y = x + \(\sqrt{a^2 + x^2}\)
\(\frac{dy}{dx}\) = 1 + \(\frac{2x}{2\sqrt{a^2+x^2}}\)
= 1 + \(\frac{x}{\sqrt{a^2+x^2}}\)
\(\frac{d^2y}{dx^2}\) = \(\frac{\sqrt{a^2+x^2}\times 1-x \times 2x/(2\sqrt{a^2+x^2})}{(a^2+x^2)}\)
= \(\frac{a^2+x^2-x^2}{(a^2+x^2)^{3/2}}\)
= \(\frac{a^2}{(a^2+x^2)^{1/2}(a^2+x^2)}\)
⇒ (a2 + x2) \(\frac{d^2y}{dx^2}\) = \(\frac{a^2}{\sqrt{a^2+x^2}}\)
Now,
(a2 + x2) \(\frac{d^2y}{dx^2}\) + x\(\frac{dy}{dx}\) - y = \(\frac{a^2}{\sqrt{a^2+x^2}}\) + x + \(\frac{x^2}{\sqrt{a^2+x^2}}\) - (x + \(\sqrt{a^2+x^2)}\)
= \(\frac{a^2+x^2}{\sqrt{a^2+x^2}}\) - \(\sqrt{a^2+x^2}\)
= \(\sqrt{a^2+x^2}\) - \(\sqrt{a^2+x^2}\) = 0
= R.H.S
Hence proved.