Question

The differential equation whose solution is Ax²+By²=1 where A and B are arbitrary constant is of:
(A) first order and first degree
(B) second order and first degree
(C) second order and second degree
(D) second order
Choose the correct answer from the options given below:

• (D) Only
• (C) and (D) Only
• (B) and (D) Only
• (A) Only

Answer 1

The Correct Option is C

The solution to the given problem involves understanding the nature of the differential equation with the solution provided as \( Ax^2 + By^2 = 1 \), where \( A \) and \( B \) are arbitrary constants.

To determine the order and the degree of the differential equation, consider the following steps:

• Given the equation \( Ax^2 + By^2 = 1 \), we need to find the differential equation whose solution corresponds to this equation.
• To eliminate the arbitrary constants \( A \) and \( B \), differentiate the given equation with respect to its independent variable \( x \).
• First differentiation: \(\frac{d}{dx}(Ax^2 + By^2) = 0\) results in \(2Ax + 2By\frac{dy}{dx} = 0\). Simplifying, we get \(Ax + By\frac{dy}{dx} = 0\).
• Second differentiation: Differentiate again to eliminate both arbitrary constants. By differentiating, we get \(\frac{d}{dx}(Ax + By\frac{dy}{dx}) = 0\).
  This results in \(A + B\frac{d}{dx}(y\frac{dy}{dx}) = 0\), further simplified to \(A + B\left(\frac{dy}{dx}\right)^2 + By\frac{d^2y}{dx^2} = 0\).
• The final differential equation involves the second derivative \( \frac{d^2y}{dx^2} \), indicating it is of second order.
• The degree of a differential equation is the power of the highest order derivative present. Since the highest order derivative, \( \frac{d^2y}{dx^2} \), is not raised to any power other than 1, the degree is 1.

Therefore, the differential equation is of second order and first degree.

Correct Answer:

(B) and (D) Only