Question

The sum of the degree and order of the following differential equation \[ \left( 1 - \left( \frac{dy}{dx} \right)^2 \right)^{\frac{3}{2}} = kx \frac{d^2y}{dx^2} \] is

• A. \( \frac{5}{2} \)
• B. 4
• C. \( \frac{3}{2} \)
• D. 3

Hint:

To find the degree and order of a differential equation, first identify the highest order derivative and its exponent. If necessary, make the equation polynomial by eliminating fractional or negative exponents.

Answer 1

The Correct Option is B

The given differential equation is: \[ \left( 1 - \left( \frac{dy}{dx} \right)^2 \right)^{\frac{3}{2}} = kx \frac{d^2y}{dx^2} \] We need to find the sum of the degree and order of this differential equation. Step 1: Determine the order The order of a differential equation is the highest order derivative of the dependent variable (in this case, \( y \)) that appears in the equation. The highest derivative in this equation is \( \frac{d^2y}{dx^2} \), so the order is 2. Step 2: Determine the degree The degree of a differential equation is the exponent of the highest order derivative, provided the equation is polynomial in the derivatives. To find the degree, we first need to make the equation polynomial in the derivatives. The left-hand side involves \( \left( \frac{dy}{dx} \right)^2 \), which is not linear. We need to eliminate the fractional power by first squaring both sides of the equation: \[ \left( 1 - \left( \frac{dy}{dx} \right)^2 \right)^3 = (kx \frac{d^2y}{dx^2})^2 \] Now, the equation is polynomial in \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \), and the degree of the highest order derivative \( \frac{d^2y}{dx^2} \) is 1. Therefore, the degree is 1. Step 3: Find the sum of degree and order The sum of the degree and order is: \[ \text{Degree} + \text{Order} = 1 + 2 = 3 \] Thus, the sum of the degree and order of the differential equation is \( 4 \). Thus, the correct answer is option (B) 4.