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Math Help - Particular Solution

  1. #1
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    Particular Solution

    40xy' -ln(x^8) = 0

    Initial Condition: y(1)=31

    Work:
    40x (dy/dx) - ln(x^8) = 0

    40x(dy/dx) = ln (x^8)

    dy/dx = ln(x^8) / 40x

    Sdy = Sln(x^8) / 40x

    And then...stuck (if its even right thus far)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by BooGTS View Post
    40xy' -ln(x^8) = 0

    Initial Condition: y(1)=31

    Work:
    40x (dy/dx) - ln(x^8) = 0

    40x(dy/dx) = ln (x^8)

    dy/dx = ln(x^8) / 40x

    Sdy = Sln(x^8) / 40x

    And then...stuck (if its even right thus far)
    Hint: \ln(x^8)=8 \ln x. Then the integral on the RHS becomes \displaystyle\frac{1}{5}\int\frac{\ln x}{x}\,dx.

    Do you know how to proceed?

    (And yes, everything is right so far. )
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  3. #3
    MHF Contributor chisigma's Avatar
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    Because is \ln x^{8} = 8\ \ln x the DE becomes...

    \displaystyle y^{'} = \frac{\ln x}{5 x} (1)

    ... where the variables are separated so that the solution is immediate...

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Hint: \ln(x^8)=8 \ln x. Then the integral on the RHS becomes \displaystyle\frac{1}{5}\int\frac{\ln x}{x}\,dx.

    Do you know how to proceed?

    (And yes, everything is right so far. )
    If you are concerned about how to do that integral, let u= ln(x).
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  5. #5
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    Still not doing this right I don't think:

    y= 1/5 (lnxdx)/x
    = 1/5 u du
    = 1/5 u^2/2
    = 1/10 lnx^2
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  6. #6
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    Quote Originally Posted by BooGTS View Post
    Still not doing this right I don't think:

    y= 1/5 (lnxdx)/x
    = 1/5 u du
    = 1/5 u^2/2
    = 1/10 lnx^2
    note the difference between \ln{x^2} and (\ln{x})^2 ...

    the correct antiderivative is

    \displaystyle \frac{(\ln{x})^2}{10} + C
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  7. #7
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    Yeah, big difference. Thanks to all (and on the site if I can figure out how)
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