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Homework Help: Integration (ln/e)

  1. Sep 12, 2011 #1


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    This is a question...

    For the following question:

    I get the solution...
    [itex]\int \frac{1}{3y} dy = \int dx[/itex]
    [itex]\frac{1}{3}ln y = x + c[/itex]
    [itex]y = e^{3x}+e^{3c}[/itex]

    However the textbook example says the solution is...
    [itex]y = ce^{3x}[/itex]

    My question is would my answer be incorrect? How should the arbitrary constant be placed in ln and e integral solutions?
  2. jcsd
  3. Sep 12, 2011 #2
    [itex]e^{x+y}=e^xe^y[/itex] and not [itex]e^{x+y}=e^x+e^y[/itex]. You made this mistake in your last line.
  4. Sep 12, 2011 #3
    You have to take into account that
    [tex] e^{x+y}=e^x \cdot e^y \neq e^x + e^y[/tex]
    To get a feeling for that relation, take for example
    [tex] 2^{3+4}=(2 \cdot 2 \cdot 2 )\cdot (2 \cdot 2 \cdot 2 \cdot 2 )=2^3 \cdot 2^4 [/tex]
  5. Sep 13, 2011 #4


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    Oh yes, that is correct. Silly mistake. Thanks.
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