Problem 59 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the differential equation \(\frac{x d x+y d y}{x d y-y d x}=\sqrt{\frac{a^{2}-x^{2}-y^{2}}{x^{2}+y^{2}}}\) is (A) \(\sqrt{x^{2}+y^{2}}=a \cos \left\\{c+\tan ^{-1} \frac{y}{x}\right\\}\) (B) \(\sqrt{x^{2}+y^{2}}=a \sin \left\\{c+\tan ^{-1} \frac{y}{x}\right\\}\) (C) \(\sqrt{x^{2}+y^{2}}=a \sin \left\\{c+\tan ^{-1} \frac{x}{y}\right\\}\) (D) None of these

Step-by-Step Solution

Answer

The solution is option (A): \( \sqrt{x^{2} + y^{2}} = a \cos \left(c + \tan^{-1} \frac{y}{x}\right) \).

Step 1: Understand the Given Differential Equation

The given equation is \( \frac{x \, dx + y \, dy}{x \, dy - y \, dx} = \sqrt{\frac{a^{2} - x^{2} - y^{2}}{x^{2} + y^{2}}} \). We need to identify the correct solution from the given options.

Step 2: Simplify the Differential Equation

First, recognize the left side of the equation as a relationship between differentials that can be reorganized. Rewrite it as: \( \frac{x}{y} \frac{dx}{dy} + \frac{y}{x} = \sqrt{\frac{a^{2} - x^{2} - y^{2}}{x^{2} + y^{2}}} \). This is a hint of separating the terms involving x and y.

Step 3: Introduce Polar Coordinates

Consider transforming the given variables into polar coordinates, where \( x = r \cos \theta \) and \( y = r \sin \theta \). Then, \( dx = dr \cos \theta - r \sin \theta \, d\theta \) and \( dy = dr \sin \theta + r \cos \theta \, d\theta \).

Step 4: Match with Polar Representation

Substitute these expressions for \( dx \) and \( dy \) into the equation. Upon substitution and after taking common factors, the expression rearranges to use polar terms: \( r = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1} \frac{y}{x} \).

Step 5: Compare to Solution Options

The form of equation in a polar coordinate context often simplifies the square root: \( r = a \cos (c + \theta) \), leading to \( \sqrt{x^2 + y^2} = a \cos (c + \tan^{-1} \frac{y}{x}) \). Compare this form to the given options.

Step 6: Select the Correct Option

By matching the derived polar form \( r = a \cos (c + \tan^{-1} \frac{y}{x}) \) with the provided options, the correct answer is (A).

Key Concepts

Polar Coordinates

Polar coordinates provide a way to understand points in a two-dimensional space using a different system than Cartesian coordinates. In this system, any point on a plane is described by the distance from a reference point and the angle from a reference direction. This can be particularly useful in dealing with problems involving circular or rotational symmetry.

In polar coordinates, a point is defined as \[ (r, \theta) \] where,

• \( r \) is the radial distance from the origin.
• \( \theta \) is the angle measured from the positive x-axis.

The conversion between Cartesian and polar coordinates is greatly useful in transforming differential equations. For example, if you have a point in Cartesian coordinates \( (x, y) \), the equivalent polar coordinates can be found using:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

This method simplifies many expressions, especially those involving equations of circles or ellipses, as it considers the symmetry directly.

Differential Equation Solution

Solving a differential equation often involves breaking down the differential components and identifying relationships between variables. In exercises like the one given, the process involves transforming expressions to a simpler form to find a solution. This typically includes:

- Identifying known forms of equations, such as Bernoulli’s or exact differential equations.
- Simplifying the expressions, often using substitution techniques like switching to polar coordinates.

In our given example, we had to rearrange and simplify the differential expression:\[ \frac{x \, dx + y \, dy}{x \, dy - y \, dx} = \sqrt{\frac{a^{2} - x^{2} - y^{2}}{x^{2} + y^{2}}} \]
By utilizing polar coordinates, where \( x = r \cos \theta \) and \( y = r \sin \theta \), this expression was devisable into polar terms making it easier to solve. This way, it becomes easier to integrate or match known solution forms, leading us to a correct option from the choices provided.

Mathematics JEE Mains

For students preparing for competitive exams like JEE Mains, it's crucial to understand the versatility of different mathematical concepts, including differential equations, polar coordinates, and problem-solving strategies.

JEE Mains often tests:

• Problem-solving ability across a wide range of topics.
• Ability to simplify complex expressions using substitutions.
• Underlying geometric and calculus principles, like using polar coordinates to simplify integrals.

In the context of our problem, recognizing that substituting with polar coordinates could simplify the differential equation was key. Such insight can save time and lead directly to the solution. Consistently practicing these transformations and understanding when they are applicable helps in mastering JEE-level math.

This highlights the importance of being able to relate different mathematical concepts and understand which approach simplifies a problem most effectively.