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 Question:
We are given the parametric equations of a curve. We need to evaluate a specific second-order differential expression involving $y$ with respect to $x$.

Step 2: Key Formula or Approach:
Find $\frac{dy}{dx}$ using the parametric rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Then, differentiate again with respect to $x$ using the chain rule: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}$.

Step 3: Detailed Explanation:
First, find the derivatives with respect to $t$:
$$x = \sin t \implies \frac{dx}{dt} = \cos t$$
$$y = \sin pt \implies \frac{dy}{dt} = p \cos pt$$
Now, find the first derivative $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{p \cos pt}{\cos t}$$
Now, differentiate $\frac{dy}{dx}$ with respect to $t$ using the quotient rule:
$$\frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{(\cos t)(-p^2 \sin pt) - (p \cos pt)(-\sin t)}{\cos^2 t}$$
$$\frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{-p^2 \sin pt \cos t + p \cos pt \sin t}{\cos^2 t}$$
Now, multiply by $\frac{dt}{dx} = \frac{1}{\cos t}$ to get $\frac{d^2y}{dx^2}$:
$$\frac{d^2y}{dx^2} = \left( \frac{-p^2 \sin pt \cos t + p \cos pt \sin t}{\cos^2 t} \right) \times \frac{1}{\cos t}$$
$$\frac{d^2y}{dx^2} = \frac{-p^2 y \cos t + \sin t (p \cos pt)}{\cos^3 t}$$
Notice that $p \cos pt = \frac{dy}{dx} \cos t$. Substitute this back in:
$$\frac{d^2y}{dx^2} = \frac{-p^2 y \cos t + \sin t \left( \frac{dy}{dx} \cos t \right)}{\cos^3 t}$$
Cancel out one $\cos t$ from the numerator and denominator:
$$\frac{d^2y}{dx^2} = \frac{-p^2 y + x \frac{dy}{dx}}{\cos^2 t}$$
Since $x = \sin t$, we know $\cos^2 t = 1 - \sin^2 t = 1 - x^2$. Substitute this:
$$\frac{d^2y}{dx^2} = \frac{-p^2 y + x \frac{dy}{dx}}{1 - x^2}$$
Multiply by $(1 - x^2)$ and rearrange the terms:
$$(1 - x^2) \frac{d^2y}{dx^2} = -p^2 y + x \frac{dy}{dx}$$
$$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$$

Step 4: Final Answer:
The entire expression evaluates exactly to 0, matching option (a).