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

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

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(e^x+1)(dy)/(dx)=y-ye^x

(e^{x} + 1) \frac{d y}{d x} = y - y e^{x}

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

Solution steps

(e^{x} + 1) \frac{d y}{d 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}

x \to 0 lim (\frac{cot ( x )}{x})

y = 3 - 2 x^{2}, (- 4, - 29)

\int - \frac{1}{x ^{2}} d x

x \to - \infty lim ((x)^{2})

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