Question

if y = log(x + root( x2 +1) then prove that (x2 + 1)dy / dx + x dy / dx =0

Answer 1

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Answer 2

The correct question is :

If $y = log(x + \sqrt{x^2+1} )$ then prove that  $(x^2 +1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} =0$

Given the equation

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

We have to prove that

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

• Given the equation

         $y = log(x + \sqrt{x^2+1} )$

• Differentiating with respect to 'x' , we get

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

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

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

        Multiplying the above equation by $\sqrt{x^2+1}$, we get

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

• Now, Differentiating again with respect to 'x' that is the second order differential of the given function, we get

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

• Multiplying the above equation by $\sqrt{x^2+1}$ we get,

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