Let y=log(x+√(x^2+1)), and a(d^2y)/(dx^2)+bdy/dx=c. Then identify the…

The correct answer is Option 3(A), (C), (D) only

Differentiable w.r.t.x.

$\frac{dy}{dx}=\frac{d[log(x+\sqrt{x^2+1})]}{d(x+\sqrt{x^2+1})}×\frac{d[x+\sqrt{x^2+1}]}{dx}$

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

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

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

Differentiable eq.(ii) w.r.t.x

$\sqrt{x^2+1}.\frac{d^2y}{dx^2}+\frac{dy}{dx}×\frac{d}{dx}\sqrt{x^2+1}=0$  $\frac{d}{dx}(u.v)=u.\frac{dv}{dx}+v.\frac{du}{dx}$

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

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

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

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

So value for a = x² + 1, b = x, c = 0