Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The differential equation \(\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}\) determines a family of circles with (A) variable radius and fixed centre (B) variable radius and variable centre (C) fixed radius and variable centre on \(x\)-axis (D) fixed radius and variable centre on \(y\)-axis

Step-by-Step Solution

Answer

The answer is (C): fixed radius and variable centre on the x-axis.

Step 1: Separate the Variables

To solve the differential equation \(\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}\), first separate the variables. Multiply both sides by \(y\) and divide by \(\sqrt{1-y^2}\) to obtain:\[ y\, dy = \sqrt{1-y^2}\, dx. \]

Step 2: Integrate Both Sides

Integrate both sides of the equation from Step 1. Start with the left side:\[ \int y\, dy = \int \sqrt{1-y^2}\, dx. \]The left side integrates to \(\frac{y^2}{2} + C_1\), and the right side requires substitution. Let \(u = \sin^{-1}(y)\), thus \(du = \frac{dy}{\sqrt{1-y^2}}\), and the integral becomes \(x + C_2\). This yields \[ \frac{y^2}{2} = x + C. \]

Step 3: Analyze the Equation

Simplify the integrated equation to relate it to the form of a circle equation. Multiply through by 2 to clear fractions:\[ y^2 = 2x + C. \]This equation resembles the general form of a circle, \((x - a)^2 + (y - b)^2 = r^2\), specifically showing a pattern of a fixed radius and variable center.

Step 4: Identify Solution Type

The derived equation \(y^2 = 2x + C\) indicates a family of circles centered on the x-axis. This is because the equation simplifies further to reveal a moving center on one of the axes with a constant radius.

Key Concepts

Separation of Variables

Separation of variables is a powerful method used to solve differential equations, especially when the equation can be algebraically manipulated to isolate one variable and its derivative from another variable and its derivative. Let's consider the given differential equation:
\( \frac{d y}{d x} = \frac{\sqrt{1-y^2}}{y} \).
The goal is to rearrange it so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other.

• Start by multiplying both sides by \( y \) to eliminate the fraction on the right: \( y\, dy = \sqrt{1-y^2}\, dx \).
• This operation leaves you with an equation where the variables are separated in terms of \( y \) and \( x \): one on each side of the equation.
• Now, it is easier to integrate both sides independently.

By separating the variables, we set the stage for subsequent integration, a crucial step in finding the solution to the original differential equation.

Integration Techniques

Integration techniques are essential for solving the separated differential equation to find a function \( y(x) \) that fits the given conditions. Once we have the separated equation:
\[ \int y\, dy = \int \sqrt{1-y^2}\, dx \]
we need to integrate both sides. Let's break it down:

• For the left side, \( \int y\, dy \), the integration is straightforward: it results in \( \frac{y^2}{2} + C_1 \).
• For the right side, \( \int \sqrt{1-y^2}\, dx \), applying a substitution such as \( u = \sin^{-1}(y) \) is useful. Thus, \( du = \frac{dy}{\sqrt{1-y^2}} \), which simplifies the integral to \( x + C_2 \).

Combining these results gives the equation:
\[ \frac{y^2}{2} = x + C \].
By mastering these integration techniques, complex and seemingly intimidating equations can be handled, revealing neat solutions like those in our problem.

Equation of a Circle

Understanding the equation of a circle is key to deducing the nature of the family of curves described by our differential equation. Typically, a circle’s equation is in the form:
\[ (x - a)^2 + (y - b)^2 = r^2 \]
where \((a, b)\) is the center and \( r \) is the radius. Let’s consider the transformed equation:
\[ y^2 = 2x + C \]

• This equation can be rearranged in terms of known circle properties, suggesting that \( (x - \frac{C}{2})^2 + y^2 = r^2 \) with adjustments, reveals properties of our solution.
• Here, it derives that the center of the equation shifts along the x-axis, represented by \( -\frac{C}{2} \) affecting the circle's center.
• The radius remains fixed based on the formulation of the differential equation.

The problem highlights that this shifted center remains on the x-axis, confirming that the center of the circle is variable, while the radius is consistent. Thus, recognizing and manipulating forms into the standard circle equation helps to identify these circle properties.