Question

If \( p \) and \( q \) are respectively the order and degree of the differential equation \( \frac{d}{dx} \left( \frac{dy}{dx} \right)^3 = 0 \), then \( (p - q) \) is:

• A. \( 0 \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( 3 \)

Hint:

The order of a differential equation is determined by the highest derivative, and the degree is the highest power of the dependent variable.

Answer 1

The Correct Option is C

To solve the problem, we need to determine the order and degree of the given differential equation:

\( \frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^3 \right) = 0 \)

1. Simplify the Differential Equation:
Let us first simplify the expression:
Let \( u = \left( \frac{dy}{dx} \right)^3 \), then the equation becomes:
\( \frac{du}{dx} = 0 \)
That implies \( u \) is constant, so:
\( \left( \frac{dy}{dx} \right)^3 = C \), where \( C \) is constant

We were given:
\( \frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^3 \right) = 0 \)
We need to look at the highest order derivative involved in the equation as written — the derivative is acting on a power of \( \frac{dy}{dx} \), making the expression effectively include a second derivative when expanded.

Differentiating: \( \frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^3 \right) = 3 \left( \frac{dy}{dx} \right)^2 \cdot \frac{d^2y}{dx^2} \)
So, the highest derivative is \( \frac{d^2y}{dx^2} \), which means:

Order (p) = 2
The equation is polynomial in its highest order derivative (linear in \( \frac{d^2y}{dx^2} \)), so:
Degree (q) = 1

2. Final Calculation:
\( p - q = 2 - 1 = 1 \)

Final Answer:
The value of \( p - q \) is 1.