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Math Help - separable equation

  1. #1
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    separable equation

    This equation is supposedly separable:

    t\frac{dy}{dx} = y + \sqrt{t^2 + y^2}

    but I can't separate it with algebra, and the substitutions I've tried (v=y/t, v=(t^2 + y^2)^1/2) haven't worked. The full problem, if it makes a difference, is to solve for the initial value y(1) = 0.

    Could you give me a hint please?
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  2. #2
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    Quote Originally Posted by Korgoth28 View Post
    This equation is supposedly separable:

    t\frac{dy}{dx} = y + \sqrt{t^2 + y^2}

    but I can't separate it with algebra, and the substitutions I've tried (v=y/t, v=(t^2 + y^2)^1/2) haven't worked. The full problem, if it makes a difference, is to solve for the initial value y(1) = 0.

    Could you give me a hint please?
    I suppose you meant that x to be a t:

    t\frac{dy}{dt} = y + \sqrt{t^2 + y^2}

    \Rightarrow \frac{dy}{dt} = \frac{y}{t} + \frac{\sqrt{t^2 + y^2}}{t} = \frac{y}{t} + \sqrt{1 + \left( \frac{y}{t}\right)^2}

    and now you make the substitution y = v t etc.
    Last edited by mr fantastic; July 4th 2010 at 04:15 PM.
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  3. #3
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    That makes sense, thank you. After substituting and rearranging I get this:
      \int \frac{dt}{t} = \int \frac{dv}{\sqrt{1+v^2}}
    And after integrating:
    \ln{t} = arcsinh (\frac{y}{t}) + c
    and plugging in the initial value gives me
    \ln{1} = arcsinh(0) + c
    so c = 0.
    However the answer in the back of the book is  \frac{t^2 - 1}{2}. Could you please tell me where I'm messing up?

    Also, out of curiosity, is there any time when it'd be useful to make a substitution other than y/t in a problem like this? For example, t/y?
    Last edited by Korgoth28; July 5th 2010 at 01:37 PM.
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  4. #4
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    I thought I hit discrete mathematics topic but I guess not. Just happened to look at this and the question that came to mind for me, and forgive my ignorance if not because I have yet to have this subject, would the sq root in the denominator have an impact in following processes?
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  5. #5
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    Quote Originally Posted by Korgoth28 View Post
    That makes sense, thank you. After substituting and rearranging I get this:
     \int \frac{dt}{t} = \int \frac{dv}{\sqrt{1+v^2}}
    And after integrating:
    \ln{t} = arcsinh (\frac{y}{t}) + c
    and plugging in the initial value gives me
    \ln{1} = arcsinh(0) + c
    so c = 0.
    However the answer in the back of the book is  \frac{t^2 - 1}{2}. Could you please tell me where I'm messing up?

    [snip]
    You have not messed up. You're expected to know that \text{arcsinh} \, (z) = \ln (z + \sqrt{1 + z^2}) and hence simplify your answer accordingly.
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