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!

Homogeneous initial value problem

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data

    4y" + 4y' + 5y = 0
    y(0) = 3
    y'(0) = 1

    2. Relevant equations

    yh = e^ax(c1cosbx + c2sinbx)

    3. The attempt at a solution

    For the roots I got -1/2 + i and -1/2 - i so my a = -1/2 and b = 1

    then I have to differentiate yh = e^(-1/2x)[c1cosx + c2sinx]

    this is where I get this overly complicated equation and I was wondering if I could do integration by parts instead.

    where I could get something like

    yh = -1/2e^(-1/2x)[c1cosx + c2sinx] + [-c1sinx + c2cosx]e^(-1/2x)
     
    Last edited: Mar 24, 2014
  2. jcsd
  3. Mar 24, 2014 #2

    Mark44

    Staff: Mentor

    Your title is misleading. Your equation is a homogeneous equation.
    ???
    You can't use integration by parts if you need to differentiate a function.
     
  4. Mar 24, 2014 #3
    Damn I'm so sorry I misled you, I've been working on these all day and well you know how that goes. What I meant to say was product rule, where f(x) = e^(-1/2x) and g(x) = [c1cosx + c2sinx]
     
  5. Mar 24, 2014 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, use product rule for y= e-x/2(c1cosx+c2sinx) to get y'.

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Homogeneous initial value problem
  1. Initial value problem (Replies: 7)

  2. Initial-value problem (Replies: 2)

Loading...