Question

The sum of the degree and order of differential equation \(\left(1 + y_1^2\right)^{\frac{2}{3}} = y_2\) is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The Correct Option is B

Given:
Differential equation: \[ \left(1 + y_1^2\right)^{\frac{2}{3}} = y_2 \] where \( y_1 = \frac{dy}{dx} \), and \( y_2 = \frac{d^2y}{dx^2} \)

Step 1: Identify the order 
The highest derivative is \( y_2 = \frac{d^2y}{dx^2} \), so:
Order = 2

Step 2: Identify the degree
Since the equation involves a fractional exponent, we must first remove the root by raising both sides to the power of 3:
\[ \left( \left(1 + y_1^2 \right)^{\frac{2}{3}} \right)^3 = y_2^3 \quad \Rightarrow \quad \left(1 + y_1^2\right)^2 = y_2^3 \] Now the equation is polynomial in the highest derivative \( y_2 \), and its highest power is 3, so:
Degree = 3

Sum of degree and order: \( 3 + 2 = 5 \)

Answer: 5

Answer 2

Given differential equation: \( \left(1 + y_1^2\right)^{\frac{2}{3}} = y_2 \) Let us rewrite using derivatives: \( y_1 = \frac{dy}{dx}, \quad y_2 = \frac{d^2y}{dx^2} \) So, the equation becomes: \( \left(1 + \left(\frac{dy}{dx}\right)^2\right)^{\frac{2}{3}} = \frac{d^2y}{dx^2} \) 

To find the degree, we need to make the equation free from fractional powers. Raise both sides to the power 3: \[ \left( \left(1 + \left(\frac{dy}{dx}\right)^2 \right)^{\frac{2}{3}} \right)^3 = \left( \frac{d^2y}{dx^2} \right)^3 \] \[ \left(1 + \left(\frac{dy}{dx}\right)^2\right)^2 = \left( \frac{d^2y}{dx^2} \right)^3 \] Now the equation is polynomial in derivatives. 

Order is the highest order derivative: \( \frac{d^2y}{dx^2} \) 

→ Order = 2 Degree is the power of highest order derivative: \( \left( \frac{d^2y}{dx^2} \right)^3 \) 

→ Degree = 3 Sum of degree and order = 3 + 2 = 5