If y = y(x), y ∈ [0, π/2) is the solution of the differential equation sec y dy/dx - sin(x + y) - si

Q: Wait, why do we use the identity $\sin(A + B) + \sin(A - B) = 2\sin A \cos B$?

Great question! In differential equations, especially in JEE problems, the goal is often to separate variables. The original expression $\sin(x + y) + \sin(x - y)$ mixes $x$ and $y$ in a way that makes separation impossible. By using this identity, we transform the sum into a product, which allows us to isolate the $x$ and $y$ terms perfectly. It is a standard technique to look for such identities when you see mixed arguments in trigonometric functions.

Q: Can I just integrate $\sec y \frac{dy}{dx}$ directly?

Not directly, because $\sec y$ is a function of $y$, and $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$. You need to have the $y$ terms with $dy$ and the $x$ terms with $dx$ before you can integrate. That is why we moved $\cos y$ to the left side to form $\sec^2 y dy$.

Q: Why do we differentiate the particular solution instead of solving for $y$ first?

Solving for $y$ would require taking the inverse tangent, which leads to $y = \arctan(2 - 2\cos x)$. Differentiating this would involve the chain rule and could get messy. It is much cleaner and less prone to error to differentiate the implicit equation $\tan y = 2 - 2\cos x$ directly with respect to $x$. This is a powerful strategy in calculus: differentiate implicitly whenever possible!

Question

If $y = y(x), y \in [0, \pi/2)$ is the solution of the differential equation $\sec y \frac{dy}{dx} - \sin(x + y) - \sin(x - y) = 0$, with $y(0) = 0$, then $5y'\left(\frac{\pi}{2}\right)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful world of differential equations. Today, we are going to tackle a problem that might look like a tangled mess of trigonometric functions, but it is actually a choreographed dance.

Consider the equation:
$$\sec y \frac{dy}{dx} - \sin(x + y) - \sin(x - y) = 0$$

The secret to mastering JEE Advanced mathematics is to see past the complexity and find the underlying structure. Let us begin by rearranging our terms to isolate the derivative $\frac{dy}{dx}$:
$$\sec y \frac{dy}{dx} = \sin(x + y) + \sin(x - y)$$

### Simplifying the Trigonometry

Take a deep breath and observe the pattern. This is a classic trigonometric setup involving the sum of two sines with mixed arguments.

We apply the product-to-sum identity:
$$\sin(A + B) + \sin(A - B) = 2\sin A \cos B$$

When we apply this, the right side of our equation collapses into something elegant:
$$\sec y \frac{dy}{dx} = 2\sin x \cos y$$

### The Art of Separation

Now that we have simplified the expression, we are ready for the next phase: separation of variables. We move the $\cos y$ to the left side and the $dx$ to the right.

Recall that $\frac{1}{\cos y} = \sec y$. Our left side becomes $\sec y \cdot \sec y \cdot dy$, which simplifies to $\sec^2 y \, dy$. The right side becomes $2\sin x \, dx$.

The separated equation is:
$$\sec^2 y \, dy = 2\sin x \, dx$$

### The Integration and the Anchor

With our variables neatly separated, we integrate both sides. The integral of $\sec^2 y$ with respect to $y$ is $	an y$, and the integral of $2\sin x$ with respect to $x$ is $-2\cos x$.

Including the constant of integration $C$, we have:
$$	an y = -2\cos x + C$$

We are given the initial condition $y(0) = 0$. Substituting $x = 0$ and $y = 0$:
$$0 = -2\cos(0) + C \Rightarrow 0 = -2(1) + C \Rightarrow C = 2$$

Thus, our particular solution is:
$$	an y = 2 - 2\cos x$$

### The Final Precision

The problem asks for the value of $5y'(\frac{\pi}{2})$. First, we differentiate our implicit equation $	an y = 2 - 2\cos x$ directly with respect to $x$:
$$\sec^2 y \frac{dy}{dx} = 2\sin x$$

At $x = \frac{\pi}{2}$, we first find the value of $\sec^2 y$. From our particular solution:
$$	an y = 2 - 2\cos\left(\frac{\pi}{2}ight) = 2 - 0 = 2$$

Using the identity $\sec^2 y = 1 + 	an^2 y$, we find:
$$\sec^2 y = 1 + (2)^2 = 5$$

Now, substitute these values into the differentiated equation:
$$5 \cdot y'\left(\frac{\pi}{2}ight) = 2\sin\left(\frac{\pi}{2}ight)$$

Since $\sin(\frac{\pi}{2}) = 1$, we obtain:
$$5 \cdot y'\left(\frac{\pi}{2}ight) = 2$$

The final answer is **2**.

Animated Solution for Mathematics - Differential Equations: If $y = y (x), y \in [0, π /2)$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ , then $5 y^{'} (\frac{π}{2})$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Apply Trigonometric Identity

Variable Separation

Integrate Both Sides

Apply Initial Condition

The Particular Solution

Find y at $x = \frac{π}{2}$

Differentiate for y'

Final Calculation

Summary and Takeaway

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Simplifying the Trigonometry

The Art of Separation

The Integration and the Anchor

The Final Precision

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