JEE Main 2026 Apr 6 Shift 1, Mathematics Q25: Methods of Solving First Order, First Degree Differential Equation

Factor the coefficient of $dy$: $x^2-x\sqrt{x^2-1}=x(x-\sqrt{x^2-1})$. The ODE rearranges to

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

Rationalize: $\frac{1}{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1}$. Linear ODE with integrating factor $\mu (x)=x$:

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

Integrate: $xy=\frac{x^3}{3}+\frac{(x^2-1{)}^{3/2}}{3}+C$.

Apply $y(1)=1$: $1=\frac{1}{3}+0+C\Rightarrow C=\frac{2}{3}$.

At $x=\sqrt{5}$: $\sqrt{5}y(\sqrt{5})=\frac{5\sqrt{5}}{3}+\frac{8}{3}+\frac{2}{3}=\frac{5\sqrt{5}+10}{3}$.

$y(\sqrt{5})=\frac{5\sqrt{5}+10}{3\sqrt{5}}=\frac{5}{3}+\frac{2\sqrt{5}}{3}=\frac{5+2\sqrt{5}}{3}\approx \frac{5+4.472}{3}\approx 3.157$.

$\therefore$ greatest integer less than $y(\sqrt{5})$ is $3$.