[Hasufel]

i would like to know what your answer to this question is:

Solve: dy/dx = Sqrt(y^2-1) with i.c. y(1)=1.

i get one answer
my book gets another mathematica gets another
maple gets another

if i exponentiate the solution, i get one for i can`t explicitly solve for y
if i take the solution as being inverse hyperbolic, i (apparently according to Maple) get the wrong answer also
(hyperbolicly, i get y(x) = Cosh(x)

what am i missing (appart from a few grey cells this time of night!)

[Hasufel]

Arccosh(y)=x-1 => y=Cosh(x-1) ? mathematica gets cosh(1-x)!!

[f1mad]

(Original post by Hasufel)
Arccosh(y)=x-1 => y=Cosh(x-1) ? mathematica gets cosh(1-x)!!

They're the same.

cosh(x-1) = 0.5(e^x-1 + e^-x+1)

cosh(1-x) = 0.5(e^-x+1 + e^x-1)

[Hasufel]

cheers!

best go to bed i suppose!