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Math Help - diff equations 2

  1. #1
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    diff equations 2

    One more diff equation. Do i still use integrating method to differentiate this one?
    Attached Thumbnails Attached Thumbnails diff equations 2-222.gif  
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  2. #2
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    Quote Originally Posted by moolimanj View Post
    One more diff equation. Do i still use integrating method to differentiate this one?
    Your DE has a t in it ..... Did you mean:

    A. \frac{dy}{dt} = \frac{1 + y^2}{y(1 + t^2)}

    B. \frac{dy}{dx} = \frac{1 + y^2}{y(1 + x^2)}

    C. None of the above. The DE is as posted.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Your DE has a t in it ..... Did you mean:

    A. \frac{dy}{dt} = \frac{1 + y^2}{y(1 + t^2)}

    B. \frac{dy}{dx} = \frac{1 + y^2}{y(1 + x^2)}

    C. None of the above. The DE is as posted.
    \frac{dy}{dt} = \frac{1 + y^2}{y(1 + t^2)} is seperable:

    \frac{y}{1 + y^2} \, dy = \frac{1}{1 + t^2} \, dt

    \Rightarrow \int \frac{y}{1 + y^2} \, dy = \int \frac{1}{1 + t^2} \, dt

    \Rightarrow \frac{1}{2} \ln (1 + y^2) = \tan^{-1} (t) + C

    and you can make y the subject if you want ......
    Last edited by mr fantastic; February 14th 2008 at 02:34 AM. Reason: Changed all the x's to t's after reading post #4
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  4. #4
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    apoligies made a mistake

    it should read dy/dt so it is A below
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  5. #5
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    Quote Originally Posted by moolimanj View Post
    apoligies made a mistake

    it should read dy/dt so it is A below
    OK. Read post #3.
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  6. #6
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    Thanks Mr Fantastic,

    just to check, if i rearrange for y I get the following:

    diff equations 2-333.gif

    If I then try and calculate the partucular solution that satisfies initial condition: y(1)=5 I get C =
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  7. #7
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    Quote Originally Posted by moolimanj View Post
    Thanks Mr Fantastic,

    just to check, if i rearrange for y I get the following:

    Click image for larger version. 

Name:	333.gif 
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ID:	5058

    If I then try and calculate the partucular solution that satisfies initial condition: y(1)=5 I get C =
    Two small mistakes I think:

    1. y = \sqrt{e^{2 (\tan^{-1} t + C)} - 1}. The square root goes over the 1 as well.

    2. It's probably easier to sub t = 1, y = 5 into \frac{1}{2} \ln (1 + y^2) = \tan^{-1} (t) + C:

    \frac{1}{2} \ln (1 + 25) = \tan^{-1} (1) + C

    \Rightarrow \ln(26) = \frac{\pi}{2} + 2C etc.
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