JEE Mathematics 2025 Question with Solution

Given: $\frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \sec^2 x$ and $y(0)=\frac{5}{4}$.

Find: $12\left(y\left(\frac{\pi}{4}\right)-e^{-2}\right)$.

This is a first-order linear differential equation of the form $y' + P(x)y = Q(x)$. The integrating factor is

$$\mu(x)=e^{\int P(x)\,dx}=e^{\int 2\sec^2 x\,dx}=e^{2\tan x}$$

Multiplying the equation by the integrating factor,

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

Now use the substitution $u=\tan x$, so that $du=\sec^2 x\,dx$. Then

$$\int e^{2\tan x}\left(2\sec^2 x+3\tan x\sec^2 x\right)dx=\int e^{2u}(2+3u)\,du$$

From the given working,

$$\int e^{2u}(2+3u)\,du=e^{2u}\left(\frac{3u}{2}+\frac{1}{4}\right)+C$$

Hence,

$$y e^{2\tan x}=e^{2\tan x}\left(\frac{3}{2}\tan x+\frac{1}{4}\right)+C$$

So,

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

Using the initial condition $y(0)=\frac{5}{4}$ and $\tan 0=0$,

$$\frac{5}{4}=\frac{1}{4}+C$$

Therefore,

$$C=1$$

and

$$y(x)=\frac{3}{2}\tan x+\frac{1}{4}+e^{-2\tan x}$$

Now at $x=\frac{\pi}{4}$, we have $\tan \frac{\pi}{4}=1$, so

$$y\left(\frac{\pi}{4}\right)=\frac{3}{2}+\frac{1}{4}+e^{-2}=\frac{7}{4}+e^{-2}$$

Thus,

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

Therefore, the correct option is A.

Note: The solution evaluates $12\left(y\left(\frac{\pi}{4}\right)-e^{-2}\right)$, while the given question text shows $e^2$. The answer follows the solution working.

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

Find: the required value after solving for $y(x)$.

Identify

$$P(x)=2\sec^2 x, \qquad Q(x)=2\sec^2 x + 3\tan x\sec^2 x$$

So the integrating factor is

$$\mu(x)=e^{\int 2\sec^2 x\,dx}=e^{2\tan x}$$

After multiplying,

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

The left-hand side is

$$\frac{d}{dx}\left(y e^{2\tan x}\right)$$

Hence,

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

Integrating both sides,

$$y e^{2\tan x}=\int e^{2\tan x}\left(2\sec^2 x+3\tan x\sec^2 x\right)dx + C$$

Put $u=\tan x$ and $du=\sec^2 x\,dx$. Then the integral becomes

$$\int e^{2u}(2+3u)\,du$$

Using the extracted result,

$$\int e^{2u}(2+3u)\,du=e^{2u}\left(\frac{3u}{2}+\frac{1}{4}\right)+C$$

Substituting back $u=\tan x$,

$$y e^{2\tan x}=e^{2\tan x}\left(\frac{3}{2}\tan x+\frac{1}{4}\right)+C$$

Divide by $e^{2\tan x}$:

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

Now apply $y(0)=\frac{5}{4}$:

$$\frac{5}{4}=\frac{3}{2}\tan 0+\frac{1}{4}+C e^0=\frac{1}{4}+C$$

So $C=1$. Therefore,

$$y(x)=\frac{3}{2}\tan x+\frac{1}{4}+e^{-2\tan x}$$

At $x=\frac{\pi}{4}$,

$$y\left(\frac{\pi}{4}\right)=\frac{3}{2}(1)+\frac{1}{4}+e^{-2}=\frac{7}{4}+e^{-2}$$

Hence,

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

Therefore, the required value is $21$, so the correct option is A.