JEE Main 2026 — Differential Equations Question with Solution

The given differential equation is:

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

Dividing the entire equation by $dx$ and rearranging, we get:

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

Dividing by $x(x-\sqrt{x^2-1})$, we obtain a linear differential equation:

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

The integrating factor (I.F.) is:

$\text{I.F.}=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$

Multiplying the differential equation by the integrating factor $x$, we get:

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

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

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

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

Integrating both sides with respect to $x$:

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

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

Given $y(1)=1$, we substitute $x=1$ and $y=1$:

$1(1)=\frac{1^3}{3}+0+C\Rightarrow C=1-\frac{1}{3}=\frac{2}{3}$

So, the particular solution is:

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

To find $y(\sqrt{5})$, we substitute $x=\sqrt{5}$:

$\sqrt{5}y(\sqrt{5})=\frac{(\sqrt{5}{)}^3}{3}+\frac{(5-1{)}^{3/2}}{3}+\frac{2}{3}$

$\sqrt{5}y(\sqrt{5})=\frac{5\sqrt{5}}{3}+\frac{4^{3/2}}{3}+\frac{2}{3}$

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

$y(\sqrt{5})=\frac{5}{3}+\frac{10}{3\sqrt{5}}=\frac{5+2\sqrt{5}}{3}$

Since $\sqrt{5}\approx 2.236$, we have $2\sqrt{5}\approx 4.472$.

$y(\sqrt{5})\approx \frac{5+4.472}{3}=\frac{9.472}{3}\approx 3.157$

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

Answer: $3$