If x=1−y1+y then x+1d2ydx2+3y+1ydydx=_____

x1+y=1−yDiff.x12ydydx+1+y1=0−12ydydx⇒ 12ydydxx+1=−1+yx+1dydx=−2y+yAgain diff.x+1d2ydx2+dydx1=−212y+1dydx⇒ x+1d2ydx2+3y+1ydydx=0

If x=1−y1+y then x+1d2ydx2+3y+1ydydx=_____

x1+y=1−yDiff.x12ydydx+1+y1=0−12ydydx⇒ 12ydydxx+1=−1+yx+1dydx=−2y+yAgain diff.x+1d2ydx2+dydx1=−212y+1dydx⇒ x+1d2ydx2+3y+1ydydx=0

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

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Detailed Solution

$x (1 + \sqrt{y}) = 1 - \sqrt{y}$
Diff.

$x [\frac{1}{2 \sqrt{y}} \frac{d y}{d x}] + (1 + \sqrt{y}) (1) = 0 - \frac{1}{2 \sqrt{y}} \frac{d y}{d x}$
$\Rightarrow \frac{1}{2 \sqrt{y}} \frac{d y}{d x} (x + 1) = - (1 + \sqrt{y})$
$(x + 1) \frac{d y}{d x} = - 2 (\sqrt{y} + y)$
Again diff.
$(x + 1) \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} (1) = - 2 (\frac{1}{2 \sqrt{y}} + 1) \frac{d y}{d x}$
$\Rightarrow (x + 1) \frac{d^{2} y}{d x^{2}} + (\frac{3 \sqrt{y} + 1}{\sqrt{y}}) \frac{d y}{d x} = 0$