Let y=y(x) be the solution of the differential equation [x√1-x^2,dy + (y√1-x^2-x^-1x),dx=0…

Question

Let $y=y(x)$ be the solution of the differential equation
$$x\sqrt{1-x^2}\,dy + \big(y\sqrt{1-x^2}-x\cos^{-1}x\big)\,dx=0,\quad x\in(0,1),\qquad \lim_{x\to1^-}y(x)=1.$$
Then $y\!\left(\dfrac12\right)$ equals:

Accepted Answer

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

I.F. $=x$. So $d(xy) = \left(\dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}\right)dx$. Integrating by parts (with $u=\cos^{-1}x$, $dv=\dfrac{x}{\sqrt{1-x^2}}dx$):
$$xy = -(\cos^{-1}x)\sqrt{1-x^2} - x + C.$$

So
$$y(x) = \frac{C}{x}-1-\frac{(\cos^{-1}x)\sqrt{1-x^2}}{x}.$$

As $x\to1^-$: $y\to C-1-0 = 1 \Rightarrow C=2$.

At $x=1/2$: $y(1/2) = 4-1-2\cdot\dfrac{\pi}{3}\cdot\dfrac{\sqrt3}{2} = 3-\dfrac{\pi}{\sqrt3}$.

Solutions · 1