If x=1-y1+y, then x+1d2ydx2+3y+1ydydx=

x=1-y1+y x1+y=1-y xy+y=1-x (x+1)y=1-x (x+1)12y dydx+y=-1 (x+1)dydx+2y=-2y (x+1)d2ydx2+dydx+2dydx=-1y dydx(x+1)d2ydx2+3+1ydydx=0 (x+1)d2ydx2+3y+1y dydx=0

If x=1-y1+y, then x+1d2ydx2+3y+1ydydx=

x=1-y1+y x1+y=1-y xy+y=1-x (x+1)y=1-x (x+1)12y dydx+y=-1 (x+1)dydx+2y=-2y (x+1)d2ydx2+dydx+2dydx=-1y dydx(x+1)d2ydx2+3+1ydydx=0 (x+1)d2ydx2+3y+1y dydx=0

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

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Detailed Solution

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

$(x + 1) \frac{d^{2} y}{{dx}^{2}} + (3 + \frac{1}{\sqrt{y}}) \frac{dy}{dx} = 0 (x + 1) \frac{d^{2} y}{{dx}^{2}} + (\frac{3 \sqrt{y} + 1}{\sqrt{y}}) \frac{dy}{dx} = 0$