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!

Where too from here?

  1. Aug 17, 2008 #1
    question attached
     

    Attached Files:

  2. jcsd
  3. Aug 17, 2008 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    You write
    [tex]\int \frac{1}{(v + 1)^2} dv = \int \frac{1}{x} dx[/tex]
    and then on the next line
    [tex]-v - 1 = \ln |v| + c[/tex]

    I think you made a writing error there, which leads to an insolvable equation.
     
  4. Aug 17, 2008 #3
    ah ok but if its lnx instead of v then i'm still on the right tracks?
     
  5. Aug 17, 2008 #4
    ok so where do i go from e-v-1=KX ?
     
  6. Aug 17, 2008 #5

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Yes, until that step you were fine.
    Remember, you want the expression for v(x), and finally y(x) = x v(x).
    So you have to solve v(x) from [itex]e^{-v(x)-1} = k x[/itex]. Try taking logarithms on both sides.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?