JEE Main 2026 — Mathematics Calculus

The given differential equation is:

$$(x^2 - x\sqrt{x^2 - 1})dy + (y(x - \sqrt{x^2 - 1}) - x)dx = 0$$(x2−xx2−1​)dy+(y(x−x2−1​)−x)dx=0

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

$$x(x - \sqrt{x^2 - 1}) \dfrac{dy}{dx} + y(x - \sqrt{x^2 - 1}) = x$$x(x−x2−1​)dxdy​+y(x−x2−1​)=x

Dividing by $x(x - \sqrt{x^2 - 1})$x(x−x2−1​), we obtain a linear differential equation:

$$\dfrac{dy}{dx} + \dfrac{1}{x}y = \dfrac{1}{x - \sqrt{x^2 - 1}}$$dxdy​+x1​y=x−x2−1​1​

The integrating factor (I.F.) is:

$$\text{I.F.} = e^{\int \dfrac{1}{x} dx} = e^{\ln x} = x$$I.F.=e∫x1​dx=elnx=x

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

$$x \dfrac{dy}{dx} + y = \dfrac{x}{x - \sqrt{x^2 - 1}}$$xdxdy​+y=x−x2−1​x​

$$\dfrac{d}{dx}(xy) = \dfrac{x(x + \sqrt{x^2 - 1})}{(x - \sqrt{x^2 - 1})(x + \sqrt{x^2 - 1})}$$dxd​(xy)=(x−x2−1​)(x+x2−1​)x(x+x2−1​)​

$$\dfrac{d}{dx}(xy) = \dfrac{x^2 + x\sqrt{x^2 - 1}}{x^2 - (x^2 - 1)}$$dxd​(xy)=x2−(x2−1)x2+xx2−1​​

$$\dfrac{d}{dx}(xy) = x^2 + x\sqrt{x^2 - 1}$$dxd​(xy)=x2+xx2−1​

Integrating both sides with respect to $x$x:

$$\int d(xy) = \int (x^2 + x\sqrt{x^2 - 1}) dx$$∫d(xy)=∫(x2+xx2−1​)dx

$$xy = \dfrac{x^3}{3} + \dfrac{(x^2 - 1)^{3/2}}{3} + C$$xy=3x3​+3(x2−1)3/2​+C

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

$$1(1) = \dfrac{1^3}{3} + 0 + C \Rightarrow C = 1 - \dfrac{1}{3} = \dfrac{2}{3}$$1(1)=313​+0+C⇒C=1−31​=32​

So, the particular solution is:

$$xy = \dfrac{x^3}{3} + \dfrac{(x^2 - 1)^{3/2}}{3} + \dfrac{2}{3}$$xy=3x3​+3(x2−1)3/2​+32​

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

$$\sqrt{5} y(\sqrt{5}) = \dfrac{(\sqrt{5})^3}{3} + \dfrac{(5 - 1)^{3/2}}{3} + \dfrac{2}{3}$$5​y(5​)=3(5​)3​+3(5−1)3/2​+32​

$$\sqrt{5} y(\sqrt{5}) = \dfrac{5\sqrt{5}}{3} + \dfrac{4^{3/2}}{3} + \dfrac{2}{3}$$5​y(5​)=355​​+343/2​+32​

$$\sqrt{5} y(\sqrt{5}) = \dfrac{5\sqrt{5}}{3} + \dfrac{8}{3} + \dfrac{2}{3} = \dfrac{5\sqrt{5} + 10}{3}$$5​y(5​)=355​​+38​+32​=355​+10​

$$y(\sqrt{5}) = \dfrac{5}{3} + \dfrac{10}{3\sqrt{5}} = \dfrac{5 + 2\sqrt{5}}{3}$$y(5​)=35​+35​10​=35+25​​

Since $\sqrt{5} \approx 2.236$5​≈2.236, we have $2\sqrt{5} \approx 4.472$25​≈4.472.

$$y(\sqrt{5}) \approx \dfrac{5 + 4.472}{3} = \dfrac{9.472}{3} \approx 3.157$$y(5​)≈35+4.472​=39.472​≈3.157

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

Answer: $3$3