Problem 103 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant is (A) \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\) (B) \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\) (C) \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\) (D) \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)

Step-by-Step Solution

Answer

The differential equation is (C) \((x^{2} - y^{2}) y' = 2xy\).

Step 1: Understanding the Problem

We are tasked with finding the differential equation for the family of circles centered at \((0, a)\) with radius \(a\), given by the equation \(x^{2} + y^{2} - 2ay = 0\). Our goal is to eliminate the arbitrary constant \(a\) to find a relationship between \(x\), \(y\), and \(y'\) (the derivative of \(y\) with respect to \(x\)).

Step 2: Find Derivative with Respect to x

To eliminate the constant \(a\), first differentiate the given equation \(x^{2} + y^{2} - 2ay = 0\) with respect to \(x\). Use implicit differentiation:\[2x + 2y\frac{dy}{dx} - 2a \frac{dy}{dx} = 0.\]

Step 3: Simplify and Isolate Terms

Rearrange the differentiated equation to isolate terms involving \(\frac{dy}{dx}\):\[2x + (2y - 2a)\frac{dy}{dx} = 0.\] Simplify to get:\[(2y - 2a)\frac{dy}{dx} = -2x.\]

Step 4: Solve for the Constant a

From the original equation \(x^{2} + y^{2} = 2ay\), solve for \(a\):\[a = \frac{x^{2} + y^{2}}{2y}.\] Substitute this back into the differentiated equation:\[\left(2y - 2\left(\frac{x^{2} + y^{2}}{2y}\right)\right)\frac{dy}{dx} = -2x.\]

Step 5: Simplify the Equation

Simplify the equation:\[2y - \left(\frac{x^{2} + y^{2}}{y}\right) = 0,\]\[2yy' - 2y\frac{x^{2} + y^{2}}{2y}y' = -2x.\]Multiply through by \(y\) to clear the fraction:\[\left(2y^{2} - x^{2} - y^{2}\right) y' = -2xy.\]This simplifies to:\[\left(y^{2} - x^{2}\right)y' = -2xy.\]

Step 6: Match with Options

Compare the simplified equation \((y^{2} - x^{2}) y' = -2xy\) with the given options. If we factor out a minus sign, this is equivalent to \((x^{2} - y^{2}) y' = 2xy\), corresponding to option (C).

Key Concepts

Implicit Differentiation

To solve the given problem, one must employ a technique called implicit differentiation. Often in calculus, we deal with equations where it's not easy or possible to isolate one variable. In these situations, implicit differentiation becomes a powerful tool.

When using implicit differentiation, you differentiate each term of the equation with respect to the same variable, often, with respect to \(x\). For example, when differentiating the equation \(x^2 + y^2 - 2ay = 0\), you apply the derivative:

• The derivative of \(x^2\) is \(2x\), as we apply the power rule directly.
• For \(y^2\), since \(y\) is a function of \(x\), we use the chain rule giving us \(2y\frac{dy}{dx}\).
• For \(-2ay\), which also involves \(y\), you get \(-2a\frac{dy}{dx}\).

After applying these rules, you obtain a new expression that relates the derivatives together. These new expressions can then be used to manipulate or simplify the original equation and eliminate unnecessary constants or solve for needed derivatives.

Eliminate Arbitrary Constant

The next step in the given solution involves eliminating the arbitrary constant, which is crucial in finding the differential equation for a family of curves. The arbitrary constant in our equation is \(a\), which originally represents the center of the circle along the y-axis.

Eliminating the constant means creating an equation that no longer includes \(a\). To do this, solve for \(a\) from the original equation, if possible:

• Using \(x^2 + y^2 = 2ay\), you can rearrange it to obtain \(a = \frac{x^2 + y^2}{2y}\).

Once \(a\) is expressed in terms of other variables, substitute this expression back into the differentiated equation. By removing \(a\), the equation now represents a relationship between \(x\), \(y\), and \(y'\) without the need for any constants, thus reflecting the behavior of the entire family of curves, rather than just a particular instance.

Family of Curves

In the realm of differential equations, a family of curves represents a set of curves that can be described by a single type of equation. An arbitrary constant within such an equation allows for different individual instances of curves, each defined by a specific value of that constant. In this problem, curves are described by circles centered on \((0, a)\) with radius \(a\).

When writing the differential equation for this family:

• The idea is to encapsulate all possible instances of these circles into one holistic formula.
• By converting through differentiation and elimination of \(a\), we obtain a generalized relationship applicable to every member of this family.

This resulting equation, post elimination, defines the broad artist's canvas on which any specific curve within the family can be drawn by assigning the appropriate conditions or values. In essence, it exemplifies the potential diversity of shapes or paths described by the differential equation.