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Given $x = \cos t$ and $y = \cos mt$. We find the derivatives with respect to $t$: $$ \frac{dx}{dt} = -\sin t $$ $$ \frac{dy}{dt} = -m \sin mt $$ Using the chain rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$: $$ \frac{dy}{dx} = \frac{-m \sin mt}{-\sin t} = \frac{m \sin mt}{\sin t} $$ Rearranging gives: $$ \sin t \frac{dy}{dx} = m \sin mt $$
Differentiate both sides with respect to $x$ using the product rule on the left side: $$ \frac{d}{dx} \left( \sin t \frac{dy}{dx} \right) = \frac{d}{dx} (m \sin mt) $$ $$ \sin t \frac{d^2y}{dx^2} + \frac{dy}{dx} \cos t \frac{dt}{dx} = m \cos mt \cdot m \frac{dt}{dx} $$ Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{-1}{\sin t}$, substitute this into the equation: $$ \sin t \frac{d^2y}{dx^2} + \frac{dy}{dx} \cos t \left( \frac{-1}{\sin t} \right) = m^2 \cos mt \left( \frac{-1}{\sin t} \right) $$
Multiply the entire equation by $-\sin t$: $$ -\sin^2 t \frac{d^2y}{dx^2} + \cos t \frac{dy}{dx} = m^2 \cos mt $$ Substitute $\sin^2 t = 1 - \cos^2 t = 1 - x^2$, $\cos t = x$, and $\cos mt = y$: $$ -(1 - x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = m^2 y $$ Rearranging the terms to the left side: $$ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2 y = 0 $$
Final Answer: The identity is proven.
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