Greatest Integer Value of Differential Equation Solution

To find the solution to the given differential equation, we begin by simplifying the expression:

$\left(x^2 - x\sqrt{x^2 - 1}\right) dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right) dx = 0, \quad x \ge 1$

Factor out $x$ from the coefficient of $dy$: $x\left(x - \sqrt{x^2 - 1}\right) dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right) dx = 0$

Dividing the entire equation by $\left(x - \sqrt{x^2 - 1}\right) dx$ gives: $x \frac{dy}{dx} + y - \frac{x}{x - \sqrt{x^2 - 1}} = 0$

Next, rationalize the denominator of the term $\frac{x}{x - \sqrt{x^2 - 1}}$: $\frac{x}{x - \sqrt{x^2 - 1}} = \frac{x\left(x + \sqrt{x^2 - 1}\right)}{x^2 - (x^2 - 1)} = x^2 + x\sqrt{x^2 - 1}$

Substituting this back into the equation, we obtain: $x \frac{dy}{dx} + y = x^2 + x\sqrt{x^2 - 1}$

Recognizing the left-hand side as the derivative of $xy$ by the product rule: $\frac{d}{dx}(xy) = x^2 + x\sqrt{x^2 - 1}$

Integrating both sides with respect to $x$: $xy = \int x^2 \, dx + \int x\sqrt{x^2 - 1} \, dx$

Using the substitution $u = x^2 - 1 \implies du = 2x \, dx$ for the second integral: $xy = \frac{x^3}{3} + \frac{1}{3}(x^2 - 1)^{3/2} + C$

We are given the initial condition $y(1) = 1$. Substituting $x = 1$ and $y = 1$: $1 \cdot 1 = \frac{1^3}{3} + \frac{1}{3}(1^2 - 1)^{3/2} + C$ $1 = \frac{1}{3} + C \implies C = \frac{2}{3}$

Thus, the exact solution function $y(x)$ is: $xy = \frac{x^3 + (x^2 - 1)^{3/2} + 2}{3}$

Now, substitute $x = \sqrt{5}$: $\sqrt{5} \cdot y\left(\sqrt{5}\right) = \frac{(\sqrt{5})^3 + (5 - 1)^{3/2} + 2}{3}$ $\sqrt{5} \cdot y\left(\sqrt{5}\right) = \frac{5\sqrt{5} + 8 + 2}{3} = \frac{5\sqrt{5} + 10}{3}$

Dividing both sides by $\sqrt{5}$: $y\left(\sqrt{5}\right) = \frac{5 + 2\sqrt{5}}{3}$

Since $2.236 < \sqrt{5} < 2.237$: $5 + 2(2.236) < 5 + 2\sqrt{5} < 5 + 2(2.237)$ $9.472 < 5 + 2\sqrt{5} < 9.474$ $3.157 < \frac{5 + 2\sqrt{5}}{3} < 3.158$

Hence, $3 < y\left(\sqrt{5}\right) < 4$.

The greatest integer less than $y\left(\sqrt{5}\right)$ is $3$.