What is the solution for xy^'+(1-x)y=2xe^x ?

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

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xy^'+(1-x)y=2xe^x

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

y = e^{x} x + \frac{c _{1}}{e ^{- x} x}

Solution steps

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

Solve linear ODE:

y = e^{x} x + \frac{c _{1}}{e ^{- x} x}

y = e^{x} x + \frac{c _{1}}{e ^{- x} x}

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