Animated Solution for Mathematics - Differential Equations: If $\frac{dy}{dx}+\frac{3}{{\cos }^{2}x}y=\frac{1}{{\cos }^{2}x},x\in (-\frac{\pi }{3},\frac{\pi }{3})$, and $y(\frac{\pi }{4})=\frac{4}{3}$, then $y(-\frac{\pi }{4})$ equals :

Welcome, future engineer. Today, we are going to peel back the layers of a problem that, at first glance, might seem like a chaotic mess of trigonometry.

You see the equation:

Your instinct might be to panic. But I want you to take a deep breath. In the world of JEE Advanced, problems are rarely designed to be brute-forced; they are designed to be recognized.

This equation is a wolf in sheep's clothing—it is a standard First-Order Linear Differential Equation hiding behind a trigonometric mask.

The Transformation

Our first step is to strip away the disguise. We know that $\frac{1}{{\cos }^{2}x}$ is simply ${\sec }^{2}x$.

By applying this identity, the equation transforms into:

Suddenly, the structure becomes crystal clear. It perfectly matches the standard form $\frac{dy}{dx}+P(x)y=Q(x)$.

Here, our $P(x)$ is $3{\sec }^{2}x$ and our $Q(x)$ is ${\sec }^{2}x$. Recognizing this form is the single most important step. Once you see it, the path forward is illuminated.

The Magic Multiplier

To solve this, we need the Integrating Factor ($I.F.$). Think of this as the 'magic multiplier' that turns the left side of our equation into the derivative of a product.

The formula is $I.F.=e^{\int P(x)dx}$. Substituting our $P(x)$, we get:

Now, don't let the integral in the exponent scare you. We know that the derivative of $\tan x$ is ${\sec }^{2}x$.

Therefore, the integral of ${\sec }^{2}x$ is simply $\tan x$. Our Integrating Factor becomes $e^{3\tan x}$. This is the key that unlocks the entire problem.

The Integration Dance

With our $I.F.$ in hand, the general solution is given by $y\cdot (I.F.)=\int Q(x)\cdot (I.F.)dx+C$.

Plugging in our values, we get:

Now, look at the integral on the right. It is a beautiful setup for substitution. Let $t=3\tan x$.

Then $dt=3{\sec }^{2}xdx$, which means ${\sec }^{2}xdx=\frac{1}{3}dt$. The integral transforms into $\frac{1}{3}\int e^{t}dt$, which is simply $\frac{1}{3}{e}^t$.

Substituting back, we get $\frac{1}{3}{e}^{3\tan x}$. Our general solution is now:

Finding the Specific Path

We are given the initial condition $y(\frac{\pi }{4})=\frac{4}{3}$. This is our anchor. It allows us to find the constant $C$.

Substituting $x=\frac{\pi }{4}$ and $y=\frac{4}{3}$ (and knowing $\tan (\frac{\pi }{4})=1$), we get:

A quick subtraction reveals that $C=e^3$. We have now found the specific curve that satisfies our conditions:

The Final Reveal

Finally, we need to find $y(-\frac{\pi }{4})$. Substituting $x=-\frac{\pi }{4}$ (where $\tan (-\frac{\pi }{4})=-1$), our equation becomes:

To isolate $y$, we multiply everything by $e^3$. The left side becomes $y$, the first term on the right becomes $\frac{1}{3}$, and the second term becomes $e^6$.

Thus, the final answer is:

There it is. The complexity dissolves, the math aligns, and we arrive at the solution. Remember, in physics and mathematics, the most daunting problems are often just simple concepts wearing complicated costumes. Keep practicing, keep observing, and you will master these patterns.