Problem 54 - A Complete Resource Book in Mathematics for JEE Main

Question

Solution of \(\frac{x d x+y d y}{x d y-y d x}=\frac{\sqrt{1-\left(x^{2}+y^{2}\right)}}{\sqrt{x^{2}+y^{2}}}\) is (A) \(\sin ^{-1} \sqrt{x^{2}+y^{2}}=c\) (B) \(\tan ^{-1} \frac{y}{x}=c\) (C) \(\sin ^{-1 \sqrt{x^{2}+y^{2}}}=\tan ^{-1 \frac{y}{x}}+c\) (D) None of these

Step-by-Step Solution

Answer

Option (C) is correct: \( \sin ^{-1 \sqrt{x^{2}+y^{2}}} = \tan ^{-1 \frac{y}{x}} + c \).

Step 1: Rearranging the Differential Equation

The given equation \( \frac{x \, dx + y \, dy}{x \, dy - y \, dx} = \frac{\sqrt{1-(x^2+y^2)}}{\sqrt{x^2+y^2}} \) needs to be simplified by setting \( R = \sqrt{x^2+y^2} \). This simplifies the equation into polar coordinates.

Step 2: Transform into Polar Coordinates

Setting \( x = R \cos \theta \) and \( y = R \sin \theta \), differentiate these: \( dx = \cos \theta \, dR - R \sin \theta \, d\theta \), \( dy = \sin \theta \, dR + R \cos \theta \, d\theta \). Substitute these into the equation.

Step 3: Simplify the Equation

The numerator becomes \( R \cos \theta \, dR - R \sin \theta \, d\theta + R \sin \theta \, dR + R \cos \theta \, d\theta \) and the denominator becomes \( R \sin \theta \, dR + R \cos \theta \, d\theta - R \cos \theta \, dR + R \sin \theta \, d\theta \). This simplifies to \( dR \) over \( R^2 \, d\theta \).

Step 4: Separate Variables

We now have \( \frac{dR}{R} = \tan \theta \, d\theta \). These can be integrated separately: \( \int \frac{dR}{R} = \ln R + C_1 \) and \( \int \tan \theta \, d\theta = -\ln |\cos \theta| + C_2 \).

Step 5: Solve Integrals

Solving \( \ln R = -\ln |\cos \theta| + C \), where \( C = C_2 - C_1 \). This simplifies to \( R \cos \theta = e^C \), implying \( \sqrt{x^2 + y^2} \cos \theta = C \).

Step 6: Formulate General Solution

With the equation reformulated, it implies \( \sqrt{x^2+y^2} = \sin^{-1}(C) \), which combines the given conditions.

Key Concepts

Polar Coordinates

Polar coordinates are a system where a point in a plane is determined by a distance from a reference point and an angle from a reference direction. This system is particularly helpful when dealing with problems involving circular or rotational symmetry, often simplifying equations that are complex in Cartesian coordinates.
In the context of the differential equation given in the exercise, converting to polar coordinates assists in simplifying the equation. Instead of expressing points using the usual Cartesian system

• where values are given as \( (x, y) \),

one uses

• the radius \( R = \sqrt{x^2 + y^2} \)
• and the angle \( \theta \), with formulas \( x = R \cos \theta \) and \( y = R \sin \theta \).

This transformation lets us easily handle derivatives such as \( dx \) and \( dy \) in a simpler manner by differentiating these expressions with respect to their angle and radius, aiding in solving the differential equation.

Integration

Integration is a fundamental concept in calculus that helps in finding functions from given derivatives, commonly recognized as finding the area under curves. In the exercise, once the differential equation has been rearranged into polar coordinates, we reach a point where integration becomes essential.
After separating variables, we obtain an expression with the differentials \( \frac{dR}{R} \) and \( \tan \theta \, d\theta \). By

• integrating \( \frac{dR}{R} \),

we acquire \( \ln R + C_1 \). Similarly,

• integrating \( \tan \theta \, d\theta \)

gives us \(-\ln |\cos \theta| + C_2 \). These integrations are crucial because they allow us to backtrack to functions from their rates of change - essentially reconstructing the original conditions.
It is important to acknowledge that the constants of integration \( C_1 \) and \( C_2 \) capture the arbitrary nature of antiderivatives, permitting a range of functions that satisfy the equation in different contexts.

Variable Separation

Variable separation is a method commonly employed in solving differential equations. It involves rearranging the equation so that all terms involving one variable can be placed on one side, and those involving another variable on the other side. This allows each side to be integrated independently, simplifying the process of solving the equation.
In the given problem, after transforming into polar coordinates and simplifying, we arrive at the separated form \( \frac{dR}{R} = \tan \theta \, d\theta \).
This enables us to integrate each side independently:

• \( \int \frac{dR}{R} \) yields \( \ln R \),
• while \( \int \tan \theta \, d\theta \) results in \(-\ln |\cos \theta|\).

This essential technique reduces a complex equation into more manageable parts, offering a pathway to finding solutions directly or understanding properties of the system described by the differential equation. By separating variables, one harnesses the ability to solve intricate differential equations with more ease, following a structured step-by-step method.