JEE Main 2026 Apr 5 Shift 2, Mathematics Q25: Methods of Solving First Order, First Degree Differential Equation

Rearranging gives the linear ODE: $\frac{dy}{dx}+y\tan x=\frac{{\sec }^{3}x}{\sqrt{\tan x}}$.

Integrating factor: $e^{\int \tan xdx}=\sec x$.

$y\sec x=\int \frac{{\sec }^{4}x}{\sqrt{\tan x}}dx=\int \frac{(1+{\tan }^{2}x){\sec }^{2}x}{\sqrt{\tan x}}dx$.

Let $t=\tan x$: $y\sec x=\int \frac{1+t^2}{\sqrt{t}}dt=2\sqrt{t}+\frac{2}{5}{t}^{5/2}+C=2\sqrt{\tan x}+\frac{2}{5}(\tan x{)}^{5/2}+C$.

Apply $y(\pi /4)=6\sqrt{2}/5$: $\frac{6\sqrt{2}}{5}\cdot \sqrt{2}=2+\frac{2}{5}+C\Rightarrow \frac{12}{5}=\frac{12}{5}+C\Rightarrow C=0$.

At $x=\pi /3$, $\tan x=\sqrt{3}$, $\sec x=2$:

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

So $y(\pi /3)=\frac{8}{5}\cdot 3^{1/4}=\frac{4}{5}\alpha \Rightarrow \alpha =2\cdot 3^{1/4}$.

${\alpha }^4=16\cdot 3=48$.