JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 25

Linear form. Dividing by $\sqrt{\tan x}$:
$$\frac{dy}{dx}+y\tan x=\frac{\sec^3x}{\sqrt{\tan x}},\qquad\text{IF}=\sec x.$$

Integrate. With $t=\tan x$:
$$y\sec x=\int\frac{\sec^4x}{\sqrt{\tan x}}\,dx=\int\left(t^{-1/2}+t^{3/2}\right)dt=2\sqrt t+\tfrac25t^{5/2}+C.$$
At $\tfrac\pi4$: $\tfrac{6\sqrt2}5\cdot\sqrt2=\tfrac{12}5=2+\tfrac25+C$, so $C=0$.

At $\tfrac\pi3$ ($t=\sqrt3$):
$$2y=3^{1/4}\left(2+\tfrac65\right)=\tfrac{16}5\,3^{1/4},$$
so $y=\tfrac85\,3^{1/4}=\tfrac45\alpha$.

So $\alpha=2\cdot3^{1/4}$ and $\alpha^4=48$.

Answer: 48