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Answer the following.
If x = sin t, y = sin pt, prove that (1 - x²)(d²y)/(dx²) - x(dy)/(dx) + p² y = 0
If $x = \sin t, y = \sin pt$, prove that $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + p^2 y = 0$
Given: $x = \sin t, y = \sin pt$
To show: $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + p^2 y = 0$
Answer
Answer
Differentiate twice, using $x=\sin t$.
- 1. $x=\sin t$ gives $\sqrt{1-x^2}=\cos t$, and $y=\sin pt$.
- 2. $\frac{dx}{dt}=\cos t$ and $\frac{dy}{dt}=p\cos pt$, so $\frac{dy}{dx}=\frac{p\cos pt}{\cos t}$ (chain rule).
- 3. Hence $\sqrt{1-x^2}\,\frac{dy}{dx}=p\cos pt$.
- 4. Differentiate both sides with respect to $x$ (product rule): $\sqrt{1-x^2}\,\frac{d^2y}{dx^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=-p^2\sin pt\cdot\frac{dt}{dx}$.
- 5. Since $\frac{dt}{dx}=\frac{1}{\sqrt{1-x^2}}$ and $\sin pt=y$: $\sqrt{1-x^2}\,\frac{d^2y}{dx^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=-\frac{p^2y}{\sqrt{1-x^2}}$.
- 6. Multiply by $\sqrt{1-x^2}$: $(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=-p^2y$.
- 7. Hence $(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0$.