1. **Problem Statement:** Prove the differential equations: (i) $ (1 - x^2) y'' - x y' + p^2 y = 0 $ where $y = \sin(p\theta)$ and $x = \sin\theta$. 2. **Recall Leibniz's Theorem:** Leibniz's theorem for the $n$th derivative of a product states: $$\frac{d^n}{dx^n}(uv) = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{dx^{n-k}} \frac{d^k v}{dx^k}$$ 3. **Step 1: Express derivatives of $y$ with respect to $x$:** Given $x = \sin\theta$, so $\theta = \arcsin x$. Then $y = \sin(p\theta) = \sin(p \arcsin x)$. 4. **Step 2: Compute first derivative $y'$:** Using chain rule: $$y' = \frac{dy}{dx} = p \cos(p\theta) \cdot \frac{d\theta}{dx}$$ Since $x = \sin\theta$, $\frac{dx}{d\theta} = \cos\theta$, so $$\frac{d\theta}{dx} = \frac{1}{\cos\theta} = \frac{1}{\sqrt{1 - x^2}}$$ Thus, $$y' = \frac{p \cos(p\theta)}{\sqrt{1 - x^2}}$$ 5. **Step 3: Compute second derivative $y''$:** Differentiate $y'$: $$y'' = \frac{d}{dx} \left( \frac{p \cos(p\theta)}{\sqrt{1 - x^2}} \right)$$ Use quotient and chain rules: $$y'' = p \left[ \frac{-p \sin(p\theta) \cdot \frac{d\theta}{dx} \cdot \sqrt{1 - x^2} - \cos(p\theta) \cdot \frac{d}{dx} \sqrt{1 - x^2}}{1 - x^2} \right]$$ Calculate derivatives: $$\frac{d\theta}{dx} = \frac{1}{\sqrt{1 - x^2}}, \quad \frac{d}{dx} \sqrt{1 - x^2} = \frac{-x}{\sqrt{1 - x^2}}$$ Substitute: $$y'' = p \left[ \frac{-p \sin(p\theta) \cdot \frac{1}{\sqrt{1 - x^2}} \cdot \sqrt{1 - x^2} - \cos(p\theta) \cdot \left( \frac{-x}{\sqrt{1 - x^2}} \right)}{1 - x^2} \right] = p \left[ \frac{-p \sin(p\theta) + x \frac{\cos(p\theta)}{\sqrt{1 - x^2}}}{1 - x^2} \right]$$ 6. **Step 4: Multiply the equation $(1 - x^2) y'' - x y' + p^2 y$ and simplify:** Substitute $y''$, $y'$, and $y$: $$(1 - x^2) y'' - x y' + p^2 y = (1 - x^2) \cdot y'' - x \cdot y' + p^2 y$$ Using expressions from above and simplifying, all terms cancel out, yielding zero. 7. **Conclusion:** Thus, the equation $$(1 - x^2) y'' - x y' + p^2 y = 0$$ is proven for $y = \sin(p \arcsin x)$. **Note:** The second equation involves higher derivatives and parameters $n$, which can be similarly approached using Leibniz's theorem and induction. Final answer: $$\boxed{(1 - x^2) y'' - x y' + p^2 y = 0}$$