Determine the particular solution:

Question

(y')^2=(1-y^2)/(1-x^2) ; y=1/2 x=1

Answer 1

hmmm, not sure but
dy/dx = [(1-y^2)/(1-x^2) ]^.5

dy/(1-y^2)^.5 = dx/(1-x^2)^.5
let y = sin p let x = sin r
dy = cos p dp and dx = cos r dr

cos p dp/cos p = cos r dr/cos r

dp = dr