Question

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 ________.

Hint:

Rearrange the differential equation into the linear form y' + P(x)y = Q(x). Rationalize the denominator of the term on the right-hand side to simplify integration.

Answer 1

Correct Answer: 3

Alternatively, we can treat this as a linear differential equation of the first order.

Step 1: Convert to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$.
Starting from $(x^2 - x\sqrt{x^2 - 1}) \frac{dy}{dx} + y(x - \sqrt{x^2 - 1}) = x$.
Divide by $(x^2 - x\sqrt{x^2 - 1})$:
$$\frac{dy}{dx} + \frac{x - \sqrt{x^2 - 1}}{x(x - \sqrt{x^2 - 1})} y = \frac{x}{x(x - \sqrt{x^2 - 1})}$$
$$\frac{dy}{dx} + \frac{1}{x} y = \frac{1}{x - \sqrt{x^2 - 1}}$$

Step 2: Find the Integrating Factor (I.F.).
$P(x) = \frac{1}{x} \implies I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.

Step 3: Solve the equation.
$y \cdot x = \int x \cdot \left( \frac{1}{x - \sqrt{x^2 - 1}} \right) dx$.
Rationalizing the integrand:
$xy = \int x (x + \sqrt{x^2 - 1}) dx = \int (x^2 + x\sqrt{x^2 - 1}) dx$.
Let $t = x^2 - 1$, then $dt = 2x dx$.
$xy = \frac{x^3}{3} + \frac{1}{2} \int \sqrt{t} dt = \frac{x^3}{3} + \frac{1}{2} \frac{t^{3/2}}{3/2} + C = \frac{x^3}{3} + \frac{1}{3}(x^2 - 1)^{3/2} + C$.

Step 4: Solve for $y(\sqrt{5})$ as in the previous solution.
Given $y(1)=1 \implies 1 = 1/3 + 0 + C \implies C = 2/3$.
$y(\sqrt{5}) = \frac{1}{\sqrt{5}} [ \frac{5\sqrt{5}}{3} + \frac{1}{3}(4)^{3/2} + \frac{2}{3} ] = \frac{5}{3} + \frac{8}{3\sqrt{5}} + \frac{2}{3\sqrt{5}} = \frac{5 + 2\sqrt{5}}{3}$.
Evaluating numerically: $y \approx 3.157$.
The greatest integer strictly less than this value is 3.