Let $$y = y(x)$$ be the solution of the differential equation $$(x^2 - x\sqrt{x^2 - 1})dy + (y(x - \sqrt{x^2 - 1}) - x)dx = 0$$, $$x \geq 1$$. If $$y(1) = 1$$, then the greatest integer less than $$y(\sqrt{5})$$ is _______.

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

$$\left(x^2 - x\sqrt{x^2 - 1}\right)\frac{dy}{dx} + \left(x - \sqrt{x^2 - 1}\right)y = x$$

$$\frac{dy}{dx} + \frac{x - \sqrt{x^2 - 1}}{x\left(x - \sqrt{x^2 - 1}\right)}y = \frac{x}{x\left(x - \sqrt{x^2 - 1}\right)}$$

$$\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x - \sqrt{x^2 - 1}}$$

$$\frac{dy}{dx} + \frac{1}{x}y = \frac{x + \sqrt{x^2 - 1}}{x^2 - (x^2 - 1)} = x + \sqrt{x^2 - 1}$$

$$\text{I.F.} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x$$

$$y \cdot x = \int x\left(x + \sqrt{x^2 - 1}\right) \, dx$$

$$xy = \int x^2 \, dx + \int x\sqrt{x^2 - 1} \, dx$$

For $$\int x\sqrt{x^2 - 1} \, dx$$, substitute $$u = x^2 - 1 \implies du = 2x\,dx$$:

$$\int \frac{1}{2}\sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3}u^{3/2} = \frac{1}{3}(x^2 - 1)^{3/2}$$

$$xy = \frac{x^3}{3} + \frac{1}{3}(x^2 - 1)^{3/2} + C$$

$$1 = \frac{1}{3} + 0 + C \implies C = \frac{2}{3}$$

$$xy = \frac{x^3}{3} + \frac{1}{3}(x^2 - 1)^{3/2} + \frac{2}{3}$$

$$\sqrt{5} \cdot y = \frac{(\sqrt{5})^3}{3} + \frac{1}{3}((\sqrt{5})^2 - 1)^{3/2} + \frac{2}{3}$$

$$y = \frac{5 + \frac{10}{\sqrt{5}}}{3} = \frac{5 + 2\sqrt{5}}{3}$$

$$y \approx \frac{5 + 2(2.236)}{3} = \frac{5 + 4.472}{3} = \frac{9.472}{3} \approx 3.157$$

$$\lfloor y(\sqrt{5}) \rfloor = 3$$