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System of linear diff eqs

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the system x'= -x+y
    y'= -2y
    Find the general solution (x(t),y(t)) by first solving for y(t) and then solving for x(t).

    2. Relevant equations

    Should I use undetermined coefficients for each of these separate diff eqs or should I substitute x'+x into the y' +2y=0?

    3. The attempt at a solution
    When I attempted this, I substituted x'+x into y'+2y=0 and got ae^-2t +be^-t where a and b are arbitraty constants. I am not sure if I need to look at a particular solution using undetermined coefficients or if I am even on the right track??
     
  2. jcsd
  3. Oct 7, 2008 #2

    gabbagabbahey

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    Well, the problem tells you to solve y'=-2y first, so what is the general solution y(t) to this homogeneous equation?
     
  4. Oct 7, 2008 #3
    I solved this and got y(t)= -(e^(t+c))/2. Is this right?
     
  5. Oct 7, 2008 #4

    gabbagabbahey

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    Well,

    [tex]y'(t)=\frac{d}{dt}\left( \frac{-1}{2} e^{t+c}\right)= \frac{-1}{2} e^{t+c} \neq -2y[/tex]

    so I'm gunna say no on that one.

    The DE y'(t)=-2y(t) tells you that the time derivative of y(t) is proportional to y(t), and that the constant of proportionality is -2. I can think of only one type of function that could account for that: Ae^(-2t)
     
    Last edited: Oct 7, 2008
  6. Oct 7, 2008 #5
    ok so y(t)=ae^(-2t). I think I understand how to get this by using undetermined coeff. where y'+2y=0.
    For x(t)... x'(t)= -x+y...what information do I have for the y part in order to slove for x(t)?
     
  7. Oct 7, 2008 #6

    gabbagabbahey

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    You have y=ae^(-2t); plug that in and use it:

    [tex]x'=-x+ae^{-2t} \Rightarrow x'+x=ae^{-2t}[/tex]

    which is an inhomogeneous DE which you can solve using undetermined coefficients. Start with the complimentary solution [itex]x_c(t)[/itex], what do you get for that?
     
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