos (x)

y^{^{''}} + 7 y^{^{'}} - 12 y = e^{3 x} - 2 e^{x} - 5 x

y^{^{''}} - 6 y^{^{'}} + 9 y = t

\frac{d ^{2} y}{d x ^{2}} + 4 y = e^{4 x} sin (3 x)

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

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