Let y = y (x) be a solution curve of the differential equation, (1...

Given,

$\left(1 - {x}^{2}{y}^{2}\right)dx = ydx + xdy$

$⇒ dx = \dfrac{ydx + xdy}{1 - {\left(xy\right)}^{2}}$

$⇒ dx = \dfrac{d\left(yx\right)}{1 - {\left(xy\right)}^{2}}$

$⇒ dx = \dfrac{1}{2}\left(\dfrac{d\left(xy\right)}{1 - xy} + \dfrac{d\left(xy\right)}{1 + xy}\right)$

Now integrating both side we get,

$⇒ 2x + c = \ln \left|\dfrac{1 + xy}{1 - xy}\right|$

$⇒ \left|\dfrac{xy + 1}{xy - 1}\right| = {e}^{c}{e}^{2x}$

Now given, $y\left(1\right) = 2$ so putting the value in above equation we get,

$3 = {e}^{c}{e}^{2}$

$⇒ {e}^{c} = \dfrac{3}{{e}^{2}}$

Hence, the equation becomes $\left|\dfrac{xy + 1}{xy - 1}\right| = 3{e}^{2x - 2}$

Now finding $y\left(2\right)$ so putting $x = 2$ in above equation we get,

\left|\dfrac{2y + 1}{2y - 1}\right| = 3{e}^{2}$</p><p>$⇒ 2y + 1 = 2 \cdot 3{e}^{2}y - 3{e}^{2}$</p><p>$⇒ 1 + 3{e}^{2} = 2y\left(3{e}^{2} - 1\right)$</p><p>$⇒ y\left(2\right) = \dfrac{1 + 3{e}^{2}}{2\left(3{e}^{2} - 1\right)}