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

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

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(1+e^x)y^'=y-ye^x

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{x}{y ^{2}}, y (0) = 2

\frac{d y}{d t} - t y = y, y (1) = 3

y^{^{'}} = y^{- \frac{1}{3}}

\frac{d y}{d x} = \frac{5 e ^{x}}{e ^{y}}

y^{^{'}} = \frac{x}{y + 1}, y (2) = - 2

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