If $y = (\log(x + \sqrt{x^2+a^2}))^2$ and $x≠\frac{1-a^2}{2}$, then $(x^2+a^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}$ is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
$y=\bigg(\log\big(x+\sqrt{x^{2}+a^{2}}\big)\bigg)^{2}$
$u=\log\big(x+\sqrt{x^{2}+a^{2}}\big)$
$y=u^{2}$
$\frac{dy}{dx}=2u\frac{du}{dx}$
$\frac{du}{dx}=\frac{1+\frac{x}{\sqrt{x^{2}+a^{2}}}}{x+\sqrt{x^{2}+a^{2}}}=\frac{1}{\sqrt{x^{2}+a^{2}}}$
$\frac{dy}{dx}=\frac{2u}{\sqrt{x^{2}+a^{2}}}$
$\frac{d^{2}y}{dx^{2}}=2\left(\frac{1}{\big(\sqrt{x^{2}+a^{2}}\big)^{2}}-u\frac{x}{\big(\sqrt{x^{2}+a^{2}}\big)^{3}}\right)$
$(x^{2}+a^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=(x^{2}+a^{2})\cdot 2\left(\frac{1}{(\sqrt{x^{2}+a^{2}})^{2}}-u\frac{x}{(\sqrt{x^{2}+a^{2}})^{3}}\right)+x\cdot\frac{2u}{\sqrt{x^{2}+a^{2}}}$
$=2\bigg(1-u\frac{x}{\sqrt{x^{2}+a^{2}}}\bigg)+2u\frac{x}{\sqrt{x^{2}+a^{2}}}=2$