If y(x) is the solution of the differential equation xdy - ( y ^ 2...

Given differential equation is $\dfrac{dy}{{y}^{2} - 4y} = \dfrac{dx}{x}$

$⇒ \dfrac{1}{4}\left(\dfrac{1}{y - 4} - \dfrac{1}{y}\right)dy = \dfrac{dx}{x}$

Integrating both sides, we get,

$\dfrac{1}{4}\ln \left|\dfrac{y - 4}{y}\right| = \ln \left|x\right| + c$

Given, $y\left(1\right) = 2$

i.e. $c = 0$

So, $\dfrac{y - 4}{y} = \pm {x}^{4}$ or, $y = \dfrac{4}{1 \pm {x}^{4}}$

Considering $y = \dfrac{4}{1 + {x}^{4}}$

So, $10y\left(\sqrt{2}\right) = 10\left(\dfrac{4}{1 + 4}\right) = 8$