Given that y=((1-x)/(1+x))^1/2 show that (1-x^2)dy/dx+y=0
y = ((1-x)/(1+x))^1/2
y^2 = (1-x)/(1+x)
Differentiating both sides wrt x,
2y.y’ = {(1+x)(-1) – (1-x)(1)} / (1+x)^2
2y.y’ = {-1 – x – 1 + x} / (1+x)^2
2y.y’ = {-2} / (1+x)^2
y.y’ = {-1} / (1+x)^2 (now multiply both sides by y)
y^2.y’ = {-y} / (1+x)^2
y^2(1+x).y’ = {-y} / (1+x)
Using y^2 = (1-x)/(1+x),
(1-x).y’ = {-y} / (1+x)
(1-x^2).y’ = -y
(1 – x^2).dy/dx + y = 0