Let $y=y(x)$ be the solution of the differential equation $(\tan x)^{1/2} dy = (\sec^3 x - (\tan x)^{3/2} y) dx$, $0 < x < \frac{\pi}{2}$, $y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}$. If $y\left(\frac{\pi}{3}\right) = \frac{4}{5}\alpha$, then $\alpha^4$ equals ______

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(tan x)^{1/2} d y = (sec^{3} x - (tan x)^{3/2} y) d x

0 < x < \frac{π}{2}

y (\frac{π}{4}) = \frac{6 2}{5}

. If

y (\frac{π}{3}) = \frac{4}{5} α

, then

α^{4}

equals ______

Answer: 48

Dividing by

tan x d x

gives the linear form

\frac{d y}{d x} + y tan x = \frac{sec ^{3} x}{tan x}

, with integrating factor

e^{\int t a n x d x} = sec x

. Then

\frac{d}{d x} (y sec x) = \frac{sec ^{4} x}{tan x} = \frac{( 1 + tan ^{2} x ) sec ^{2} x}{tan x}

, and with

t = tan x

this integrates to

y sec x = \int \frac{1 + t ^{2}}{t} d t = 2 t + \frac{2}{5} t^{5/2} + c = 2 tan x + \frac{2}{5} (tan x)^{5/2} + c

. At

x = \frac{π}{4}

\frac{6 2}{5} \cdot 2 = \frac{12}{5} = 2 + \frac{2}{5} + c

, so

c = 0

. At

x = \frac{π}{3}

tan x = 3

and

sec x = 2

, giving

2 y = 2 \cdot 3^{1/4} (1 + \frac{3}{5}) = \frac{16}{5} 3^{1/4}

, so

y (\frac{π}{3}) = \frac{8}{5} 3^{1/4} = \frac{4}{5} α

and

α = 2 \cdot 3^{1/4}

. Therefore

α^{4} = 16 \cdot 3 = 48

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Integer answer