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: Second Diff Eq question of the day

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    10. If y(t) = -te^(3t) is a solution to y'' + py' + qy = 0, then what are p and q?

    2. Relevant equations

    3. The attempt at a solution

    so it's equal to r^2 + pr +q = 0, but then how can i find the two numbers?

  2. jcsd
  3. Feb 8, 2009 #2
    You don't need to do any algebra. Remember there are some formulas for solving different equations? This one looks like e^at (At + B) form. Just look over them.
  4. Feb 8, 2009 #3


    Staff: Mentor

    If your solutions were y1 = e^(3t) and y2 = e^(4t), your characteristic equation would be r^2 - 7r + 12 = 0, or (r - 3)(r - 4) = 0, right?

    Under what circumstances related to the work above will you get a solution y = te^(3t)?
  5. Feb 9, 2009 #4


    User Avatar
    Science Advisor

    You don't need to look at the characteristic equation at all. If [itex]y= -te^{3t}[/itex] then [itex]y'= -e^{3t}- 3te^{3t}= -(1+ 3t)e^{3t}[/itex] and y"= [itex]-3e^{3t}- 3(1+ 3t)e^{3t}[/itex][itex]= -(6+ 9t)e^{3t}[/itex]. Putting those into the equation we have [itex]-(6+ 9t)e^{3t}- p(1+ 3t)e^{3t}- qte^{3t}= ((-6- p)+(-9-3p-q)t)e^{3t}= 0[/itex] for all t. Since [itex]e^{3t}[/itex] is never 0, we must have (-6- p)+ (-9-3p-q)t= 0 for all t which means we must have -6-p= 0 and -9- 3p- q= 0.

    Of course, it is also true that eat will be a solution if and only if a is a root of the characteristic equation and that teat will be a solution if and only if a is a double root of the characteristic equation, which means that the characteristic equation must reduce to (x-a)2= 0.
    Last edited by a moderator: Feb 9, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook