JEE Main 2026 — Differential Equations Question with Solution

The given differential equation is:

$x\sqrt{1-x^2}dy+\left(y\sqrt{1-x^2}-x{\cos }^{-1}x\right)dx=0$

Rearranging the terms, we get:

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

Dividing the entire equation by $\sqrt{1-x^2}$:

$xdy+ydx=\frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}dx$

The left side is the exact differential of $xy$:

$d(xy)=\frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}dx$

Integrating both sides:

$xy=\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}dx+C$

To evaluate the integral, use integration by parts. Let $u={\cos }^{-1}x$ and $dv=\frac{x}{\sqrt{1-x^2}}dx$.

$du=-\frac{1}{\sqrt{1-x^2}}dx$ and $v=-\sqrt{1-x^2}$

$\int udv=uv-\int vdu$

$\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2}{\cos }^{-1}x-\int \left(-\sqrt{1-x^2}\right)\left(-\frac{1}{\sqrt{1-x^2}}\right)dx$

$=-\sqrt{1-x^2}{\cos }^{-1}x-\int 1dx$

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

Substituting this back into the equation:

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

It is given that ${\lim }_{x\to 1^{-}}y(x)=1$. Substituting $x=1$ and $y=1$:

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

$1=0-1+C\Rightarrow C=2$

The particular solution is:

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

To find $y\left(\frac{1}{2}\right)$, substitute $x=\frac{1}{2}$:

$\frac{1}{2}y\left(\frac{1}{2}\right)=2-\frac{1}{2}-\sqrt{1-{\left(\frac{1}{2}\right)}^2}{\cos }^{-1}\left(\frac{1}{2}\right)$

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

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

Multiplying by $2$:

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