JEE Mathematics 2025 Question with Solution

Given:

• $\frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x$
• $y(0) = \frac{1}{3} + e^3$

Find: $y\left(\frac{\pi}{4}\right)$

First rewrite the differential equation using the identity $\tan^2 x + 1 = \sec^2 x$:

So,

This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 3\sec^2 x$ and $Q(x) = \sec^2 x$.

The integrating factor is

Using the standard formula,

Hence,

Let

Then

so

Therefore,

Substituting back,

Now use the initial condition $y(0) = \frac{1}{3} + e^3$. Since $\tan 0 = 0$,

So,

which gives

Thus the particular solution is

Now substitute $x = \frac{\pi}{4}$. Since $\tan\left(\frac{\pi}{4}\right) = 1$,

Dividing by $e^3$,

Therefore, $y\left(\frac{\pi}{4}\right) = \frac{4}{3}$, so the correct option is B.

Given:

• $\frac{dy}{dx} + 3(\tan^2 x)y + 3y = \sec^2 x$
• $y(0) = \frac{1}{3} + e^3$

Find: $y\left(\frac{\pi}{4}\right)$

From the worked solution,

Dividing throughout by $e^{3\tan x}$ gives

Using $y(0) = \frac{1}{3} + e^3$ and $\tan 0 = 0$,

Hence,

So the particular solution becomes

Now put $x = \frac{\pi}{4}$. Since $\tan\left(\frac{\pi}{4}\right) = 1$,

Therefore, the correct option is B.