Key Concepts and Formulas

• Order of a Differential Equation: The order of a differential equation is the order of the highest derivative present in the equation.
• Degree of a Differential Equation: The degree of a differential equation is the power of the highest order derivative, provided the equation is a polynomial in its derivatives (i.e., free from radicals and fractions involving derivatives).
• Elimination of Arbitrary Constants: To form a differential equation from a given relation involving arbitrary constants, differentiate the relation successively with respect to the independent variable until you have enough equations to eliminate the arbitrary constants. If there are n arbitrary constants, the order of the differential equation will be n.

Step-by-Step Solution

Step 1: Understanding the Given Equation

We are given the equation: $Ax^2 + By^2 = 1 \quad \ldots(1)$ Here, $A$ and $B$ are arbitrary constants. Our goal is to eliminate these constants by differentiating the equation and combining the results to form a differential equation. Since we have two arbitrary constants, we expect the resulting differential equation to be of the second order.

Step 2: First Differentiation with respect to $x$

Differentiate equation (1) with respect to $x$, remembering that $y$ is a function of $x$ and applying the chain rule: $\frac{d}{dx}(Ax^2) + \frac{d}{dx}(By^2) = \frac{d}{dx}(1)$ $2Ax + 2By\frac{dy}{dx} = 0$ Divide the entire equation by 2: $Ax + By\frac{dy}{dx} = 0 \quad \ldots(2)$

Step 3: Second Differentiation with respect to $x$

Differentiate equation (2) with respect to $x$, using the product rule where necessary: $\frac{d}{dx}(Ax) + \frac{d}{dx}\left(By\frac{dy}{dx}\right) = 0$ $A + B\left(\frac{dy}{dx}\cdot\frac{dy}{dx} + y\frac{d^2y}{dx^2}\right) = 0$ $A + B\left(\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2}\right) = 0 \quad \ldots(3)$

Step 4: Eliminating the Arbitrary Constant $A$

Solve equation (2) for $A$: $Ax = -By\frac{dy}{dx}$ $A = -\frac{By}{x}\frac{dy}{dx} \quad \ldots(4)$ Substitute this expression for $A$ into equation (3): $-\frac{By}{x}\frac{dy}{dx} + B\left(\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2}\right) = 0$ Since $B$ is an arbitrary constant and cannot be zero (otherwise $Ax^2=1$, which is a simpler family of curves, not the general one), divide the entire equation by $B$: $-\frac{y}{x}\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} = 0$

Step 5: Eliminating the Fraction

Multiply the entire equation by $x$ to eliminate the fraction: $-y\frac{dy}{dx} + x\left(\frac{dy}{dx}\right)^2 + xy\frac{d^2y}{dx^2} = 0$

Step 6: Rearranging Terms

Rearrange the terms to get the differential equation in a standard form: $xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0$

Step 7: Determining the Order and Degree

• Order: The highest order derivative in the equation is $\frac{d^2y}{dx^2}$, which is a second derivative. Therefore, the order of the differential equation is 2.
• Degree: The power of the highest order derivative $\frac{d^2y}{dx^2}$ is 1. Therefore, the degree of the differential equation is 1.

Common Mistakes & Tips

• Chain Rule and Product Rule: Be extremely careful when applying the chain rule and product rule during differentiation. These are frequent sources of error.
• Dividing by Constants: Ensure that you are not dividing by zero when eliminating arbitrary constants. Consider cases where the constant might be zero separately if necessary.
• Identifying Order and Degree: Remember that the degree is the power of the highest order derivative, not just the highest power of any derivative.

Summary

We started with the equation $Ax^2 + By^2 = 1$ and, by differentiating twice and eliminating the arbitrary constants $A$ and $B$, derived the differential equation $xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0$. This differential equation is of the second order and the first degree.

Final Answer

The differential equation is of second order and first degree, which corresponds to option (A). The final answer is \boxed{A}.