Question

If $x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$, then $(x+1)\frac{d^{2}y}{d{x}^2}+\left(\frac{3\sqrt{y}+1}{\sqrt{y}}\right)\frac{dy}{dx}$ equals

• A. $-2y$
• B. 0
• C. $-y$
• D. $y$

Answer 1

Answer: (B)

We have,
$x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$
$\Rightarrow x(1+\sqrt{y})=1-\sqrt{y}$
$\Rightarrow x+x\sqrt{y}=1-\sqrt{y}$
$\Rightarrow x+(x+1)\sqrt{y}=1$
On differentiating both sides w.r.t. $x$, we get
$1+\frac{(x+1)}{2\sqrt{y}}{y}_1+\sqrt{y}=0$
$\Rightarrow (x+1)y_1+2y=-2\sqrt{y}$
Again differentiating both sides w.r.t. $X$, we get
$(x+1)y_2+y_1+2y_1=\frac{-2}{2\sqrt{y}}{y}_1$
$\Rightarrow (x+1)y_2+3y_1+\frac{1}{\sqrt{y}}{y}_1=0$
$\Rightarrow (x+1)y_2+\left(\frac{3\sqrt{y}+1}{\sqrt{y}}\right)y_1=0$