Let y = y(x) be the solution of the differential equation dy/dx = (tan x + y)/(sin x(sec x - sin x t

Q: Why did we convert everything to $\sin x$ and $\cos x$? Couldn't we have just used the identities directly?

That is a great question! While you *could* try to manipulate the secant and tangent terms directly, it often leads to algebraic traps. Converting to $\sin x$ and $\cos x$ is a universal 'safety net' in trigonometry. It reduces the expression to a common denominator, which almost always reveals cancellations that aren't immediately obvious. In this problem, it turns a messy denominator into the simple product $\sin x \cos x$.

Q: How do I know for sure that this is a Linear Differential Equation?

A Linear Differential Equation (LDE) of the first order always follows the structure $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$ only. Once we split the fraction $\frac{\tan x + y}{\sin x \cos x}$ into $\frac{\tan x}{\sin x \cos x} + \frac{y}{\sin x \cos x}$, we see that the second term is a function of $x$ multiplied by $y$. That is the hallmark of an LDE. If you see $y$ appearing linearly (not $y^2$ or $\sin y$), it is almost certainly an LDE.

Q: Why does the constant $C$ matter so much? Can't I just ignore it until the end?

You absolutely cannot ignore $C$! The constant of integration represents the entire family of solutions to the differential equation. The initial condition $y(\frac{\pi}{4}) = 2$ is what 'pins' the solution to a specific curve in that family. If you forget $C$ or calculate it incorrectly, you are essentially picking the wrong curve, and your final answer will be completely off. Always solve for $C$ as soon as you have the general solution.

Question

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = \frac{(	an x)+y}{\sin x(\sec x-\sin x \ tan x)}, x \in (0, \frac{\pi}{2})$ satisfying the condition $y(\frac{\pi}{4}) = 2$. Then, $y(\frac{\pi}{3})$ is

Options:

(A) $\sqrt{3} (2 + \log_e \sqrt{3})$

(B) $\frac{\sqrt{3}}{2} (2 + \log_e 3)$

(C) $\sqrt{3} (1 + 2 \log_e 3)$

(D) $\sqrt{3} (2 + \log_e 3)$

Sigma Percentile · Accepted Answer

### The Art of Simplifying the Chaos

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of trigonometric functions.

You see an expression like

$$\frac{dy}{dx} = \frac{	an x + y}{\sin x (\sec x - \sin x 	an x)}$$

and your instinct might be to panic. But let me tell you a secret: in the world of JEE Advanced, complexity is often just a mask for elegance. Our job is to peel back that mask.

### Phase 1: The Trigonometric Cleanup

Look at that denominator: $D = \sin x (\sec x - \sin x 	an x)$. It is a mess of secants and tangents.

Whenever you face such a situation, do not try to force a complex identity. Instead, go back to the roots. Convert everything to $\sin x$ and $\cos x$.

By substituting $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\cos x}$, the denominator becomes:

$$D = \sin x \left( \frac{1}{\cos x} - \frac{\sin^2 x}{\cos x} ight)$$

Suddenly, the fog clears. We have a common denominator of $\cos x$, leading us to:

$$D = \sin x \left( \frac{1 - \sin^2 x}{\cos x} ight)$$

And what is $1 - \sin^2 x$? It is the fundamental identity $\cos^2 x$. The expression simplifies to:

$$D = \sin x \left( \frac{\cos^2 x}{\cos x} ight) = \sin x \cos x$$

Just like that, the chaos has vanished, replaced by a clean, manageable product.

### Phase 2: The Linear Differential Equation

Now that we have our simplified denominator, our equation looks like this:

$$\frac{dy}{dx} = \frac{	an x + y}{\sin x \cos x}$$

To see the structure, we split the fraction:

$$\frac{dy}{dx} = \frac{	an x}{\sin x \cos x} + \frac{y}{\sin x \cos x}$$

Let's tackle the first term:

$$\frac{	an x}{\sin x \cos x} = \frac{\sin x / \cos x}{\sin x \cos x} = \frac{1}{\cos^2 x} = \sec^2 x$$

For the second term, we have $\frac{y}{\sin x \cos x}$. Multiplying the numerator and denominator by $2$, we get:

$$\frac{2y}{2 \sin x \cos x} = 2y \csc 2x$$

Our equation is now $\frac{dy}{dx} = \sec^2 x + 2y \csc 2x$. Rearranging this into the standard form:

$$\frac{dy}{dx} - (2 \csc 2x)y = \sec^2 x$$

We have successfully identified a Linear Differential Equation where $P(x) = -2 \csc 2x$ and $Q(x) = \sec^2 x$.

### Phase 3: The Integrating Factor

This is where the magic happens. To solve an LDE, we need the Integrating Factor (IF), defined as:

$$IF = e^{\int P(x) dx}$$

Substituting our $P(x)$, we get $IF = e^{\int -2 \csc 2x dx}$. Using the standard integral $\int \csc 	heta d	heta = \ln |	an \frac{	heta}{2}|$, we find that:

$$\int -2 \csc 2x dx = -2 \left( \frac{1}{2} \ln |	an x| ight) = -\ln |	an x|$$

Thus, our IF is:

$$IF = e^{-\ln |	an x|} = e^{\ln |\cot x|} = \cot x$$

The integrating factor is simply $\cot x$. How elegant is that?

### Phase 4: The Final Stretch

With the IF in hand, the general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$. Substituting our values, we get:

$$y \cot x = \int \sec^2 x \cot x dx + C$$

The integrand $\sec^2 x \cot x$ simplifies to:

$$\frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = \frac{1}{\sin x \cos x} = 2 \csc 2x$$

The integral of $2 \csc 2x$ is $\ln |	an x|$. So, $y \cot x = \ln |	an x| + C$.

Using the initial condition $y(\frac{\pi}{4}) = 2$, we find:

$$2 \cot\left(\frac{\pi}{4}ight) = \ln\left(	an \frac{\pi}{4}ight) + C$$

Since $\cot(\frac{\pi}{4}) = 1$ and $\ln(1) = 0$, we get $C = 2$. Our particular solution is:

$$y \cot x = \ln |	an x| + 2 \implies y = 	an x (\ln |	an x| + 2)$$

Finally, evaluating at $x = \frac{\pi}{3}$, we get:

$$y\left(\frac{\pi}{3}ight) = 	an\left(\frac{\pi}{3}ight) \left( \ln\left(	an \frac{\pi}{3}ight) + 2 ight) = \sqrt{3} (\ln \sqrt{3} + 2)$$

This matches our target perfectly. You see? With patience and systematic steps, even the most daunting problems yield to the beauty of mathematics.

Select Answer:

Visualized Solution

The Given Equation

Simplifying the Denominator

Substituting Sine and Cosine

Applying Trigonometric Identity

Rearranging to Standard Form

Simplifying the Terms

Standard Linear Differential Equation

Calculating the Integrating Factor

Integrating $P (x)$

The General Solution Setup

Integrating the Right Hand Side

Using the Initial Condition

Finding the Constant $C$

Final Substitution for $y (\frac{π}{3})$

Final Calculation

The Sigma Insight: Linear Differential Equations

The Art of Simplifying the Chaos

Phase 1

Phase 2

Phase 3

Phase 4

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