Let y = y(x) be the solution of the differential equation (x^2 - x...

The given differential equation is:

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

Dividing the entire equation by $dx$ and rearranging, we get:

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

Dividing by $x(x - \sqrt{x^2 - 1})$, we obtain a linear differential equation:

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

The integrating factor (I.F.) is:

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

Multiplying the differential equation by the integrating factor $x$, we get:

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

$\dfrac{d}{dx}(xy) = \dfrac{x(x + \sqrt{x^2 - 1})}{(x - \sqrt{x^2 - 1})(x + \sqrt{x^2 - 1})}$

$\dfrac{d}{dx}(xy) = \dfrac{x^2 + x\sqrt{x^2 - 1}}{x^2 - (x^2 - 1)}$

$\dfrac{d}{dx}(xy) = x^2 + x\sqrt{x^2 - 1}$

Integrating both sides with respect to $x$:

$\int d(xy) = \int (x^2 + x\sqrt{x^2 - 1}) dx$

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

Given $y(1) = 1$, we substitute $x = 1$ and $y = 1$:

$1(1) = \dfrac{1^3}{3} + 0 + C \Rightarrow C = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

So, the particular solution is:

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

To find $y(\sqrt{5})$, we substitute $x = \sqrt{5}$:

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

$\sqrt{5} y(\sqrt{5}) = \dfrac{5\sqrt{5}}{3} + \dfrac{4^{3/2}}{3} + \dfrac{2}{3}$

$\sqrt{5} y(\sqrt{5}) = \dfrac{5\sqrt{5}}{3} + \dfrac{8}{3} + \dfrac{2}{3} = \dfrac{5\sqrt{5} + 10}{3}$

$y(\sqrt{5}) = \dfrac{5}{3} + \dfrac{10}{3\sqrt{5}} = \dfrac{5 + 2\sqrt{5}}{3}$

Since $\sqrt{5} \approx 2.236$, we have $2\sqrt{5} \approx 4.472$.

$y(\sqrt{5}) \approx \dfrac{5 + 4.472}{3} = \dfrac{9.472}{3} \approx 3.157$

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

Answer: $3$