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