Question

If $x = \sin t$ and $y = \sin pt$, then the value of $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = \dots$

• A. 0
• B. 1
• C. -1
• D. $\sqrt{2}$

Hint:

This is the standard Chebyshev differential equation. Any function of the form $y = \sin(p \sin^{-1} x)$ or $y = \cos(p \cos^{-1} x)$ will always perfectly satisfy the equation $(1-x^2)y'' - xy' + p^2y = 0$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a problem of second-order differentiation of parametric functions. We need to find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$.

Step 2: Formula Application:
$\frac{dx}{dt} = \cos t$, $\frac{dy}{dt} = p \cos pt$. $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p \cos pt}{\cos t}$.

Step 3: Explanation:
$\cos t \frac{dy}{dx} = p \cos pt$. Differentiating w.r.t $x$: $-\sin t \frac{dt}{dx} \frac{dy}{dx} + \cos t \frac{d^2y}{dx^2} = -p^2 \sin pt \frac{dt}{dx}$ $-\sin t (\frac{1}{\cos t}) \frac{dy}{dx} + \cos t \frac{d^2y}{dx^2} = -p^2 \sin pt (\frac{1}{\cos t})$ Multiply by $\cos t$: $-x \frac{dy}{dx} + (1-x^2) \frac{d^2y}{dx^2} = -p^2 y$ Rearranging: $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$.

Step 4: Final Answer:
The value is 0.