JEE Main 2026 — Mathematics Calculus

The given differential equation can be rewritten as:

$$\dfrac{dy}{dx} + (\tan x)y = \dfrac{\sec^3 x}{(\tan x)^{1/2}}$$dxdy​+(tanx)y=(tanx)1/2sec3x​

This is a linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$dxdy​+P(x)y=Q(x).

Integrating Factor (IF) = $\displaystyle e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x$e∫tanxdx=eln(secx)=secx

Multiplying both sides by the IF and integrating, we get:

$$y \sec x = \int \dfrac{\sec^4 x}{(\tan x)^{1/2}} \, dx + C$$ysecx=∫(tanx)1/2sec4x​dx+C

$$y \sec x = \int \dfrac{(1 + \tan^2 x)\sec^2 x}{(\tan x)^{1/2}} \, dx + C$$ysecx=∫(tanx)1/2(1+tan2x)sec2x​dx+C

Let $\tan x = t$tanx=t, then $\sec^2 x \, dx = dt$sec2xdx=dt.

$$y \sec x = \int \dfrac{1 + t^2}{t^{1/2}} \, dt + C$$ysecx=∫t1/21+t2​dt+C

$$y \sec x = \int (t^{-1/2} + t^{3/2}) \, dt + C$$ysecx=∫(t−1/2+t3/2)dt+C

$$y \sec x = 2t^{1/2} + \dfrac{2}{5}t^{5/2} + C$$ysecx=2t1/2+52​t5/2+C

$$y \sec x = 2(\tan x)^{1/2} + \dfrac{2}{5}(\tan x)^{5/2} + C$$ysecx=2(tanx)1/2+52​(tanx)5/2+C

Given $y\left(\dfrac{\pi}{4}\right) = \dfrac{6\sqrt{2}}{5}$y(4π​)=562​​, substituting $x = \dfrac{\pi}{4}$x=4π​:

$$\dfrac{6\sqrt{2}}{5} \cdot \sqrt{2} = 2(1)^{1/2} + \dfrac{2}{5}(1)^{5/2} + C$$562​​⋅2​=2(1)1/2+52​(1)5/2+C

$$\dfrac{12}{5} = 2 + \dfrac{2}{5} + C \Rightarrow C = 0$$512​=2+52​+C⇒C=0

The solution is $y \sec x = 2(\tan x)^{1/2} + \dfrac{2}{5}(\tan x)^{5/2}$ysecx=2(tanx)1/2+52​(tanx)5/2.

Substituting $x = \dfrac{\pi}{3}$x=3π​:

$$y\left(\dfrac{\pi}{3}\right) \cdot 2 = 2(\sqrt{3})^{1/2} + \dfrac{2}{5}(\sqrt{3})^{5/2}$$y(3π​)⋅2=2(3​)1/2+52​(3​)5/2

$$y\left(\dfrac{\pi}{3}\right) = 3^{1/4} + \dfrac{1}{5} \cdot 3^{5/4} = 3^{1/4}\left(1 + \dfrac{3}{5}\right) = \dfrac{8}{5} \cdot 3^{1/4}$$y(3π​)=31/4+51​⋅35/4=31/4(1+53​)=58​⋅31/4

Given $y\left(\dfrac{\pi}{3}\right) = \dfrac{4}{5}\alpha$y(3π​)=54​α, we have:

$$\dfrac{4}{5}\alpha = \dfrac{8}{5} \cdot 3^{1/4} \Rightarrow \alpha = 2 \cdot 3^{1/4}$$54​α=58​⋅31/4⇒α=2⋅31/4

Therefore, $\alpha^4 = (2 \cdot 3^{1/4})^4 = 16 \cdot 3 = 48$α4=(2⋅31/4)4=16⋅3=48

Answer: $48$48