If $y=(x+\sqrt{x+1})^{11}$ satisfies the relation $(x^2+1)\frac{d^y}{dx^2}+x\frac{dy}{dx}-Ay=0$. Then, value of A is :

121

$y=(x+\sqrt{x^2+1})^{11}.$

$\text{Let }u=x+\sqrt{x^2+1}.$

$\ln y=11\ln u.$

$\frac{1}{y}\frac{dy}{dx}=11\frac{1}{u}\frac{du}{dx}.$

$\frac{du}{dx}=1+\frac{x}{\sqrt{x^2+1}}=\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}=\frac{u}{\sqrt{x^2+1}}.$

$\frac{dy}{dx}=11\frac{y}{\sqrt{x^2+1}}.$

$\frac{d^2y}{dx^2}=11\left[\frac{1}{\sqrt{x^2+1}}\frac{dy}{dx}-\frac{x}{(x^2+1)^{3/2}}y\right].$

$=\frac{11}{x^2+1}\left[11y-xy\right].$

$\frac{d^2y}{dx^2}=\frac{11(11-x)}{x^2+1}y.$

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

$\Rightarrow 11(11-x)y+11xy-Ay=0.$

$\Rightarrow (121-A)y=0.$

$A=121.$

$A=121.$