Let a solution y = y(x) of the differential equation x√(x²-1) dy - y√(y²-1) dx = 0 satisfy y(2) = 2/

Q: Wait, why is the integral of $\frac{1}{t\sqrt{t^2-1}}$ equal to $\sec^{-1} t$? I always forget if it's $\sec^{-1}$ or $\csc^{-1}$.

That is a very common point of confusion! Think about the derivative of $\sec^{-1} t$. It is $\frac{1}{|t|\sqrt{t^2-1}}$. Since we are dealing with positive values in this domain, we can safely write $\sec^{-1} t$. Just remember: Secant is the one that 'secures' the $t$ outside the root!

Q: Why do we need to expand $\cos(\sec^{-1} x - \frac{\pi}{6})$? Can't we just look at the expression and see if it matches Statement 2?

Great question! In JEE Advanced, 'eyeballing' is the enemy. Statement 2 is a specific algebraic form. To prove or disprove it, you must perform the expansion using the compound angle identity. Only by simplifying the expression into a form involving $x$ and $\sqrt{1 - \frac{1}{x^2}}$ can you definitively compare it to the statement provided.

Q: Is there a sign error in Statement 2? My calculation gave a plus sign, but the statement has a minus sign.

You are absolutely right to be suspicious! That is exactly the trap. By expanding the cosine identity carefully, you will find that the signs do not match the statement. This is why we verify every step—never assume the statement is true just because it looks complex!

Question

Let a solution $y = y(x)$ of the differential equation $x\sqrt{x^2-1} dy - y\sqrt{y^2-1} dx = 0$ satisfy $y(2) = \frac{2}{\sqrt{3}}$.

STATEMENT-1 : $y(x) = \sec(\sec^{-1} x - \frac{\pi}{6})$ and

STATEMENT-2 : $y(x)$ is given by $\frac{1}{y} = \frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}}$

Options:

(A) STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

(B) STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

(C) STATEMENT - 1 is True, STATEMENT - 2 is False

(D) STATEMENT - 1 is False, STATEMENT - 2 is True

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to dissect a problem that looks intimidating at first glance but unfolds with the elegance of a well-choreographed dance. We are dealing with the differential equation:

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

When you see an equation like this, your first instinct might be to panic. Don't. Instead, take a deep breath and look at the structure. We have $x$ and $y$ terms mixed together, and our mission is to separate them.

### Phase 1: The Separation

We rearrange the terms to get:

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

Now, we perform the magic of separation:

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

See how beautiful that is? We have isolated the $y$ world on the left and the $x$ world on the right. This is the heart of the **variable separable** method.

### Phase 2: The Integration

Now that the variables are separated, we integrate both sides:

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

Do you recognize these integrals? They are standard forms! The integral of $\frac{dt}{t\sqrt{t^2-1}}$ is simply $\sec^{-1} t$. So, our equation becomes:

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

This is our general solution. But we are not done yet; we have an initial condition: $y(2) = \frac{2}{\sqrt{3}}$. Let's plug these values in to find our constant $C$.

Substituting $x = 2$ and $y = \frac{2}{\sqrt{3}}$, we get:

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

Since $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$ and $\sec^{-1}(2) = \frac{\pi}{3}$, we have:

$$\frac{\pi}{6} = \frac{\pi}{3} + C \implies C = -\frac{\pi}{6}$$

### Phase 3: The Verification

Now we have our specific solution:

$$\sec^{-1} y = \sec^{-1} x - \frac{\pi}{6}$$

Taking the secant of both sides, we get $y = \sec\left(\sec^{-1} x - \frac{\pi}{6}\right)$. This matches Statement 1 perfectly, so **Statement 1 is true!**

Now, for the final test: Statement 2. We need to check if $\frac{1}{y} = \frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}}$. Since $y = \sec\left(\sec^{-1} x - \frac{\pi}{6}\right)$, then:

$$\frac{1}{y} = \cos\left(\sec^{-1} x - \frac{\pi}{6}\right)$$

Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$, where $A = \sec^{-1} x$ and $B = \frac{\pi}{6}$, we expand this to:

$$\frac{1}{y} = \cos(\sec^{-1} x) \cos\left(\frac{\pi}{6}\right) + \sin(\sec^{-1} x) \sin\left(\frac{\pi}{6}\right)$$

Substituting the values $\cos(\sec^{-1} x) = \frac{1}{x}$ and $\sin(\sec^{-1} x) = \sqrt{1 - \frac{1}{x^2}}$, we get:

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

Comparing this to Statement 2, we see it is clearly different. The signs and coefficients do not match. Therefore, **Statement 2 is false.**

Select Answer:

Visualized Solution

The Differential Equation

Separating Variables

Integration Setup

Standard Integral Formula

Applying Initial Condition

Solving for Constant $C$

Verifying Statement 1

Checking Statement 2

Trigonometric Expansion

Substituting Values

Final Comparison

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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