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Math Help - Separable equation

  1. #1
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    Separable equation

    I have another problem with separable equations, can anyone help?

    I need to know how I end up with PI and tan in the answers to these differential equations, its just not making much sense to me at the minute!

    so I start with equation

    dy/dx = 1+y^2/2+4x

    which satisfies y(1) = 1


    so this becomes

    1/1+y^2 dy = 1/2+4x dx

    ??
    Am I right so far?

    then does the LHS
    become
    tan-1 and do you take the log of the RHS?
    can anyone help me through the stages??

    I have the solution
    i just need to see how this works! especially with PI coming into it, how??

    tan(PI/4+1/4ln(|2+4x/6|)).....
    Any help would be gr8!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dankelly07 View Post
    I have another problem with separable equations, can anyone help?

    I need to know how I end up with PI and tan in the answers to these differential equations, its just not making much sense to me at the minute!

    so I start with equation

    dy/dx = 1+y^2/2+4x

    which satisfies y(1) = 1


    so this becomes

    1/1+y^2 dy = 1/2+4x dx

    ??
    Am I right so far?

    then does the LHS
    become
    tan-1 and do you take the log of the RHS?
    can anyone help me through the stages??

    I have the solution
    i just need to see how this works! especially with PI coming into it, how??

    tan(PI/4+1/4ln(|2+4x/6|)).....
    Any help would be gr8!
    \frac{dy}{1 + y^2} = \frac{dx}{4x + 2}

    Integrating both sides gives
    tan^{-1}(y) = \frac{1}{4} ln(4x + 2) + C

    I'm not clear on this: Are you asking how to do the integral for the RHS?

    -Dan
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  3. #3
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    Krizalid's Avatar
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    Same problem posted and solved here.

    Does really make sense to change the numbers?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Same problem posted and solved here.

    Does really make sense to change the numbers?
    (Growls) A very good question.

    Please review the rule on double posting. (Technically this isn't one, however you are making yourself a tempting target for ThePerfectHacker...)

    -Dan
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  5. #5
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    sorry about that, I'm just having trouble with a similar kind of problem... well I get the integration and the separating, I was just wondering really how tan and PI get involved in the equation? I understand tan-1 now because
    the integral of 1/1+y^2 = tan-1, but still dont understand where Pi is coming from...
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  6. #6
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    mr fantastic's Avatar
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    Quote Originally Posted by dankelly07 View Post
    sorry about that, I'm just having trouble with a similar kind of problem... well I get the integration and the separating, I was just wondering really how tan and PI get involved in the equation? I understand tan-1 now because
    the integral of 1/1+y^2 = tan-1, but still dont understand where Pi is coming from...
    You now know that <br />
\tan^{-1}(y) = \frac{1}{4} \ln |4x + 2| + C<br />
.

    You're given y(1) = 1. You plug that in to get the value for C:

    \tan^{-1}(1) = \frac{1}{4} \ln |4 + 2| + C

    \Rightarrow \frac{\pi}{4} = \frac{1}{4} \ln(6) + C

    \Rightarrow \pi = \ln(6) + 4C

    \Rightarrow C = \frac{\pi - \ln(6)}{4}.

    Therefore:

    \tan^{-1}(y) = \frac{1}{4} \ln |4x + 2| + \frac{\pi - \ln(6)}{4}

    \Rightarrow \tan^{-1} (y) = \frac{1}{4} \left( \ln |4x + 2| + \pi - \ln (6) \right) = \frac{1}{4} \left( \ln\left| \frac{4x + 2}{6}\right| + \pi \right) = \frac{\pi}{4} + \frac{1}{4} \ln\left| \frac{2x + 1}{3}\right|.

    If you're working at this level, where the \pi comes from really shouldn't be an issue .....
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