Question

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

• A. $-2 y$
• 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 $
$\Rightarrow (x+1) y_{1}+2 y=-2 \sqrt{y}$
Again differentiating both sides w.r.t. $X$, we get
$(x+1) y_{2}+y_{1}+2 y_{1}=\frac{-2}{2 \sqrt{y}} y_{1}$
$\Rightarrow (x+1) y_{2}+3 y_{1}+\frac{1}{\sqrt{y}} y_{1}=0$
$\Rightarrow (x+1) y_{2}+\left(\frac{3 \sqrt{y}+1}{\sqrt{y}}\right) y_{1}=0$