Question

If $y={\left[\log \left(x+\sqrt{x^2+1}\right)\right]}^2$.
Prove that $\left(1+x^2\right)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=2$.

Answer 1

Given,

$y={\left[\log (x+\sqrt{x^2+1})\right]}^2$

$\frac{dy}{dx}$

substitute $u=\log (x+\sqrt{x^2+1})$

$\therefore \frac{dy}{dx}=\frac{2\log (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}$

$\frac{d^{2}y}{d{x}^2}$

$=\frac{d}{dx}\left(\frac{2\log (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}\right)$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{2(\sqrt{x^2+1}-x\log (x+\sqrt{x^2+1}))}{(x^2+1)\sqrt{x^2+1}}$

Given,

$(1+x^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}$

substituting the above values, we get,

$=(1+x^2)\left(\frac{2(\sqrt{x^2+1}-x\log (x+\sqrt{x^2+1}))}{(x^2+1)\sqrt{x^2+1}}\right)+x\left(\frac{2\log (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}\right)$

$=(1+x^2)(\frac{2}{1+x^2}-\frac{2x\log (x+\sqrt{x^2+1})}{(1+x^2)\sqrt{x^2+1}})+\frac{2x\log (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}$

$=2-\frac{2x\log (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}+\frac{2x\log (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}$

$=2$

Hence proved.