JEE Main 2026 Apr 8 Shift 2, Mathematics Q17: Methods of Solving First Order, First Degree Differential Equation

Dividing by $x\sqrt{1-x^2}$ rewrites the ODE in linear form:

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

Integrating factor: ${\mathrm{e}}^{\int \frac{1}{x}dx}=x$. So $\frac{d}{dx}(xy)=\frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}$.

Evaluate the RHS integral. Substitute $u={\cos }^{-1}x$, $x=\cos u$, $dx=-\sin udu$, $\sqrt{1-x^2}=\sin u$:

$\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}dx=\int \frac{\cos u\cdot u}{\sin u}(-\sin u)du=-\int u\cos udu=-(u\sin u+\cos u)+C$

$=-\sqrt{1-x^2}{\cos }^{-1}x-x+C$

So $xy=-\sqrt{1-x^2}{\cos }^{-1}x-x+C$.

Apply ${\lim }_{x\to 1^{-}}y(x)=1$: at $x=1$, $\sqrt{1-x^2}=0$ and ${\cos }^{-1}1=0$, giving $1\cdot 1=0-1+C⟹C=2$.

Plug in $x=\frac{1}{2}$: $\sqrt{1-1/4}=\frac{\sqrt{3}}{2}$, ${\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}$.

$\frac{1}{2}y=-\frac{\sqrt{3}}{2}\cdot \frac{\pi }{3}-\frac{1}{2}+2=-\frac{\pi }{2\sqrt{3}}+\frac{3}{2}$

$y\left(\frac{1}{2}\right)=3-\frac{\pi }{\sqrt{3}}$.