Question

Determine the order and degree of the following differential equation: \(\left(\dfrac{d^2y}{dx^2}\right)^3 + \sqrt{1+\left(\dfrac{dy}{dx}\right)^2=0.\)}

• A. Order 3, Degree 2
• B. Order 2, Degree 3
• C. Order 2, Degree 1
• D. Order 1, Degree 2

Hint:

Order is decided by the highest derivative. Degree is decided by the power of the highest order derivative.

Answer 1

The Correct Option is B

Concept:
The order of a differential equation is the order of the highest derivative present in the equation. The degree is the power of the highest order derivative when the equation is expressed in polynomial form with respect to derivatives.

Step 1: Write the given differential equation. \[ \left(\frac{d^2y}{dx^2}\right)^3 + \sqrt{1+\left(\frac{dy}{dx}\right)^2}=0 \]

Step 2: Identify the derivatives present.
The equation contains: \[ \frac{dy}{dx} \] and \[ \frac{d^2y}{dx^2} \]

Step 3: Find the order.
The highest order derivative present is: \[ \frac{d^2y}{dx^2} \] This is a second order derivative. \[ \text{Order} = 2 \]

Step 4: Find the degree.
The highest order derivative is: \[ \frac{d^2y}{dx^2} \] It appears with power \(3\): \[ \left(\frac{d^2y}{dx^2}\right)^3 \] Therefore, \[ \text{Degree} = 3 \] Thus, the order and degree are: \[ \text{Order } 2,\ \text{Degree } 3 \] \[ \therefore \text{Correct Answer is (B)} \]