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

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

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

Solution steps

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

y = \frac{400}{x} + 7 + \frac{1}{5} x

x \to 4 lim (2 x - 6)

\int_{5}^{\infty} e^{- \frac{x}{2}} d x

\frac{d}{d x} (\frac{2}{3} x^{3} - 4 x^{2} + 10 x + 6)

\frac{d y}{d x} = \frac{y}{x}, y (1) = 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}