Q50 / 75 · JEE Main 2026 (05 April Shift 2) · Mathematics
Let $y = y(x)$ be the solution of the differential equation $(\tan x)^{1/2}\,dy = (\sec^3 x - (\tan x)^{3/2} y)\,dx$, $0 < x < \dfrac{\pi}{2}$, $y\left(\dfrac{\pi}{4}\right) = \dfrac{6\sqrt{2}}{5}$. If $y\left(\dfrac{\pi}{3}\right) = \dfrac{4}{5}\alpha$, then $\alpha^4$ equals _______.
Correct answer: 48
Solution
The given differential equation can be rewritten as:
$\dfrac{dy}{dx} + (\tan x)y = \dfrac{\sec^3 x}{(\tan x)^{1/2}}$
This is a linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$.
Integrating Factor (IF) = $e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x$
Multiplying both sides by the IF and integrating, we get:
$y \sec x = \int \dfrac{\sec^4 x}{(\tan x)^{1/2}} \, dx + C$
$y \sec x = \int \dfrac{(1 + \tan^2 x)\sec^2 x}{(\tan x)^{1/2}} \, dx + C$
Let $\tan x = t$, then $\sec^2 x \, dx = dt$.
$y \sec x = \int \dfrac{1 + t^2}{t^{1/2}} \, dt + C$
$y \sec x = \int (t^{-1/2} + t^{3/2}) \, dt + C$
$y \sec x = 2t^{1/2} + \dfrac{2}{5}t^{5/2} + C$
$y \sec x = 2(\tan x)^{1/2} + \dfrac{2}{5}(\tan x)^{5/2} + C$
Given $y\left(\dfrac{\pi}{4}\right) = \dfrac{6\sqrt{2}}{5}$, substituting $x = \dfrac{\pi}{4}$:
$\dfrac{6\sqrt{2}}{5} \cdot \sqrt{2} = 2(1)^{1/2} + \dfrac{2}{5}(1)^{5/2} + C$
$\dfrac{12}{5} = 2 + \dfrac{2}{5} + C \Rightarrow C = 0$
The solution is $y \sec x = 2(\tan x)^{1/2} + \dfrac{2}{5}(\tan x)^{5/2}$.
Substituting $x = \dfrac{\pi}{3}$:
$y\left(\dfrac{\pi}{3}\right) \cdot 2 = 2(\sqrt{3})^{1/2} + \dfrac{2}{5}(\sqrt{3})^{5/2}$
$y\left(\dfrac{\pi}{3}\right) = 3^{1/4} + \dfrac{1}{5} \cdot 3^{5/4} = 3^{1/4}\left(1 + \dfrac{3}{5}\right) = \dfrac{8}{5} \cdot 3^{1/4}$
Given $y\left(\dfrac{\pi}{3}\right) = \dfrac{4}{5}\alpha$, we have:
$\dfrac{4}{5}\alpha = \dfrac{8}{5} \cdot 3^{1/4} \Rightarrow \alpha = 2 \cdot 3^{1/4}$
Therefore, $\alpha^4 = (2 \cdot 3^{1/4})^4 = 16 \cdot 3 = 48$
Answer: $48$
$\dfrac{dy}{dx} + (\tan x)y = \dfrac{\sec^3 x}{(\tan x)^{1/2}}$
This is a linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$.
Integrating Factor (IF) = $e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x$
Multiplying both sides by the IF and integrating, we get:
$y \sec x = \int \dfrac{\sec^4 x}{(\tan x)^{1/2}} \, dx + C$
$y \sec x = \int \dfrac{(1 + \tan^2 x)\sec^2 x}{(\tan x)^{1/2}} \, dx + C$
Let $\tan x = t$, then $\sec^2 x \, dx = dt$.
$y \sec x = \int \dfrac{1 + t^2}{t^{1/2}} \, dt + C$
$y \sec x = \int (t^{-1/2} + t^{3/2}) \, dt + C$
$y \sec x = 2t^{1/2} + \dfrac{2}{5}t^{5/2} + C$
$y \sec x = 2(\tan x)^{1/2} + \dfrac{2}{5}(\tan x)^{5/2} + C$
Given $y\left(\dfrac{\pi}{4}\right) = \dfrac{6\sqrt{2}}{5}$, substituting $x = \dfrac{\pi}{4}$:
$\dfrac{6\sqrt{2}}{5} \cdot \sqrt{2} = 2(1)^{1/2} + \dfrac{2}{5}(1)^{5/2} + C$
$\dfrac{12}{5} = 2 + \dfrac{2}{5} + C \Rightarrow C = 0$
The solution is $y \sec x = 2(\tan x)^{1/2} + \dfrac{2}{5}(\tan x)^{5/2}$.
Substituting $x = \dfrac{\pi}{3}$:
$y\left(\dfrac{\pi}{3}\right) \cdot 2 = 2(\sqrt{3})^{1/2} + \dfrac{2}{5}(\sqrt{3})^{5/2}$
$y\left(\dfrac{\pi}{3}\right) = 3^{1/4} + \dfrac{1}{5} \cdot 3^{5/4} = 3^{1/4}\left(1 + \dfrac{3}{5}\right) = \dfrac{8}{5} \cdot 3^{1/4}$
Given $y\left(\dfrac{\pi}{3}\right) = \dfrac{4}{5}\alpha$, we have:
$\dfrac{4}{5}\alpha = \dfrac{8}{5} \cdot 3^{1/4} \Rightarrow \alpha = 2 \cdot 3^{1/4}$
Therefore, $\alpha^4 = (2 \cdot 3^{1/4})^4 = 16 \cdot 3 = 48$
Answer: $48$
Hence, the correct answer is 48.