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 $$10y(\sqrt{2})$$ is _______.

The given differential equation is

$$x\,dy-(y^2-4y)\,dx=0.$$

Isolate $$dy/dx$$:

$$x\,dy=(y^2-4y)\,dx \;\;\Longrightarrow\;\; \frac{dy}{dx}=\frac{y^2-4y}{x}=\frac{y(y-4)}{x}.$$

This is a separable equation. Bring all $$y$$-terms to the left and the $$x$$-term to the right:

$$\frac{dy}{y(y-4)}=\frac{dx}{x}.$$

Use partial fractions for the left side. Write

$$\frac{1}{y(y-4)}=\frac{A}{y}+\frac{B}{y-4}.$$

Matching numerators gives $$1=A(y-4)+By=(A+B)y-4A.$$
Therefore $$A+B=0,\;-4A=1 \;\Longrightarrow\; A=-\frac14,\;B=\frac14.$$

Hence

$$\frac{1}{y(y-4)}=-\frac14\frac1y+\frac14\frac1{\,y-4\,}.$$

Integrate both sides:

$$\int\left(-\frac14\frac1y+\frac14\frac1{y-4}\right)dy = \int\frac{dx}{x}.$$

$$-\frac14\ln|y|+\frac14\ln|y-4|=\ln|x|+C.$$

Multiply by 4 to simplify:

$$-\ln|y|+\ln|y-4| = 4\ln|x| + C_1.$$

Combine the logarithms:

$$\ln\left|\frac{y-4}{y}\right| = \ln\!\bigl(x^4\bigr)+C_1.$$

Exponentiate:

$$\frac{y-4}{y} = Kx^4,\quad\text{where }K=e^{C_1}.$$

Apply the initial condition $$y(1)=2$$:

$$\frac{2-4}{2}=K\,(1)^4 \;\Longrightarrow\; -1=K.$$

Thus

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

Solve for $$y$$:

$$y-4 = -x^4y \;\Longrightarrow\; y(1+x^4)=4 \;\Longrightarrow\; y(x)=\frac{4}{1+x^4}.$$

The slope $$dy/dx=\dfrac{y(y-4)}{x}$$ is never zero because $$y=0$$ or $$y=4$$ would make it zero; however $$y=\dfrac{4}{1+x^4}$$ satisfies $$0\lt y\lt4$$ for all $$x\gt0.$$/

Evaluate at $$x=\sqrt2$$:

$$y(\sqrt2)=\frac{4}{1+(\sqrt2)^4}=\frac{4}{1+4}=\frac45.$$

Therefore

$$10\,y(\sqrt2)=10\left(\frac45\right)=8.$$

Final Answer: 8