1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration (ln/e)

  1. Sep 12, 2011 #1

    Air

    User Avatar

    This is a question...

    For the following question:
    [itex]y^{'}=\frac{dy}{dx}=3y[/itex]

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    [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

    Air

    User Avatar

    Oh yes, that is correct. Silly mistake. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration (ln/e)
  1. Integration (ln) (Replies: 3)

  2. Integrate ln (Replies: 5)

  3. Integral e^x ln(x)dx (Replies: 9)

  4. Integral of ln (Replies: 14)

Loading...