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

$\text{Given: } x\,dy - (y^2 - 4y)\,dx = 0,\quad x > 0,\quad y(1) = 2$ $\text{slope of } y = y(x) \text{ is never zero}$ $\text{To find: } 10\,y(\sqrt{2})$ $1.\ \text{separate the variables}$ $2.\ \text{partial fractions, integrate}$ $3.\ \text{use } y(1)=2,\ \text{then } y(\sqrt2)$ $x\,dy = (y^2 - 4y)\,dx \Rightarrow \dfrac{dy}{y^2 - 4y} = \dfrac{dx}{x}$ $\dfrac{1}{y(y-4)} = \dfrac14\left(\dfrac{1}{y-4} - \dfrac{1}{y}\right)$ $\dfrac14\big(\ln|y-4| - \ln|y|\big) = \ln x + C$ $\ln\left|\dfrac{y-4}{y}\right| = 4\ln x + C'$ $\dfrac{y-4}{y} = C\,x^4$ $x=1,\ y=2:\quad \dfrac{2-4}{2} = C \Rightarrow C = -1$ $y - 4 = -x^4 y \Rightarrow y = \dfrac{4}{1 + x^4}$ $y' = -\dfrac{16x^3}{(1+x^4)^2} < 0 \Rightarrow \text{slope never zero}$ $y(\sqrt2) = \dfrac{4}{1 + (\sqrt2)^4} = \dfrac{4}{1+4} = \dfrac{4}{5}$ $10\,y(\sqrt2) = 10 \times \dfrac{4}{5} = 8$ $\boxed{8}$