Let $y=y(x)$ be a solution curve of the differential equation.

$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$.

If the line $x=1$ intersects the curve $y=y(x)$ at $y=2$ and the line $x=2$ intersects the curve $y=y(x)$ at $y=\alpha$, then a value of $\alpha$ is :

We have, <br/><br/>$$ \begin{aligned} & \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\\\ & d x=\frac{y d x+x d y}{1-(x y)^2} \end{aligned} $$ <br/><br/>On integrating both sides, we get <br/><br/>$$ \begin{aligned} \int d x & =\int \frac{d(x y)}{1-(x y)^2} \\\\ x & =\frac{1}{2} \log \left|\frac{1+x y}{1-x y}\right|+C \end{aligned} $$ <br/><br/>As, $y(1)=2$ <br/><br/>$$ \begin{aligned} 1 & =\frac{1}{2} \log \left|\frac{1+2}{1-2}\right|+C \\\\ \Rightarrow C & =1-\frac{1}{2} \log 3 \end{aligned} $$ <br/><br/>Now, substitute $x=2$ as $y(2)=\alpha$ <br/><br/>$$ 2=\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \log 3 $$ <br/><br/>$$ \begin{aligned} 1+\frac{1}{2} \log 3 & =\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\\\ 2+\log 3 & =\log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \end{aligned} $$ <br/><br/>$$ \begin{aligned} &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha} \mid =3 e^2 \\\\ &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha}= \pm \frac{3 e^2}{1} \\\\ & \Rightarrow\frac{1}{2 \alpha} =\frac{ \pm 3 e^2+1}{ \pm 3 e^2-1} \\\\ &\Rightarrow 2 \alpha=\frac{ \pm 3 e^2-1}{ \pm 3 e^2+1} \end{aligned} $$ <br/><br/>$$ \begin{array}{ll} \Rightarrow \alpha=\frac{1}{2}\left(\frac{ \pm 3 e^2-1}{1 \pm 3 e^2}\right) \\\\ \therefore \alpha=\frac{3 e^2-1}{2\left(1+3 e^2\right)} \text { or } \frac{3 e^2+1}{2\left(3 e^2-1\right)} \end{array} $$