Given that y=((1-x)/(1+x))½ show that  (1-x²)dy/dx+y=0
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y^2=(1-x)/(1+x); differentiating:

2yy'=(-(1+x)-(1-x))/(1+x)^2=(-1-x-1+x)/(1+x)^2=-2/(1+x)^2.

yy'+1/(1+x)^2=0; yy'(1+x)^2+1=0. 

But y=√(1-x^2)/(1+x) when we multiply ((1-x)/(1+x))^(1/2) by ((1+x)/(1+x))^(1/2).

So we have yy'(1+x)^2+1=y'(1+x)√(1-x^2)+1=0. But 1+x=√(1-x^2)/y, so y'(1-x^2)/y+1=0 and (1-x^2)dy/dx+y=0 QED.

 

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