Question

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

Hint:

Find dy/dx = m sin(mt)/sin t, then d2y/dx2 using dx/dt = -sin t, and use 1 - x^2 = sin^2 t.

Answer 1

Step 1: Write y in terms of x:
Since $x = \cos t$, we have $t = \cos^{-1}x$, so $y = \cos(m\cos^{-1}x)$.
We will differentiate with respect to $x$ directly. We use $\frac{dt}{dx} = -\frac{1}{\sqrt{1-x^2}}$.

Step 2: First derivative:
By the chain rule, $\frac{dy}{dx} = -m\sin(mt)\cdot\frac{dt}{dx}$.
\[ \frac{dy}{dx} = \frac{m\sin mt}{\sqrt{1-x^2}} \]
Multiply both sides by the root to clear the denominator.
\[ \sqrt{1-x^2}\,\frac{dy}{dx} = m\sin mt \]

Step 3: Differentiate again:
Differentiate both sides with respect to $x$. The left side needs the product rule.
\[ \sqrt{1-x^2}\,\frac{d^2y}{dx^2} - \frac{x}{\sqrt{1-x^2}}\,\frac{dy}{dx} = m^2\cos mt\cdot\frac{dt}{dx} = -\frac{m^2\cos mt}{\sqrt{1-x^2}} \]

Step 4: Clear the root:
Multiply the whole equation by $\sqrt{1-x^2}$.
\[ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = -m^2\cos mt \]
Since $y = \cos mt$, the right side is $-m^2y$.

Step 5: Rearrange:
Move $-m^2y$ to the left side.
\[ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + m^2y = 0 \]

Final Answer:
The relation is proved. \[ \boxed{(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + m^2y = 0} \]