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+2 votes
70.5k views
in Continuity and Differentiability by (36.2k points)
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If y = log [x + \(\sqrt{x^2+1}\)], then prove that (x2 + 1) \(\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0\)

2 Answers

+1 vote
by (17.0k points)
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Best answer

\(y = \log(x + \sqrt{x^2 + 1})\)    ......(1)

Diff. equation (1) w.r.t. x

\(\frac{dy}{dx} = \frac{d[\log(x + \sqrt{x^2 + 1})]}{d(x + \sqrt{x^2 + 1})}\times \frac{d[x + \sqrt{x^2 + 1}]}{dx}\)     (chain rule)

\(\frac{d(\log x)}{dx} = \frac 1x\)

\(\frac{dy}{dx} = \frac 1{x + \sqrt {x^2 + 1} } \times \left[1 + \frac 1{2\sqrt{x^2 +1}}\times2x\right]\)

\(\frac{dy}{dx} = \frac 1{x + \sqrt{x^2 + 1}} \times \frac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}\)

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

\(\sqrt{x^2 + 1}. \frac{dy}{dx} = 1\)   ......(2)

Diff. equation (2) w.r.t. x

\(\sqrt{x^2 + 1}.\frac{d^2y}{dx^2} + \frac{dy}{dx}\times \frac d{dx}(\sqrt {x^2 + 1}) = 0\) 

\(\sqrt{x^2 + 1}.\frac{d^2y}{dx^2} + \frac{dy}{dx}\times \frac 1{2\sqrt{x^2 + 1}}\times 2x = 0\)

\(\sqrt{x^2 + 1}.\frac{d^2y}{dx^2} + \frac{x.\frac{dy}{dx}}{\sqrt{x^2 + 1}} = 0\)

\(\frac{(x^2 + 1).\frac{d^2y}{dx^2} + x.\frac{dy}{dx}}{\sqrt{x^2 + 1}} = 0\)

\({(x^2 + 1).\frac{d^2y}{dx^2} + x.\frac{dy}{dx}} = 0\)

Hence proved.

+2 votes
by (33.4k points)

Given,

y = log [x + \(\sqrt{x^2+1}\)],

Differentiating with respect to x, we get

Differentiating again with respect to x, we get

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