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Math Help - Separation of Variables

  1. #1
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    Separation of Variables

    Use separation of variables to solve the IVP

    How would you go about doing this? So far I've done this:

    dy/\sqrt(1-y^2) = dt/t

    arcsin(y) = ln(t) +c

    I don't know if my work is correct, but would the solution just be y=sin(ln(t))?
    Last edited by cdlegendary; June 6th 2010 at 01:48 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by cdlegendary View Post
    Use separation of variables to solve the IVP
    ,,

    How would you go about doing this? So far I've done this:

    dy/\sqrt(1-y^2) = dt/t

    arcsin(y) = ln(t) +c

    I don't know if my work is correct, but would the solution just be y=sin(ln(t))?
    Almost but not quite it does not satisfy the IC.

    y(1)=\sin(\ln(1))=\sin(0)=0

    You forgot the C. The solution should look like

    y=\sin(\ln(t)+c) Now use the IC to solve for c.

    1=\sin(c) \implies c=\frac{\pi}{2}
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    Almost but not quite it does not satisfy the IC.

    y(1)=\sin(\ln(1))=\sin(0)=0

    You forgot the C. The solution should look like

    y=\sin(\ln(t)+c) Now use the IC to solve for c.

    1=\sin(c) \implies c=\frac{\pi}{2}
    Ah I see, thanks. So essentially the \pi/2 makes it cos(lnt)
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