What is the solution for y^{''}-y^'=4xe^x ?

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

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y^{''}-y^'=4xe^x

y^{^{''}} - y^{^{'}} = 4 x e^{x}

y = c_{1} e^{x} + c_{2} + 2 e^{x} x^{2} - 4 (e^{x} x - e^{x})

Solution steps

y^{''} (x) - y^{'} (x) = 4 x e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + 2 e^{x} x^{2} - 4 (e^{x} x - e^{x})

y = c_{1} e^{x} + c_{2} + 2 e^{x} x^{2} - 4 (e^{x} x - e^{x})

d, d (q) = \frac{300}{0.4 q + 8}

y^{^{''}} + 4 y^{^{'}} + 3 y = 1 - H (t - 2) + H (t - 6)

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

y^{^{''}} + 4 y = \frac{1}{sin ( 2 t )}

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{- 2 x}

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