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

Q: Wait, why do we get a constant $C = \frac{\pi}{2}$? Is it always that simple?

That is a great observation! In this specific problem, the initial condition $y(\frac{1}{2}) = \frac{\sqrt{3}}{2}$ leads to $\sin^{-1}(\frac{\sqrt{3}}{2}) + \sin^{-1}(\frac{1}{2}) = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}$. It is not always $\frac{\pi}{2}$, but in JEE problems, constants often simplify to standard angles like $\frac{\pi}{2}$, $\pi$, or $0$ to make the final algebraic manipulation elegant. Always trust your integration and substitution!

Q: Why did we get $y = \sqrt{1-x^2}$ and not $\pm \sqrt{1-x^2}$?

Excellent question! When we have $\sin^{-1}(y) = \cos^{-1}(x)$, we are looking for a function $y(x)$. Since $\sin(\cos^{-1}(x))$ is always positive for the range of $x$ given ($|x| < 1$), the positive root is the only one that satisfies the equation. The initial condition $y(\frac{1}{2}) = \frac{\sqrt{3}}{2}$ (which is positive) further confirms that we are on the upper branch of the semi-circle.

Q: Can we use substitution like $x = \sin \theta$ and $y = \sin \phi$ from the start?

Yes, absolutely! That is a very powerful way to think about it. If you substitute $x = \sin \theta$ and $y = \sin \phi$, the differential equation becomes $\cos \theta \frac{d(\sin \phi)}{dx} + \cos \phi = 0$, which simplifies beautifully. However, the variable separable method is generally more direct for this specific structure. Both paths lead to the same beautiful result!

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 $y(\frac{1}{2}) = \frac{\sqrt{3}}{2}$, then $y(\frac{-1}{\sqrt{2}})$ is equal to :

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The differential equation provided is:

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

At first glance, the $x$ and $y$ terms exhibit a striking symmetry. This suggests that the relationship between the variables is deeply geometric in nature.

### The Art of Separation

To solve this, we first isolate the derivative term:

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

Next, we separate the variables by moving all $y$-terms to the left and all $x$-terms to the right:

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

Note that the negative sign is preserved. In competitive mathematics, maintaining precision with signs is essential to avoid common pitfalls.

### The Integration

We now integrate both sides of the equation:

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

This yields the following inverse trigonometric relationship:

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

Rearranging the terms gives us the general solution:

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

### Finding the Constant

We are given the initial condition $y(\frac{1}{2}) = \frac{\sqrt{3}}{2}$. Substituting $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$ into our general solution:

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

Since $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ and $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$, we find:

$$C = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}$$

Thus, our specific curve is defined by $\sin^{-1}(y) + \sin^{-1}(x) = \frac{\pi}{2}$.

### The Geometric Revelation

Recall the fundamental trigonometric identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$. Comparing this to our specific solution, it is clear that:

$$\sin^{-1}(y) = \cos^{-1}(x)$$

Taking the sine of both sides, we obtain $y = \sin(\cos^{-1}(x))$. Using the identity $\sin(\cos^{-1}(x)) = \sqrt{1 - x^2}$, we identify the curve as:

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

This equation represents the upper half of a unit circle.

### The Final Step

We are tasked with finding $y$ when $x = -\frac{1}{\sqrt{2}}$. Substituting this value into our derived equation:

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

$$y = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}}$$

The final result is:

**$y = \frac{1}{\sqrt{2}}$**

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

Separate the Variables

Integrate Both Sides

General Solution

Apply Initial Condition

Calculate Constant $C$

Simplify the Relation

Algebraic Form of the Curve

Setup for Target Value

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration

Finding the Constant

The Geometric Revelation

The Final Step

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