Q.If $x = \cos t, y = \cos mt$, then prove that $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2y = 0$.

Concept: Parametric Second Derivative — we compute $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ using the parameter $t$, then substitute into the given expression.

Step 1: First derivatives with respect to $t$:

Hence

Step 2: Second derivative:

First compute

Then divide by $-\sin t$:

Step 3: Substitute into $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2 y$.

Note $1 - x^2 = 1 - \cos^2 t = \sin^2 t$, $x = \cos t$, $y = \cos(mt)$.

First term: …

When a curve is given parametrically — here $x = \cos t$, $y = \cos mt$ — we cannot directly write $y$ as a function of $x$ in a simple closed form. But we can still find derivatives using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided $dx/dt \neq 0$. For the second derivative, we differentiate $\frac{dy}{dx}$ with respect to $t$ and then divide by $dx/dt$ again. This is the standard parametric second derivative formula.

The problem asks us to prove a differential equation involving $y$, $dy/dx$, and $d^2y/dx^2$, with coefficients in $x$. So we will compute these derivatives in terms of $t$, then replace $x$ and $y$ with their parametric forms, and simplify to zero.

Let's go step by step.

1. First derivatives with respect to $t$

   Given $x = \cos t$, so $\frac{dx}{dt} = -\sin t$.

   Given $y = \cos mt$, so $\frac{dy}{dt} = -m \sin mt$.

2. First derivative $\frac{dy}{dx}$

   Using the parametric formula:

Notice the negatives cancel neatly. This is valid wherever $\sin t \neq 0$, i.e., $t \neq n\pi$.

3. Second derivative $\frac{d^2y}{dx^2}$ We differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $dx/dt$:

First compute $\frac{d}{dt}\left( m \frac{\sin mt}{\sin t} \right)$:

using the quotient rule. So:

Now divide by $\frac{dx}{dt} = -\sin t$:

4. Substitute into the left-hand side of the given equation We need to verify:

Substitute $x = \cos t$, $y = \cos mt$, and the derivatives we found.

First, $1 - x^2 = 1 - \cos^2 t = \sin^2 t$.

So the first term becomes:

Simplify the numerator:

$= - \frac{ m^2 \cos mt \sin t - m \sin mt \cos t }{\sin t} = - m^2 \cos mt + m \frac{ \sin mt \cos t }{\sin t}.$ …