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 :

Question

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 :

$\frac{1+3 e^{2}}{2\left(3 e^{2}-1\right)}$

$\frac{3 e^{2}}{2\left(3 e^{2}-1\right)}$

$\frac{1-3 e^{2}}{2\left(3 e^{2}+1\right)}$

$\frac{3 e^{2}}{2\left(3 e^{2}+1\right)}$

Solution

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