JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 25

Simplify. Divide by $x-\sqrt{x^2-1}$, and use $\dfrac x{x-\sqrt{x^2-1}}=x\left(x+\sqrt{x^2-1}\right)$:
$$x\,dy+y\,dx=\left(x^2+x\sqrt{x^2-1}\right)dx.$$

Integrate.
$$xy=\frac{x^3}3+\frac{(x^2-1)^{3/2}}3+C,$$
and $y(1)=1$ gives $C=\tfrac23$.

At $\sqrt5$.
$$\sqrt5\,y=\frac{5\sqrt5}3+\frac83+\frac23,\qquad y=\frac{5+2\sqrt5}3\approx3.16.$$

Answer: 3