Let y=y(x) be the solution of the differential equation [(x^2-xx^2-1),dy + (y(x-x^2-1)-x),…

Question

Let $y=y(x)$ be the solution of the differential equation
$$(x^2-x\sqrt{x^2-1})\,dy + \big(y(x-\sqrt{x^2-1})-x\big)\,dx=0,\quad x\ge 1,\qquad y(1)=1.$$
Then $[y(\sqrt5)]$ is \_\_\_\_\_\_ (greatest integer function).

Answer: 3

Accepted Answer

Standard form:
$$\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x-\sqrt{x^2-1}} = x+\sqrt{x^2-1}\ \ \text{(rationalising)}.$$

Integrating factor $=e^{\int dx/x}=x$. So
$$yx = \int (x+\sqrt{x^2-1})x\,dx = \frac{x^2}{2}+\frac{(x^2-1)^{3/2}}{3}+C.$$

Using $y(1)=1$: $1=\frac12+C \implies C=\frac12$. So
$$y = \frac{x}{2}+\frac{(x^2-1)^{3/2}}{3x}+\frac{1}{2x}.$$

At $x=\sqrt5$:
$$y(\sqrt5) = \frac{\sqrt5}{2}+\frac{8}{3}+\frac{1}{2\sqrt5} \approx 3.36,\qquad [y(\sqrt5)] = 3.$$

Solutions · 1