If y x is the solution of the differential equation x d y - y 2 - 4 y d x = 0 for x > 0 ,   y 1 = 2…

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

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

Integrating both sides, we get, 

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

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

i.e. $c=0$

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

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

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