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!

Differential equation problem

  1. May 28, 2005 #1
    i'm trying to find what the intermediate steps were used in solving this differential equation:

    [tex]y'' + y' -3y =0[/tex]

    [tex]m=\frac{-1\pm\sqrt{13}}{2}[/tex] now we have

    [tex]y=C_1e^{\frac{-.5+\sqrt{13}}{2}}t + C_2e^{\frac{-.5 -\sqrt{13}}{2}}t[/tex]

    note that the .5 are 1\2. i couldn't get the latex for this to work
     
  2. jcsd
  3. May 28, 2005 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Do you know where "m" came from?
     
  4. May 28, 2005 #3

    that m came from the auxillary equation of the differential
    [tex]r^2 +r -3=0[/tex]
     
  5. May 28, 2005 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    So what's bugging you, then?
     
  6. May 28, 2005 #5

    BobG

    User Avatar
    Science Advisor
    Homework Helper

    As you noted, you created the auxiliary (or characteristic) equation. The auxiliary equation was solved via the quadratic formula.

    Note: You should have:
    [tex]y=C_1e^{(-.5 + \frac{\sqrt{13}}{2})t} + C_2e^{(-.5-\frac{\sqrt{13}}{2})t}[/tex]
     
  7. May 29, 2005 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you were wondering where the characteristic equation came from, assume a solution ofthe form y= ert, plug it into the differential equation and see what r must be in order to make the equation true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Differential equation problem
Loading...