If \(y = \left(\log\left(x + \sqrt{x^2+a^2}\right)\right)^2\) and \(x \ne \displaystyle\frac{1-a^2}{2}\), then \((x^2+a^2)\displaystyle\frac{d^2y}{dx^2} + x\displaystyle\frac{dy}{dx}\) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
Let \(u = \log\left(x + \sqrt{x^2+a^2}\right)\), so \(y = u^2\). By the chain rule:
\( \displaystyle \frac{du}{dx} = \frac{1}{x + \sqrt{x^2+a^2}}\left(1 + \frac{x}{\sqrt{x^2+a^2}}\right) \)
\( \displaystyle = \frac{1}{x + \sqrt{x^2+a^2}} \cdot \frac{\sqrt{x^2+a^2} + x}{\sqrt{x^2+a^2}} \)
\( \displaystyle = \frac{1}{\sqrt{x^2+a^2}} \)
\( \displaystyle \frac{dy}{dx} = 2u\,\frac{du}{dx} = \frac{2u}{\sqrt{x^2+a^2}} \)
Before differentiating again, clear the root:
\( \displaystyle \sqrt{x^2+a^2}\,\frac{dy}{dx} = 2u \)
Differentiating both sides with respect to x (product rule on the left):
\( \displaystyle \sqrt{x^2+a^2}\,\frac{d^2y}{dx^2} + \frac{x}{\sqrt{x^2+a^2}}\,\frac{dy}{dx} = \frac{2}{\sqrt{x^2+a^2}} \)
Multiplying by \(\sqrt{x^2+a^2}\):
\( \displaystyle (x^2+a^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = 2 \)
This holds for every x; the condition \(x \ne \frac{1-a^2}{2}\) given in the question plays no part.