The solution curve of (1+e^-x)(1+y^2)dydx=y^2, through (0,1), is:

Question

The solution curve of $(1+e^{-x})(1+y^2)\dfrac{dy}{dx}=y^2$, through $(0,1)$, is:

Accepted Answer

$$\int\frac{y^2+1}{y^2}\,dy = \int\frac{e^x}{e^x+1}\,dx \implies y-\frac1y = \log_e(e^x+1)+c.$$

Using $(0,1)$: $c=-\log_e2$. So $y^2-1=y\log_e\dfrac{e^x+1}{2} \implies y^2 = 1+y\log_e\left(\dfrac{e^x+1}{2}\right)$.

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