Let y = y(x) be the solution of the differential equation cosec² x dy + 2dx = (1 + y cos 2x) cosec²

Q: Wait, why did we divide by $\text{cosec}^2 x$ at the very beginning?

Great question! In differential equations, our primary goal is to isolate the derivative term, $\frac{dy}{dx}$. The original equation was cluttered with $\text{cosec}^2 x$ attached to $dy$ and $dx$. By dividing by $\text{cosec}^2 x$ and $dx$, we clean up the equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, which allows us to use the powerful Integrating Factor method.

Q: How do I know for sure that this is a linear differential equation?

A first-order linear differential equation always follows the pattern $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$ only. After our simplification, we arrived at $\frac{dy}{dx} - y \cos 2x = \cos 2x$. Here, $P(x) = -\cos 2x$ and $Q(x) = \cos 2x$. Since both are functions of $x$, it fits the definition perfectly!

Q: Why does the constant $C$ end up being $e^{-1/2}$?

That comes from applying the initial condition $y(\frac{\pi}{4}) = 0$. When we substitute $x = \frac{\pi}{4}$ into our general solution, the term involving $y$ becomes zero. We are left with an equation where $C$ is isolated. Since $\sin(2 \cdot \frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$, the exponential term becomes $e^{-1/2}$. Solving for $C$ gives us exactly $e^{-1/2}$.

Question

Let $y = y(x)$ be the solution of the differential equation $	ext{cosec}^2 x dy + 2dx = (1 + y \cos 2x) 	ext{cosec}^2 x dx$, with $y(\frac{\pi}{4}) = 0$. Then, the value of $(y(0) + 1)^2$ is equal to:

Options:

(A) $e^{1/2}$

(B) $e^{-1/2}$

(C) $e^{-1}$

(D) $e$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a seemingly intimidating differential equation. In the world of JEE Advanced, problems are rarely just about calculation; they are about pattern recognition and the elegance of transformation.

Let us look at our given equation:
$$	ext{cosec}^2 x \, dy + 2 \, dx = (1 + y \cos 2x) 	ext{cosec}^2 x \, dx$$

At first glance, it looks like a mess of trigonometric functions and differentials. But remember, every complex problem is just a simple one waiting to be revealed.

### Phase 1: The Cleanup

Our first mission is to standardize. We want to see this in the form $\frac{dy}{dx} + P(x)y = Q(x)$.

To get there, we divide the entire equation by $dx \cdot 	ext{cosec}^2 x$. The term $	ext{cosec}^2 x \, dy$ becomes $\frac{dy}{dx}$.

The term $2 \, dx$ becomes $\frac{2}{	ext{cosec}^2 x}$, which is $2 \sin^2 x$. On the right side, the $	ext{cosec}^2 x$ terms cancel out, leaving us with $1 + y \cos 2x$.

So, we have:
$$\frac{dy}{dx} + 2 \sin^2 x = 1 + y \cos 2x$$

### Phase 2: The Linear Reveal

Now, let us group the $y$ terms. Moving $y \cos 2x$ to the left, we get:
$$\frac{dy}{dx} - y \cos 2x = 1 - 2 \sin^2 x$$

Here is where the mathematical intuition kicks in. Do you recognize $1 - 2 \sin^2 x$? It is the classic double angle identity for cosine: $\cos 2x$.

Our equation simplifies beautifully to:
$$\frac{dy}{dx} - y \cos 2x = \cos 2x$$

This is a textbook Linear Differential Equation where $P(x) = -\cos 2x$ and $Q(x) = \cos 2x$.

### Phase 3: The Integrating Factor

To solve this, we need the 'magic key'—the Integrating Factor ($I.F.$). The formula is $I.F. = e^{\int P(x) \, dx}$.

Substituting our $P(x)$, we calculate $e^{\int -\cos 2x \, dx}$. The integral of $-\cos 2x$ is $-\frac{\sin 2x}{2}$.

Thus, our $I.F.$ is:
$$I.F. = e^{-\frac{\sin 2x}{2}}$$

This factor is the bridge that allows us to integrate the left side as a product rule derivative.

### Phase 4: The Integration

The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$. Plugging in our values, we get:
$$y e^{-\frac{\sin 2x}{2}} = \int \cos 2x \, e^{-\frac{\sin 2x}{2}} \, dx + C$$

This integral might look scary, but substitution is our best friend here. Let $t = -\frac{\sin 2x}{2}$. Then $dt = -\cos 2x \, dx$, or $\cos 2x \, dx = -dt$.

The integral becomes:
$$\int e^t (-dt) = -e^t = -e^{-\frac{\sin 2x}{2}}$$

Our general solution is now:
$$y e^{-\frac{\sin 2x}{2}} = -e^{-\frac{\sin 2x}{2}} + C$$

### Phase 5: The Final Stretch

We are almost there! We use the initial condition $y(\frac{\pi}{4}) = 0$. Substituting $x = \frac{\pi}{4}$ and $y = 0$, we find:
$$0 = -e^{-1/2} + C \implies C = e^{-1/2}$$

Now we have the particular solution:
$$y e^{-\frac{\sin 2x}{2}} = -e^{-\frac{\sin 2x}{2}} + e^{-1/2}$$

Dividing by the exponential term, we get:
$$y(x) = -1 + e^{-1/2} e^{\frac{\sin 2x}{2}}$$

At $x = 0$, $\sin 0 = 0$, so $y(0) = -1 + e^{-1/2}$. The question asks for $(y(0) + 1)^2$.

Substituting our value, we get:
$$(-1 + e^{-1/2} + 1)^2 = (e^{-1/2})^2 = e^{-1}$$

And there it is—the elegance of the final answer! Keep practicing, and you will see this beauty in every problem. The final result is **$e^{-1}$**.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $cosec^{2} x d y + 2 d x = (1 + y cos 2 x) cosec^{2} x d x$ , with $y (\frac{π}{4}) = 0$ . Then, the value of $(y (0) + 1)^{2}$ is equal to:

Select Answer:

Visualized Solution

Problem Setup

Standardizing the Equation

Rearranging and Simplifying

Linear Differential Equation

Integrating Factor (I.F.)

General Solution Setup

Solving the Integral

Applying Initial Condition

Evaluating $y (0)$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

If $y = y (x)$ is the solution of the differential equation $4 - x^{2} \frac{d y}{d x} = ((sin^{- 1} (\frac{x}{2}))^{2} - y) sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2$ , $y (2) = \frac{π ^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to

Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ is equal to :

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ is

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

Let $y = f (x)$ is the solution of the differential equation $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to: