What is the solution for xy^'+(1+x)y=e^{-x} ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (e^{2} x^{2} + 4 e^{2} x)

\int x + x^{2} + y^{2} d x

\int sinh (3 x) d x

\int \frac{x - 7 x ^{4}}{x} d x

\frac{s + 2}{s ^{2} + 9}

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