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: System of Differential Equations

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data

    So my friend asked me for help because he assumed having a math degree meant I knew math :rolleyes:

    [PLAIN]http://courses.webwork.maa.org:8080/wwtmp/equations/0e/c5a05957810916cfdff379ee0642fc1.png [Broken]

    (dx/dt=−4y and dy/dt = −4x)

    Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation.

    Solve the equation you obtained for y as a function of t; hence find x as a function of t. If we also require x(0) = 4 and y(0) = 1, what are x and y?



    2. Relevant equations

    y(t)=C1e^(at)cos(Bt) + C2e^(at)sin(Bt)

    3. The attempt at a solution

    So I've tried a few approaches but I've failed. This is the approach I got the farthest with.


    d^2y/d^2t = -4

    you can then say r^2 + 4 = 0

    b^2-4ac = 0 - 4*4 = -16

    r = 0 +/- sqrt(-16) / 2 = +/- 2i
    y = Ae^(0t)cos(2t) + Be^(0t)sin(2t)
    y=Acos(2t)+Bsin(2t)
    y' =2Bcos(2t) + 2Asin(2t)

    If I solve for x(t) I end up with the same equation as y(t).
    If you set up for x(0)=4 and y(0)=1, you end up with:

    1=A
    4=A?
    I know this is easy but I'm just not seeing it. Any help would be appreciated. Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 24, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No, [itex]\frac{d}{dx}\left(\frac{dy}{dt}\right)=-4[/itex] but [tex]\frac{d^2y}{dt^2}=\frac{d}{dt}\left(-4x\right)=-4\frac{dx}{dt}[/itex]
     
  4. Sep 24, 2010 #3

    Ah ok, I see.

    Now I get [itex]\frac{-1}{4}\frac{d^2y}{dt^2}[/itex] = -4y
    [itex]\frac{-1}{4}\frac{d^2y}{dt^2}[/itex] +4y = 0
    so,
    [itex]\frac{-1}{4}r^2 + 4y = 0 [/itex]

    y(t) = C1 e^(4t) + C2e^(-4t)

    Now doing the same for x(t), I end up with the same equation of

    x(t) = C1 e^(4t) + C2e^(-4t),

    however given my initial values that can't be.
     
  5. Sep 24, 2010 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Good..

    [itex]x(t)[/itex] is related to [itex]y(t)[/itex] via [itex]y'(t)=-4x(t)[/itex], once you've found [itex]y(t)[/itex], you need only differentiate it once and divide by -4 to obtain [itex]x(t)[/itex]
     
  6. Sep 24, 2010 #5
    Thank you for leading me down the right path! I got it now :)

    The solution was
    y(t) = (5/2)e^(-4t)+(-3/2)e^(4t)
    x(t) =(5/2)e^(-4t)-(-3/2)e^(4t)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook