Solving the Differential Equation

The given differential equation is:

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

This can be rearranged into the standard form of a first-order linear differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$.

Step 1: Rearrange the ODE

Divide the equation by $dx$ and then by $\left(x^2 - x\sqrt{x^2 - 1}\right)$:

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

Simplify the terms:

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

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

Rearranging gives the standard form:

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

Here, $P(x) = \frac{1}{x}$ and $Q(x) = \frac{1}{x - \sqrt{x^2 - 1}}$.

Step 2: Find the Integrating Factor

The integrating factor $I(x)$ is given by $e^{\int P(x) dx}$.

$I(x) = e^{\int \frac{1}{x} dx} = e^{\ln|x|}$

Since $x \geq 1$, $|x| = x$. Thus,

$I(x) = x$

Step 3: Solve the ODE

Multiply the standard form equation by the integrating factor $I(x) = x$:

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

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

The left side is the derivative of the product $I(x)y = xy$:

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

Integrate the right side. First, rationalize the fraction:

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

Now, integrate:

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

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

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

The general solution is:

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

Step 4: Apply the Initial Condition

We are given $y(1) = 1$. Substitute $x=1$ into the general solution:

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

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

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

The specific solution is:

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

Step 5: Evaluate the Solution at $x=\sqrt{5}$

Substitute $x = \sqrt{5}$ into the specific solution:

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

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

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

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

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

Simplify further:

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

Step 6: Find the Greatest Integer

Estimate the value of $y(\sqrt{5})$:

We know $2.23 < \sqrt{5} < 2.24$.

$5 + 2(2.23) < 5 + 2\sqrt{5} < 5 + 2(2.24)$

$5 + 4.46 < 5 + 2\sqrt{5} < 5 + 4.48$

$9.46 < 5 + 2\sqrt{5} < 9.48$

Divide by 3:

$\frac{9.46}{3} < \frac{5 + 2\sqrt{5}}{3} < \frac{9.48}{3}$

$3.153... < y(\sqrt{5}) < 3.16$

The value $y(\sqrt{5})$ is approximately 3.157.

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