Let y = y(x) be a solution of the differential equation, √1 - x² dy/dx + √1 - y² = 0, |x|

Q: Wait, why can't I just integrate both sides immediately without moving the terms?

That is a fantastic question! If you try to integrate $\sqrt{1-x^2} \frac{dy}{dx}$ directly with respect to $x$, you are mixing variables. The differential operator $\frac{dy}{dx}$ implies that $y$ is a function of $x$. To integrate successfully, the differential $dx$ must only be associated with functions of $x$, and $dy$ with functions of $y$. Separating them is not just a rule; it is the physical requirement for the integration to be valid.

Q: Why does the constant $C$ end up being $\frac{\pi}{2}$?

Great observation. When we substitute the initial condition $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$, we get $\sin^{-1}(\frac{1}{2}) + \sin^{-1}(\frac{\sqrt{3}}{2}) = C$. Since $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ and $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$, their sum is $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. It is a beautiful coincidence of the specific values chosen for this problem!

Q: Is there a geometric way to think about this?

Absolutely! The equation $\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}$ is actually equivalent to $x^2 + y^2 = 1$ for the first quadrant. You are essentially tracing a path along a unit circle. When you solve for $y$ at a specific $x$, you are simply finding the corresponding coordinate on that circle.

Question

Let $y = y(x)$ be a solution of the differential equation, $\sqrt{1 - x^2} \frac{dy}{dx} + \sqrt{1 - y^2} = 0$, $|x| < 1$. If $y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$, then $y\left(\frac{1}{\sqrt{2}}\right)$ is equal to :

Options:

(A) $-\frac{1}{\sqrt{2}}$

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

(C) $\frac{1}{\sqrt{2}}$

(D) $\frac{1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a problem that, at first glance, might look like a tangled mess of square roots and derivatives.

In the world of JEE Advanced, the most intimidating problems are often just simple concepts wearing a disguise. Our equation is:

$$\sqrt{1 - x^2} \frac{dy}{dx} + \sqrt{1 - y^2} = 0$$

It looks like a wall, but we are going to walk right through it.

### The Art of Separation

The first instinct of a master problem solver is to look for symmetry. Notice how the $x$ terms and $y$ terms are neatly tucked away in their own square roots? This is a massive hint that we are dealing with a **Variable Separable** differential equation.

We start by shifting the $y$ term to the right side:

$$\sqrt{1 - x^2} \frac{dy}{dx} = -\sqrt{1 - y^2}$$

Now, we perform the cross-multiplication. We bring the $dx$ to the right and push the square root of $1-y^2$ to the left. Suddenly, the equation transforms into something elegant:

$$\frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}}$$

Look at that! The variables are perfectly separated. We have successfully isolated the chaos.

### The Calculus Bridge

Now that we have separated our variables, we need to bridge the gap between the differential world and the algebraic world. We apply the integral sign to both sides:

$$\int \frac{dy}{\sqrt{1 - y^2}} = -\int \frac{dx}{\sqrt{1 - x^2}}$$

Do you recognize these forms? These are the standard integrals that every JEE aspirant should have etched into their memory. The integral of $\frac{1}{\sqrt{1-t^2}}$ is simply $\sin^{-1} t$.

Applying this, we get:

$$\sin^{-1} y = -\sin^{-1} x + C$$

We have arrived at the general solution. However, we still have a constant $C$ that is currently a mystery.

### The Constant of Destiny

We are given an initial condition: $y(\frac{1}{2}) = \frac{\sqrt{3}}{2}$. This is our anchor, telling us exactly where our curve sits in the coordinate plane.

Let's substitute $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$ into our equation:

$$\sin^{-1}\left(\frac{1}{2}\right) + \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = C$$

Now, recall your trigonometry. At what angle is the sine value $\frac{1}{2}$? That is $\frac{\pi}{6}$. For $\frac{\sqrt{3}}{2}$, the angle is $\frac{\pi}{3}$.

Adding these together, $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. Our constant $C$ is exactly $\frac{\pi}{2}$. Our particular solution is now locked in:

$$\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}$$

### The Final Reveal

The problem asks us to find $y$ when $x = \frac{1}{\sqrt{2}}$. We simply plug this value into our particular solution:

$$\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + \sin^{-1} y = \frac{\pi}{2}$$

We know that $\sin^{-1}(\frac{1}{\sqrt{2}})$ is $\frac{\pi}{4}$. So, the equation becomes:

$$\frac{\pi}{4} + \sin^{-1} y = \frac{\pi}{2}$$

Subtracting $\frac{\pi}{4}$ from both sides, we get $\sin^{-1} y = \frac{\pi}{4}$. Finally, taking the sine of both sides, we find:

**$$y = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$**

And there it is! We started with a complex differential equation and ended with a clean, elegant result. It wasn't about brute force; it was about recognizing the patterns and following the logic.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0$ , $∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{1}{2})$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Identify the Method

Rearrange the Terms

Separate the Variables

Set Up the Integration

Perform the Integration

Rearrange to Standard Form

Apply the Initial Condition

Evaluate Inverse Trigonometric Values

Calculate the Constant $C$

Write the Particular Solution

Set Up for the Final Goal

Evaluate the Known Term

Solve for $sin^{- 1} y$

Find the Final Value of $y$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Calculus Bridge

The Constant of Destiny

The Final Reveal

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