If $y\left(x\right)$ is the solution of the differential equation : $xdy-(y^2-4y)dx=0$ for $x>0$, $y\left(1\right)=2$, and the slope of the curve $y=y\left(x\right)$ is never zero, then the value of $10y\left(\sqrt{2}\right)$ is _____________ .

The correct answer is 8.

Let us solve the given differential equation step-by-step:
$xdy-(y^2-4y)dx=0$

Rearranging the terms to separate the variables $x$ and $y$:
$\frac{dy}{y^2-4y}=\frac{dx}{x}$

We can rewrite the left side by splitting it using partial fractions:
$\frac{1}{y(y-4)}=\frac{1}{4}\left(\frac{1}{y-4}-\frac{1}{y}\right)$

Substituting this back into our differential equation gives:
$\frac{1}{4}\left(\frac{1}{y-4}-\frac{1}{y}\right)dy=\frac{dx}{x}$

Multiplying both sides by 4:
$\left(\frac{1}{y-4}-\frac{1}{y}\right)dy=4\frac{dx}{x}$

Now, integrate both sides:
$\int \left(\frac{1}{y-4}-\frac{1}{y}\right)dy=4\int \frac{dx}{x}$

$\ln |y-4|-\ln \left|y\right|=4\ln \left|x\right|+\ln \left|C\right|$

Using logarithmic properties:
$\ln \left|\frac{y-4}{y}\right|=\ln \left|C{x}^4\right|$

Taking exponential on both sides:
$\frac{y-4}{y}=C{x}^4$

We are given the initial condition $y\left(1\right)=2$. Substituting $x=1$ and $y=2$:
$\frac{2-4}{2}=C{\left(1\right)}^4$

$\frac{-2}{2}=C\Rightarrow C=-1$

Now, substitute $C=-1$ back into the relation:
$\frac{y-4}{y}=-x^4$

$1-\frac{4}{y}=-x^4$

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

$y\left(x\right)=\frac{4}{1+x^4}$

Now, we need to evaluate $y\left(\sqrt{2}\right)$:
$y\left(\sqrt{2}\right)=\frac{4}{1+{\left(\sqrt{2}\right)}^4}=\frac{4}{1+4}=\frac{4}{5}=0.8$

Finally, we calculate the required value of $10y\left(\sqrt{2}\right)$:
$10y\left(\sqrt{2}\right)=10\times \frac{4}{5}=8$