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

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: Identify the Differential Form

Given the differential equation:\[\frac{x \, dx + y \, dy}{x \, dy - y \, dx} = \sqrt{\frac{a^{2} - x^{2} - y^{2}}{x^{2} + y^{2}}}\]We observe that the left side can be written as the differential form associated with polar coordinates.

Step 2: Convert to Polar Coordinates

In polar coordinates, we have: \[ x = r \cos \theta, \quad y = r \sin \theta \]Then, \[ dx = \cos \theta \, dr - r \sin \theta \, d\theta, \quad dy = \sin \theta \, dr + r \cos \theta \, d\theta \]Substitute these into the differential equation. The left side becomes\[\frac{x \, dx + y \, dy}{x \, dy - y \, dx} = \frac{r \, dr}{r^{2} \, d\theta} = \frac{dr}{r \, d\theta}\]

Step 3: Simplify the Expressions

Simplify the expression using the polar conversion:\[\frac{dr}{r \, d\theta} = \sqrt{\frac{a^{2} - r^{2}}{r^{2}}}\]This reduces to:\[\frac{dr}{r \, d\theta} = \frac{\sqrt{a^{2} - r^{2}}}{r}\]

Step 4: Separate Variables

Separate the variables to integrate:\[\frac{dr}{\sqrt{a^{2} - r^{2}}} = r \, d\theta\]The left side only involves \(r\), and the right side only involves \(\theta\).

Step 5: Integrate Both Sides

Integrate each side independently:\[\int \frac{dr}{\sqrt{a^{2} - r^{2}}} = \int r \, d\theta\]The left side integrates to:\[\sin^{-1} \left(\frac{r}{a}\right)\]and the right side integrates to:\[\frac{r^{2}}{2} + C\]where \(C\) is the constant of integration.

Step 6: Solve for the General Solution

From the integration result:\[\sin^{-1} \left(\frac{r}{a}\right) = \frac{r^{2}}{2} + C\]Reorganize to pull out \(r\):\[\sqrt{x^{2} + y^{2}} = a \cos \left(c + \tan^{-1} \left(\frac{y}{x}\right)\right)\]Matching this with the given options, we find the solution corresponds to option (A).

Key Concepts

Polar Coordinates

Understanding polar coordinates is essential when tackling this problem involving a differential equation. Polar coordinates transform Cartesian coordinates,

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

where \( r \) stands for the radius (or distance from the origin), and \( \theta \) is the angle. This approach is particularly useful in problems concerning circles or rotational symmetry.
In the context of the given differential equation, converting to polar coordinates helps simplify the equation by expressing \( x \) and \( y \) via \( r \) and \( \theta \), and allows the equation to be more manageable in terms of integration per variable. Understanding these basic conversions can greatly ease solving complex geometry-based problems.

Variable Separation

Variable separation is a powerful method for solving differential equations where each side of the equation can be written in terms of a single variable. In our example, after converting to polar coordinates,

• \( \frac{dr}{\sqrt{a^{2} - r^{2}}} = r \, d\theta \)

we see that the left-hand side contains only \( r \) while the right-hand side contains only \( \theta \). This setup is ideal for separation of variables.
The method involves isolating one variable on each side of the equation, enabling separate integration. This setup is crucial as it allows us to solve each piece independently, integrating one side with respect to \( r \) and the other with respect to \( \theta \), helping solve the differential equation more systematically.

Integration Techniques

For this differential equation, effective integration techniques make the difference between success and struggle. Once the variables are separated,

• the integral \( \int \frac{dr}{\sqrt{a^{2} - r^{2}}} \) corresponds to the inverse sine
• the integral \( \int r \, d\theta \) becomes a straightforward polynomial integration.

Understanding when and how to use these techniques is key.
The equation \( \sin^{-1} (\frac{r}{a}) \) emerges from the left side, a standard integral resulting from the substitution involving \( a \) and \( r \). On the right, the integration results in \( \frac{r^{2}}{2} + C \), a typical polynomial outcome where \( C \) is the constant of integration. This foundational skill is pivotal in handling more complex integration scenarios found in calculus.