Question

Let y = y(x) be the solution of the differential equation $(1-x^2)\frac{dy}{dx}-xy=1, x \in (-1, 1)$. If y(0) = 0, then $y\left(\frac{1}{2}\right)$ is equal to

• A. $\frac{\pi}{3\sqrt{3}}$
• B. $\frac{\pi}{\sqrt{3}}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (A)

$\frac{\pi}{3\sqrt{3}}$

Answer 2

Given the differential equation $(1-x^2)\frac{dy}{dx} - xy = 1,$ we first rewrite it in standard linear form by dividing by $1-x^2$ (valid for $x\in(-1,1)$): $\frac{dy}{dx} - \frac{x}{1-x^2} y = \frac{1}{1-x^2}.$

The integrating factor $\mu(x)$ is computed as: $\mu(x) = \exp\left(\int -\frac{x}{1-x^2}\,dx\right).$ Let $u = 1-x^2$ so that $du = -2x\,dx$. Then: $\int -\frac{x}{1-x^2}\,dx = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| = \frac{1}{2}\ln|1-x^2|.$ Thus, $\mu(x)=e^{\frac{1}{2}\ln|1-x^2|} = \sqrt{1-x^2}.$

Multiplying the differential equation by $\sqrt{1-x^2}$ gives: $\sqrt{1-x^2}\frac{dy}{dx} - \frac{x}{\sqrt{1-x^2}} y = \frac{1}{\sqrt{1-x^2}}.$

Notice that the left-hand side is the derivative of $y\sqrt{1-x^2}$: $\frac{d}{dx}\left(y\sqrt{1-x^2}\right) = \frac{1}{\sqrt{1-x^2}}.$

Integrate both sides: $y\sqrt{1-x^2} = \int \frac{dx}{\sqrt{1-x^2}} = \arcsin{x} + C.$

Using the initial condition $y(0)=0$: $0\cdot\sqrt{1-0^2} = \arcsin{0} + C \quad \Rightarrow \quad C=0.$

Thus, the solution is: $y(x) = \frac{\arcsin x}{\sqrt{1-x^2}}.$

To find $y\left(\frac{1}{2}\right)$: $y\left(\frac{1}{2}\right) = \frac{\arcsin\left(\frac{1}{2}\right)}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = \frac{\frac{\pi}{6}}{\sqrt{1-\frac{1}{4}}} = \frac{\frac{\pi}{6}}{\sqrt{\frac{3}{4}}} = \frac{\pi/6}{\sqrt3/2} = \frac{\pi}{3\sqrt{3}}.$