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Math Help - Tricky seperable DE. At least it's tricky to me

  1. #1
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    Tricky seperable DE. At least it's tricky to me

    Solve the sepereable D.E.

    xdy = y*ln(y)dx , x = 2 , y = e

    I started by dividing both sides by y*ln(y) and x to wind up with

    dx/x = dy/(y*ln(y))

    Then I go to integrate and end up with

    ln(x) + C = ln(ln(y)) + C


    And then I'm stuck. I'm entirely unsure of where to go with this. Any ideas? Thank you

    - Charles
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  2. #2
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    Quote Originally Posted by Charles Dexter Ward View Post

    Then I go to integrate and end up with

    ln(x) + C = ln(ln(y)) + C
    Just consider one constant, make it k, hence \ln|x|=\ln|\ln y|+k.

    We can also set \ln|x|=\ln|\ln y|+\ln|k_1|, and from there you can easily isolate y.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    Just consider one constant, make it k, hence \ln|x|=\ln|\ln y|+k.

    We can also set \ln|x|=\ln|\ln y|+\ln|k_1|, and from there you can easily isolate y.
    Ok. So I can ln the constant because it doesn't change anything, it's still a constant, right? So from

    ln|x| = ln|lny| + ln|k|

    I e everything right? e^ln|x| = e^ln|lny| + e^ln|k|

    which leaves

    x = lny + k

    lny = x - k

    y = e^x - e^k => e^x - k

    k = e^x - y , x = 2, y = e

    k = e^2 - e^1 = 4.6708

    Is that correct?

    Am I allowed to e everything like that?
    Last edited by Charles Dexter Ward; May 29th 2008 at 04:46 PM.
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    Anyone able to confirm if my solution is correct or not? Thank you.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Charles Dexter Ward View Post
    Solve the sepereable D.E.

    xdy = y*ln(y)dx , x = 2 , y = e
    \frac{dy}{y\ln(y)}=\frac{dx}{x}\implies \ln|\ln(y)|+C_1=\ln|x|+C_2\implies \ln|\ln(y)|=\ln|x|+C; C=C_2-C_1
    \implies \ln(y)=Ae^{\ln|x|}; A=\pm e^C
    \therefore y=e^{A|x|}

    When x=2, y=e:

    e=e^{2A}\implies 1=2A\implies A=\frac{1}{2}

    Therefore, \color{red}\boxed{y=e^{\frac{1}{2}|x|}}
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  6. #6
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    I see. My method also gets the same answer to both ways are correct! Thank you!

    Quote Originally Posted by Chris L T521 View Post
    \frac{dy}{y\ln(y)}=\frac{dx}{x}\implies \ln|\ln(y)|+C_1=\ln|x|+C_2\implies \ln|\ln(y)|=\ln|x|+C; C=C_2-C_1
    \implies \ln(y)=Ae^{\ln|x|}; A=\pm e^C
    \therefore y=e^{A|x|}

    When x=2, y=e:

    e=e^{2A}\implies 1=2A\implies A=\frac{1}{2}

    Therefore, \color{red}\boxed{y=e^{\frac{1}{2}|x|}}
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