If $y\left(x\right)$ is the solution of the differential equation $xdy-\left(y^2-4y\right)dx=0$ for $x>0,y\left(1\right)=2,$ and the slope of the curve $y=y\left(x\right)$ is never zero, then the positive value of $15.y\left(\sqrt{2}\right)$ is _________

Given differential equation $xdy-(y^2-4y)dx=0$

$\frac{dy}{y^2-4y}=\frac{dx}{x}$

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

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

$\Rightarrow \text{Put\hspace{0.17em}\hspace{0.17em}}x=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}}y=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}we get }c=0$

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

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

When $y=\frac{4}{1+x^4}\Rightarrow y\left(\sqrt{2}\right)=\frac{4}{5}$

So, $15y\left(\sqrt{2}\right)=15\times \frac{4}{5}=12$