If y = y ( x ) is the solution curve of the differential equation (...

Given,

$\left({x}^{2} - 4\right)dy - \left({y}^{2} - 3y\right)dx = 0$

$⇒ ∫ \dfrac{dy}{{y}^{2} - 3y} = ∫ \dfrac{dx}{{x}^{2} - 4}$

$⇒ \dfrac{1}{3} ∫ \dfrac{y - ( y - 3 )}{y ( y - 3 )}dy = ∫ \dfrac{dx}{{x}^{2} - 4}$

$⇒ \dfrac{1}{3} ∫ \left(\dfrac{1}{y - 3} - \dfrac{1}{y}\right)dy = ∫ \dfrac{dx}{{x}^{2} - {2}^{2}}$

$⇒ \dfrac{1}{3}\left(log | y - 3 | - log | y |\right) = \dfrac{1}{4}\log \left|\dfrac{x - 2}{x + 2}\right| + C$

$⇒ \dfrac{1}{3}\log \left|\dfrac{y - 3}{y}\right| = \dfrac{1}{4}\log \left|\dfrac{x - 2}{x + 2}\right| + C$

At $x = 4, y = \dfrac{3}{2}$

$⇒ \dfrac{1}{3}\log \left|\dfrac{\dfrac{3}{2} - 3}{\dfrac{3}{2}}\right| = \dfrac{1}{4}\log \left|\dfrac{4 - 2}{4 + 2}\right| + C$

$⇒ 0 = \dfrac{1}{4}\log \left|\dfrac{1}{3}\right| + C$

$⇒ 0 = - \log \left|3\right|\dfrac{1}{4} + C$

$⇒ C = \dfrac{1}{4}\log \left(3\right)$

$⇒ \dfrac{1}{3}\log \left|\dfrac{y - 3}{y}\right| = \dfrac{1}{4}\log \left|\dfrac{x - 2}{x + 2}\right| + \dfrac{1}{4}\log ( 3 )$

Putting, $x = 10$

$⇒ \dfrac{1}{3}\log \left|\dfrac{y - 3}{y}\right| = \dfrac{1}{4}\log \left|\dfrac{2}{3}\right| + \dfrac{1}{4}\log ( 3 )$

$⇒ \dfrac{1}{3}\log \left|\dfrac{y - 3}{y}\right| = \dfrac{1}{4}\log \left|2\right|$

$⇒ \log \left|\dfrac{y - 3}{y}\right| = \dfrac{3}{4}\log 2$

$⇒ \log \left|\dfrac{y - 3}{y}\right| = \log \left(2\dfrac{3}{4}\right)$

$⇒ - y + 3 = 8\dfrac{1}{4} \cdot y$

$⇒ 3 = \left(8\dfrac{1}{4} + 1\right)y$

$⇒ \dfrac{3}{1 + \left(8\right)\dfrac{1}{4}} = y$