Question

y = log[x+root(1+x2)]. Prove that (1+x2)d2y/dx2 + x.dy/dx = 0

Answer 1

Please refer image...hope this helps u

Answer 2

Solution:

Given

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

Differentiate with respect to x,

we get

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

= $\frac{1}{x+\sqrt{1+x^{2}}(\frac{(\sqrt1+x^{2}+x}{\sqrt{1+x^{2}})$

$\implies \frac{dy}{dx}=\frac{1}{sqrt{1+x^{2}}\\</p><p>= (1+x°{2})^\frac{-1}{2}$

Differentiate with respect to x,

we get

$\frac{d^{2}y}{dx^{2}} = \frac{-1}{2}\times (1+x^{2})^{\frac{-1}{2}-1}\times (0+2x)\\</p><p>= \frac{-1}{2}\times(1+x^{2})^{\frac{-3}{2}} \times 2x$

=\frac{-x}{(1+x^{2})^{\frac{3}{2}}

$(1+x^{2})\frac{d^{2}y}{dx^{2}}= \frac{-x(1+x^{2})}{(1+x^{2})^{\frac{3}{2}}\\=\frac{-x}{\sqrt{1+x^{2}}$---(1)

multiply equation (1) by x , we get

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

Add equation (1) and (2), we get

(1+x²)d²y/dx² + xdy/dx = 0

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