The solution of the differential equation $\left(1+e^{-x}\right)\left(1+y^2\right) \frac{d y}{d x}=y^2$…
The solution of the differential equation $\left(1+e^{-x}\right)\left(1+y^2\right) \frac{d y}{d x}=y^2$ which passes through the point $(0,1)$ is
- $y^2+1=y\left(\log \left(\frac{1+e^x}{2}\right)+2\right)$
- $y^2+1=y\left(\log \left(\left(\frac{1+e^{-x}}{2}\right)+2\right)\right)$
- $y^2=1+y \log \left(\frac{1+e^{-x}}{2}\right)$
- $y^2=1+y \log \left(\frac{1+e^x}{2}\right)$
Solution
$\begin{aligned} & \left(1+e^{-x}\right)\left(1+y^2\right) \frac{d y}{d x}=y^2 \\ & \Rightarrow \int \frac{1+y^2}{y^2} d y=\int \frac{e^x}{1+e^x} d x \\ & \Rightarrow-\frac{1}{y}+y=\log \left(1+e^x\right)+\log C \\ & \Rightarrow-1+y^2=y \log c\left(1+e^x\right)\end{aligned}$
Putting $x=0$ and $y=1$ we get $c=\frac{1}{2}$
$\begin{aligned} & \Rightarrow-1+y^2=y \log \left(\frac{1+e^x}{2}\right) \\ & \Rightarrow y^2=1+y \log \left(\frac{1+e^x}{2}\right)\end{aligned}$
Asked in: MHT CET 2022 (06 Aug Shift 1)
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