JEE Main 2025 Apr 2 Shift 2, Mathematics Q21: Methods of Solving First Order, First Degree Differential Equation

The integrating factor is $\mu =e^{\int 2{\sec }^{2}xdx}=e^{2\tan x}$.

Multiplying through:

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

Substitute $u=\tan x$, $du={\sec }^{2}xdx$. RHS becomes $(2+3u)e^{2u}du$.

Integrate: $\int 2e^{2u}du=e^{2u}$ and $\int 3u{e}^{2u}du=\frac{3u{e}^{2u}}{2}-\frac{3e^{2u}}{4}$ (using parts).

So $y{e}^{2\tan x}=e^{2\tan x}+\frac{3\tan x{e}^{2\tan x}}{2}-\frac{3e^{2\tan x}}{4}+C\Rightarrow y=\frac{1}{4}+\frac{3\tan x}{2}+C{e}^{-2\tan x}$.

Apply $y(0)=\frac{5}{4}$: $\frac{5}{4}=\frac{1}{4}+0+C\Rightarrow C=1$.

At $x=\frac{\pi }{4}$ ($\tan x=1$): $y=\frac{1}{4}+\frac{3}{2}+e^{-2}=\frac{7}{4}+e^{-2}$.

$\therefore 12\left(y(\pi /4)-e^{-2}\right)=12\cdot \frac{7}{4}=21$.