1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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
    [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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...