If y=root1-x/1+x.then prove that (1-X)^2dy/dx+y=0?

Question

If $y = \sqrt{\frac{1 - x}{1 + x}}$ $y=sqrt((1-x)/(1+x))$ , then prove that $(1 - x^{2}) \frac{d y}{d x} + y = 0$ $(1-x^2)(dy)/(dx)+y=0$

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Answer 1 · 2018-04-13T15:34:58

Please see below.

$y = \sqrt{\frac{1 - x}{1 + x}}$ $y=sqrt((1-x)/(1+x))$

$y = \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}}$ $y=sqrt((1-x)/(1+x)xx(1-x)/(1-x))$

$y = \sqrt{\frac{{(1 - x)}^{2}}{1 - x^{2}}}$ $y=sqrt((1-x)^2/(1-x^2)$

$y=(1-x)/sqrt(1-x^2)...to(A)$

$"Using "color(blue)"Quotient Rule"$

$(dy)/(dx)=((sqrt(1-x^2))(0-1)-(1-x)(1/(2sqrt(1-x^2)) (-2x)))/((sqrt(1-x^2))^2)$

$(dy)/(dx)=(-sqrt(1-x^2)+((1-x)x)/sqrt(1-x^2))/(1-x^2)$

$(1-x^2)(dy)/(dx)=(-(1-x^2)+(1-x)x)/sqrt(1-x^2)$

$(1-x^2)(dy)/(dx)=(-1+x^2+x-x^2)/sqrt(1-x^2)$

$(1-x^2)(dy)/(dx)=(-1+x)/sqrt(1-x^2)$

$(1-x^2)(dy)/(dx)=-(1-x)/sqrt(1-x^2)...toUse(A)$

$(1-x^2)(dy)/(dx)=-y$

$(1-x^2)(dy)/(dx)+y=0$