Asked by Kenzo
Determine the particular solution:
(y')^2=(1-y^2)/(1-x^2) ; y=1/2 x=1
(y')^2=(1-y^2)/(1-x^2) ; y=1/2 x=1
Answers
Answered by
Damon
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
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
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