JEE Main 2025 Apr 3 Shift 2, Mathematics Q14: Methods of Solving First Order, First Degree Differential Equation

Use ${\tan }^{2}x+1={\sec }^{2}x$ to combine: $3{\tan }^{2}x+3=3{\sec }^{2}x$. The ODE becomes

$\frac{dy}{dx}+3{\sec }^{2}x\cdot y={\sec }^{2}x$.

Integrating factor: $\mu =e^{\int 3{\sec }^{2}xdx}=e^{3\tan x}$.

$\frac{d}{dx}\left(y{e}^{3\tan x}\right)={\sec }^{2}x\cdot e^{3\tan x}$.

Integrate the right side using $u=3\tan x$, $du=3{\sec }^{2}xdx$:

$\int {\sec }^{2}x{e}^{3\tan x}dx=\frac{e^{3\tan x}}{3}+C$.

So $y{e}^{3\tan x}=\frac{e^{3\tan x}}{3}+C$.

Apply $y(0)=\frac{1}{3}+e^3$: at $x=0$, $\tan x=0$, so $\frac{1}{3}+e^3=\frac{1}{3}+C\Rightarrow C=e^3$.

At $x=\pi /4$: $\tan x=1$, $e^{3\tan x}=e^3$.

$y\cdot e^3=\frac{e^3}{3}+e^3\Rightarrow y=\frac{1}{3}+1=\frac{4}{3}$.