Finding the Differential Equation of Parabolas \( y^2 = 4a(x-b) \)

To find the differential equation of a family of curves, we need to eliminate the arbitrary constants present in the equation by differentiating it. The given equation for the family of parabolas is:

$$ y^2 = 4a(x-b) $$

This equation contains two arbitrary constants, 'a' and 'b'. Therefore, we need to differentiate the equation twice with respect to \( x \) to eliminate these constants.

Step-by-Step Derivation

First Differentiation

Differentiate the given equation with respect to \( x \). Remember to use the chain rule for \( y^2 \) and treat 'a' and 'b' as constants.

$$ \frac{d}{dx}(y^2) = \frac{d}{dx}(4a(x-b)) $$ $$ 2y \frac{dy}{dx} = 4a \frac{d}{dx}(x-b) $$ $$ 2y \frac{dy}{dx} = 4a (1 - 0) $$ $$ 2y \frac{dy}{dx} = 4a $$

We can simplify this equation:

$$ y \frac{dy}{dx} = 2a $$

This equation is free from the constant 'b'. Now we need to eliminate 'a'.

Second Differentiation

Differentiate the new equation \( y \frac{dy}{dx} = 2a \) with respect to \( x \). Use the product rule on the left side.

$$ \frac{d}{dx}\left(y \frac{dy}{dx}\right) = \frac{d}{dx}(2a) $$ $$ y \frac{d}{dx}\left(\frac{dy}{dx}\right) + \left(\frac{dy}{dx}\right) \frac{d}{dx}(y) = 0 $$ $$ y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right) \left(\frac{dy}{dx}\right) = 0 $$ $$ y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 $$

This resulting equation is the differential equation for the given family of parabolas. It involves \( y \), the first derivative \( \frac{dy}{dx} \), and the second derivative \( \frac{d^2y}{dx^2} \), and importantly, it does not contain the arbitrary constants 'a' or 'b'.

Comparing with Options

Let's compare the derived differential equation with the given options:

• Option 1: \( \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 \) (Incorrect)
• Option 2: \( \frac{d^2 y}{d x^2}+x^2\left(\frac{d y}{d x}\right)^2=0\) (Incorrect)
• Option 3: \(y^2 \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 \) (Incorrect)
• Option 4: \(y \frac{d^2 y}{d x^2}+\left(\frac{dy}{dx}\right)^2=0\) (Matches the derived equation)

The differential equation matching our result is \(y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0\).

Revision Table: Forming Differential Equations

Additional Information: Families of Curves and Differential Equations

A differential equation represents a relationship between a function and its derivatives. When we talk about the differential equation of a "family of curves," we are looking for a single equation that describes a property common to all members of that family, regardless of the specific values of the arbitrary constants that define individual members.

For the family of parabolas \( y^2 = 4a(x-b) \), the constants 'a' and 'b' determine the width and the horizontal position of the parabola's vertex, respectively. By differentiating and eliminating 'a' and 'b', we found a relationship \( y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 \) that must hold true for any parabola defined by the original form, regardless of the specific values of 'a' and 'b'. This differential equation captures the fundamental geometric or physical property that all these parabolas share. The order of the differential equation is equal to the number of arbitrary constants in the original equation of the family of curves, in this case, two.