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    The equation is: \frac{dy}{dx} = \sqrt (y^2 - 1)

    Okay, here's what I did so far... I'm not sure if it's mathematically valid introducing an imaginary term, but it was all I could think of..

    \frac{dy}{\sqrt (y^2 - 1)} = dx
    \frac{dy}{\sqrt(-1) \sqrt (1 - y^2)}= dx
    \frac{1}{i} \frac{dy}{\sqrt (1 - y^2)} = dx
    \frac{1}{i} \int \frac{1}{\sqrt (1 - y^2)}dy = \int dx
    \frac{1}{i} arcsiny = x + C
    y = sin(ix + iC)

    The answer given by Wolfram is
    y = 1 + 2sinh^2 (0.5(C+x))

    ... Help anyone? How on earth are you supposed to get that answer?
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    You introduced the i term to make the integral look like an arcsin. There's a related standard integral ..

    Wolfram has given you a big hint - if you know about hyperbolics you're OK.
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    (Original post by ian.slater)
    You introduced the i term to make the integral look like an arcsin. There's a related standard integral ..

    Wolfram has given you a big hint - if you know about hyperbolics you're OK.
    This a question from my textbook (the answer given in the back of the book is COMPLETELY different by the way), but the hyperbolic functions aren't actually covered in my course... hence why I was pretty shocked by this answer

    Would you mind elaborating on your point please? I don't understand, isn't the derivative of sinh(x) just cosh(x)? So the derivative of sinh^2(x) would just be 2sinh(x)cosh(x)... I know pretty much nothing about hyperbolics, sorry :/
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    (Original post by innerhollow)
    This a question from my textbook (the answer given in the back of the book is COMPLETELY different by the way), but the hyperbolic functions aren't actually covered in my course... hence why I was pretty shocked by this answer

    Would you mind elaborating on your point please? I don't understand, isn't the derivative of sinh(x) just cosh(x)? So the derivative of sinh^2(x) would just be 2sinh(x)cosh(x)...
    Try substituting y = cosh(t).

    You're right about the derivative.

    There are lots of interesting similarities between trig functions and the hyperbolic trigs - which is why your original approach was so good.
 
 
 
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