Question

Let a solution \(y=y(x)\) of the differential equation $$ x \sqrt{x^{2}-1} d y-y \sqrt{y^{2}-1} d x=0 $$ satisfy \(y(2)=\frac{2}{\sqrt{3}}\) STATEMENT-1: \(\quad y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)\) and STATEMENT-2: \(\quad y(x)\) is given by $$ \frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^{2}}} $$ (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

Short answer

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

Step-by-step solution

- Separating Variables

To solve the given differential equation, separate the variables by dividing both sides by \((x\sqrt{x^2-1})(y\sqrt{y^2-1})\): \[\frac{dy}{y\sqrt{y^2-1}} = \frac{dx}{x\sqrt{x^2-1}}\]

- Integrate Both Sides

Integrate both sides of the equation to find the form of \(\fy(x)\). On the left side, use the substitution \(\fw = \sqrt{y^2-1}\) and on the right side use \(\fu = \sqrt{x^2-1}\) to simplify the integrands. \[\int \frac{dy}{y\sqrt{y^2-1}} = \int \frac{dx}{x\sqrt{x^2-1}}\]

- Find the General Solution

The integrals on both sides will result in inverse secant functions, leading to the general solution: \[\sec^{-1}(y) = \sec^{-1}(x) + C\]

- Use Initial Condition

Apply the initial condition \(\fy(2) = \frac{2}{\sqrt{3}}\) to find the value of the constant \(\fC\). \[\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \sec^{-1}(2) + C\]

- Solve for \(\fC\) and \(\fy(x)\)

Use the identity \(\f\sec(\sec^{-1}(x)) = x\) to express \(\fy(x)\) in terms of \(\fx\). \[\sec\left(\sec^{-1}(y)\right) = \sec\left(\sec^{-1}(x) - \frac{\pi}{6}\right)\]

- Verify Statement-2

Verify the alternative form given by Statement-2 by transforming the general solution and finding an expression for \(\f1/y\).

- Evaluation of the Statements and Correct Explanation

Evaluate if Statement-1 and Statement-2 are true and if Statement-2 is a correct explanation for Statement-1 by comparing the derived expressions from the previous steps.

Key concepts

Separable Differential Equations

Separable differential equations are a class of differential equations that are particularly solvable because they allow the separation of variables. In these equations, the variables involved can be separated on opposite sides of the equation, each associated only with its derivative. This technique simplifies the process of finding the solution to the problem.

To tackle such an equation, one would rearrange the terms to isolate each variable's derivative on one side and then integrate to find the general solution. An example from the exercise is the equation \(\frac{dy}{y\sqrt{y^2-1}} = \frac{dx}{x\sqrt{x^2-1}}\), which is explicitly designed for separation of variables. The integration of both sides with respect to their respective variables leads to a solution involving inverse trigonometric functions, which brings us to the next critical concept of integrating these functions.

Integrating Inverse Trigonometric Functions

Integrating inverse trigonometric functions is often encountered when solving differential equations, particularly after separating variables. The inverse trigonometric functions, like \( \arcsin(x) \) and \( \arcsec(x) \), can be integrated by recognizing them from standard forms that you gradually become familiar with through practice.

In the exercise provided, we see integrals that lead to inverse secant functions. These functions arise from integrals like \( \int \frac{dx}{x\sqrt{x^2-1}} \) and are part of a standard set of integrals often taught in calculus courses. Proficiency in integrating these functions is crucial for students preparing for competitive exams like the JEE Advanced, where such problems are frequently posed.

Initial Condition in Differential Equations

The initial condition in differential equations is used to find the particular solution that specifically satisfies the differential equation at a given point. After obtaining the general solution, it often involves an arbitrary constant that must be determined. The initial condition gives a value of the function, or its derivatives, at a particular point.

With this information, you can solve for the constant, ensuring that your solution is not just general but also fits the specific scenario given in the problem. In our example, the initial condition \( y(2)=\frac{2}{\sqrt{3}} \) is applied to find the constant of integration \( C \) in Step 4, leading to the particular solution that satisfies the given differential equation not just for a range of values but specifically for \( x=2 \).

Verification of Differential Equation Solutions

Verification of differential equation solutions is the final step in confirming that the obtained function indeed satisfies the given differential equation. Once a potential solution is found, either through integration or other means, you must substitute it back into the original differential equation to ensure that both sides balance.

This step is crucial to validate your working process and your answer. It's similar to checking your work in arithmetic by reversing the operations. In the context of our given problem, after obtaining the potential solution, it is translated into different forms which are then verified against the given statements. This practice not only confirms the correctness of the solution but also deepens the understanding of the relationship between different representations of the solution and the original differential equation.