If dy/dx + 2y tan x = sin x, 0

Q: Wait, why do we use the Integrating Factor method here? Can't we just separate variables?

That is a fantastic question! If you look closely at the equation $\frac{dy}{dx} + 2y \tan x = \sin x$, you will notice that the $y$ term is tied to a function of $x$ in a way that makes separation of variables impossible. The $2y \tan x$ term prevents us from isolating $dy$ and $dx$ on opposite sides. The Integrating Factor method is specifically designed for this 'linear' structure, acting as a mathematical 'glue' that allows us to collapse the left side into a single derivative of a product.

Q: Why does $\int \tan x dx$ become $\ln(\sec x)$? I thought it was $-\ln(\cos x)$.

You are absolutely correct! Both are mathematically equivalent. Since $\cos x = \frac{1}{\sec x}$, then $-\ln(\cos x) = -\ln((\sec x)^{-1}) = -(-1)\ln(\sec x) = \ln(\sec x)$. Using $\ln(\sec x)$ is often preferred in these problems because it avoids the negative sign, making the subsequent algebra with the Integrating Factor much cleaner.

Q: How do I know the maximum value of $y(x)$ occurs within the interval $(0, \frac{\pi}{2})$?

Great intuition! We transformed the problem into a quadratic $y = t - 2t^2$ where $t = \cos x$. Since $x$ is in $(0, \frac{\pi}{2})$, $t$ must be in $(0, 1)$. Our critical point was $t = \frac{1}{4}$. Since $\frac{1}{4}$ is clearly within the interval $(0, 1)$, the maximum we found is valid and exists within the specified domain of $x$.

Question

If $\frac{dy}{dx} + 2y 	an x = \sin x, 0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$, then the maximum value of $y(x)$ is

Options:

(A) $\frac{1}{8}$

(B) $\frac{3}{4}$

(C) $\frac{1}{4}$

(D) $\frac{3}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of JEE mastery. Today, we are going to dissect a problem that, at first glance, might seem like a chaotic mess of trigonometric functions.

But I want you to take a deep breath. In the world of differential equations, chaos is often just order in disguise. Let us peel back the layers of this problem together.

Look at the equation:

$$\frac{dy}{dx} + 2y 	an x = \sin x$$

Does it look familiar? It should! This is the classic, elegant structure of a first-order Linear Differential Equation:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

In this arena, our $P(x) = 2 	an x$ and our $Q(x) = \sin x$. Identifying these two components is like finding the North Star in a dark sky. Once you have them, the path forward becomes clear.

We are not just solving an equation; we are building a bridge between the derivative of a function and the function itself.

### The Magic of the Integrating Factor

Now, we need our secret weapon: the Integrating Factor (I.F.). The formula is defined as:

$$	ext{I.F.} = e^{\int P(x) dx} = e^{\int 2 	an x dx}$$

Here is where the beauty of calculus shines. We know that $\int 	an x dx = \ln(\sec x)$.

So, our exponent becomes $2 \ln(\sec x)$. Using the laws of logarithms, that coefficient of $2$ jumps up to become an exponent: $\ln(\sec^2 x)$.

And because $e$ and $\ln$ are inverse functions, they vanish, leaving us with the elegant result:

$$	ext{I.F.} = \sec^2 x$$

This factor is the key that unlocks the entire equation.

### The Integration

With our I.F. in hand, the general solution is given by:

$$y \cdot (	ext{I.F.}) = \int Q(x) \cdot (	ext{I.F.}) dx + C$$

Substituting our values, we get:

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

Do not be intimidated by that integral! If you rewrite $\sec^2 x$ as $\frac{1}{\cos^2 x}$, the integral becomes:

$$\int \frac{\sin x}{\cos^2 x} dx = \int 	an x \sec x dx$$

We know this integral by heart—it is $\sec x$. Thus, our general solution is:

$$y \sec^2 x = \sec x + C$$

### Finding the Constant and the Curve

We are given the initial condition $y(\frac{\pi}{3}) = 0$. This is our anchor.

Plugging in $x = \frac{\pi}{3}$ and $y = 0$, we find:

$$0 = \sec\left(\frac{\pi}{3}ight) + C$$

Since $\sec(\frac{\pi}{3}) = 2$, we get $0 = 2 + C$, which means $C = -2$.

Our specific solution is $y \sec^2 x = \sec x - 2$. Dividing by $\sec^2 x$, we get the beautiful function:

$$y = \cos x - 2 \cos^2 x$$

### The Quadratic Transformation

We are almost there. We need to maximize $y = \cos x - 2 \cos^2 x$ for $x \in (0, \frac{\pi}{2})$.

Instead of wrestling with derivatives of trigonometric functions, let us use a substitution. Let $t = \cos x$.

As $x$ ranges from $0$ to $\frac{\pi}{2}$, $t$ ranges from $1$ to $0$. Our function becomes a simple downward-opening parabola:

$$y = t - 2t^2$$

To find the maximum, we take the derivative with respect to $t$:

$$\frac{dy}{dt} = 1 - 4t$$

Setting this to zero, we find $t = \frac{1}{4}$. Since the second derivative is $-4$, we know this is a maximum.

Finally, substituting $t = \frac{1}{4}$ back into our quadratic, we get:

$$y = \frac{1}{4} - 2\left(\frac{1}{4}ight)^2 = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$$

And there it is. The **maximum value is $\frac{1}{8}$**. You have navigated the differential equation, mastered the integration, and conquered the optimization. Well done!

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} + 2 y tan x = sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is

Select Answer:

Visualized Solution

Identifying the Linear Differential Equation

Calculating the Integrating Factor

Setting up the General Solution

Solving the Integral

Finding the Constant $C$

Expressing $y$ as a Function of $x$

Maximizing the Function

Finding the Critical Point

Final Calculation of Maximum Value

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Integration

Finding the Constant and the Curve

The Quadratic Transformation

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​+2ytanx=sinx,0<x<2π​ and y(3π​)=0, then the maximum value of y(x) is