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Prove the following.
If y = (x + √(1 + x²))ⁿ, then prove that (1 + x²) (d²y)/(dx²) + x (dy)/(dx) = n² y.
If $y = (x + \sqrt{1 + x^2})^n$, then prove that $(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n^2 y$.
Given: $y = (x + \sqrt{1 + x^2})^n$
To show: $(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n^2 y$
Answer
Answer
Differentiate $y = (x + \sqrt{1+x^2})^n$ twice and eliminate to reach the result.
- $y = (x + \sqrt{1+x^2})^n$. Differentiating by the chain rule: $\frac{dy}{dx} = n(x + \sqrt{1+x^2})^{n-1}\left(1 + \frac{x}{\sqrt{1+x^2}}\right)$.
- Simplifying: $1 + \frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}$, so $\frac{dy}{dx} = \frac{n(x + \sqrt{1+x^2})^n}{\sqrt{1+x^2}} = \frac{ny}{\sqrt{1+x^2}}$.
- Hence $\sqrt{1+x^2}\,\frac{dy}{dx} = ny$.
- Differentiating both sides with respect to $x$ (product rule): $\sqrt{1+x^2}\,\frac{d^2y}{dx^2} + \frac{x}{\sqrt{1+x^2}}\frac{dy}{dx} = n\frac{dy}{dx}$.
- Multiplying by $\sqrt{1+x^2}$: $(1+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = n\sqrt{1+x^2}\,\frac{dy}{dx}$.
- Substituting $\sqrt{1+x^2}\,\frac{dy}{dx} = ny$ from step 3: $(1+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = n \cdot ny = n^2 y$. Hence proved.