Value of Solution to Differential Equation at Square Root of Two

To find the required value, we start by solving the given differential equation for $x > 0$: $x dy - (y^2 - 4y) dx = 0$

Separating the variables, we get: $\frac{dy}{y^2 - 4y} = \frac{dx}{x}$

Using partial fractions on the left-hand side: $\frac{1}{y(y - 4)} = \frac{1}{4} \left( \frac{1}{y - 4} - \frac{1}{y} \right)$

Substituting this back into the equation, we obtain: $\frac{1}{4} \left( \frac{1}{y - 4} - \frac{1}{y} \right) dy = \frac{dx}{x}$

Integrating both sides with respect to their corresponding variables gives: $\frac{1}{4} \int \left( \frac{1}{y - 4} - \frac{1}{y} \right) dy = \int \frac{dx}{x}$ $\frac{1}{4} \ln \left| \frac{y - 4}{y} \right| = \ln x + C$

Using the initial condition $y(1) = 2$: $\frac{1}{4} \ln \left| \frac{2 - 4}{2} \right| = \ln(1) + C$ $\frac{1}{4} \ln|-1| = 0 + C \implies C = 0$

Thus, the equation simplifies to: $\frac{1}{4} \ln \left| \frac{y - 4}{y} \right| = \ln x$ $\ln \left| \frac{y - 4}{y} \right| = \ln(x^4)$ $\left| \frac{y - 4}{y} \right| = x^4$

Since $\frac{y(1) - 4}{y(1)} = \frac{-2}{2} = -1 < 0$, we remove the absolute value sign with a negative sign: $\frac{y - 4}{y} = -x^4$

Solving for $y(x)$: $1 - \frac{4}{y} = -x^4$ $\frac{4}{y} = 1 + x^4$ $y(x) = \frac{4}{1 + x^4}$

We check the condition on the slope of $y(x)$: $\frac{dy}{dx} = -\frac{16x^3}{(1 + x^4)^2}$ For $x > 0$, $\frac{dy}{dx} < 0$, which guarantees that the slope of the curve is never zero.

Now, substituting $x = \sqrt{2}$ into $y(x)$: $y(\sqrt{2}) = \frac{4}{1 + (\sqrt{2})^4} = \frac{4}{1 + 4} = \frac{4}{5}$

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