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Prove the following.
If y = e^(msin-1 x), prove that (1 - x²)(d²y)/(dx²) - x(dy)/(dx) = m² y.
If $y = e^{m\sin^{-1} x}$, prove that $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = m^2 y$.
Given: $y = e^{m\sin^{-1} x}$
To show: $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = m^2 y$
Answer
Answer
Hence $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = m^2 y$.
- $y = e^{m\sin^{-1}x}$. Differentiate by the chain rule: $\frac{dy}{dx} = e^{m\sin^{-1}x}\cdot\frac{m}{\sqrt{1 - x^2}} = \frac{my}{\sqrt{1 - x^2}}$.
- Rearrange: $\sqrt{1 - x^2}\,\frac{dy}{dx} = my$.
- Differentiate again, by the product rule: $\sqrt{1 - x^2}\,\frac{d^2y}{dx^2} - \frac{x}{\sqrt{1 - x^2}}\frac{dy}{dx} = m\frac{dy}{dx}$.
- Multiply by $\sqrt{1 - x^2}$: $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = m\sqrt{1 - x^2}\,\frac{dy}{dx}$.
- Using step 2, $\sqrt{1 - x^2}\,\frac{dy}{dx} = my$, so the right side is $m \cdot my = m^2y$.
- Hence $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = m^2 y$.