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.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.