Understanding the Problem: Forming a Differential Equation

The question asks us to find the differential equation that represents the given relation involving \(x\) and \(y\). The relation is \( \rm y = A- \frac{B}{x} \). A differential equation is an equation that relates a function with its derivatives. The key here is that the given relation contains arbitrary constants, \(A\) and \(B\). To find the differential equation, we need to eliminate these arbitrary constants by differentiating the given equation.

The number of arbitrary constants in the given relation tells us the order of the differential equation we expect to form. In this case, there are two constants, \(A\) and \(B\), so we anticipate a second-order differential equation.

Step-by-Step Solution: Eliminating Arbitrary Constants

Let's start with the given equation:

\( y = A - \frac{B}{x} \)

We can rewrite the term \( \frac{B}{x} \) using negative exponents for easier differentiation:

\( y = A - Bx^{-1} \)

Step 1: First Differentiation

Differentiate the equation with respect to \(x\). Remember that \(A\) and \(B\) are constants.

\( \frac{dy}{dx} = \frac{d}{dx}(A) - \frac{d}{dx}(Bx^{-1}) \)

\( \frac{dy}{dx} = 0 - B \cdot (-1)x^{-1-1} \)

\( \frac{dy}{dx} = Bx^{-2} \)

Using the notation \(y_1 = \frac{dy}{dx}\), we have:

\( y_1 = \frac{B}{x^2} \)

From this equation, we can isolate \(B\):

\( B = x^2 y_1 \)

This expression for \(B\) will be useful later to eliminate \(B\).

Step 2: Second Differentiation

Now, differentiate the first derivative (\(y_1\)) with respect to \(x\) to get the second derivative, \(y_2\). We differentiate \( y_1 = Bx^{-2} \).

\( \frac{d^2y}{dx^2} = \frac{d}{dx}(Bx^{-2}) \)

Since \(B\) is a constant, we can pull it out of the differentiation:

\( \frac{d^2y}{dx^2} = B \frac{d}{dx}(x^{-2}) \)

\( \frac{d^2y}{dx^2} = B \cdot (-2)x^{-2-1} \)

\( \frac{d^2y}{dx^2} = -2Bx^{-3} \)

Using the notation \(y_2 = \frac{d^2y}{dx^2}\), we have:

\( y_2 = -2Bx^{-3} \)

\( y_2 = \frac{-2B}{x^3} \)

Step 3: Eliminating the Constant B

We have an expression for \(B\) from Step 1: \( B = x^2 y_1 \). Now substitute this expression for \(B\) into the equation from Step 2:

\( y_2 = \frac{-2(x^2 y_1)}{x^3} \)

Simplify the expression:

\( y_2 = \frac{-2x^2 y_1}{x^3} \)

\( y_2 = -2 y_1 x^{2-3} \)

\( y_2 = -2 y_1 x^{-1} \)

\( y_2 = \frac{-2y_1}{x} \)

Step 4: Rearranging into the Differential Equation

Now, rearrange the equation \( y_2 = \frac{-2y_1}{x} \) to get the differential equation in a standard form, usually with all terms on one side.

Multiply both sides by \(x\):

\( x \cdot y_2 = x \cdot \frac{-2y_1}{x} \)

\( xy_2 = -2y_1 \)

Move the term \(-2y_1\) to the left side:

\( xy_2 + 2y_1 = 0 \)

This is the differential equation corresponding to the given relation \( y = A - \frac{B}{x} \).

Verifying Options

Let's look at the possible options provided and compare them with our derived differential equation \( xy_2 + 2y_1 = 0 \).

Our derived equation matches Option 2.

Revision Table: Key Steps in Forming Differential Equations

Additional Information: Concepts in Differential Equations

Forming differential equations from given general solutions is a fundamental concept. The general solution contains arbitrary constants, and the process of differentiation helps eliminate these constants to arrive at a unique differential equation that represents the entire family of curves given by the general solution.

• Order: The order of a differential equation is the order of the highest derivative appearing in the equation. If the given relation has \(n\) arbitrary constants, the resulting differential equation will be of order \(n\). In our problem, there were two constants \(A\) and \(B\), and the resulting differential equation \( xy_2 + 2y_1 = 0 \) is of order 2 (because of the \(y_2\) term).
• Degree: The degree of a differential equation is the power of the highest order derivative, provided the equation is a polynomial in derivatives. In \( xy_2 + 2y_1 = 0 \), the highest order derivative is \(y_2\), and its power is 1. So the degree is 1.
• General Solution vs. Particular Solution: The given relation \( y = A - \frac{B}{x} \) is the general solution because it contains arbitrary constants. A particular solution is obtained by assigning specific values to these constants. The differential equation represents the family of all possible particular solutions derived from the general solution.