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Homework Help: System of Differential Equations

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{dx}{dt} +\frac{dy}{dt} + y - x = e^{2t}[/tex]
    [tex]\frac{d^2x}{dt^2} +\frac{dy}{dt} = 3e^{2t}[/tex]


    I am a little stumped on this one. I tried writing it using D as the differential operator d/dt:

    D(x+y) +x - y = e2t
    D2x+Dy = 3e2t

    but that doesn't do anything but change the variables to something else... if I subtract the 1st EQ from second I get

    [tex]\frac{d^2x}{dt^2}-\frac{dx}{dt} -y +x = 2e^{2t}[/tex]

    But I still have x AND y in here.

    Any hints on this?
     
  2. jcsd
  3. Sep 29, 2010 #2

    Mark44

    Staff: Mentor

    It seems to me that both x and y will have to involve e2t.
     
  4. Sep 29, 2010 #3
    Hi Mark44 :smile: I am not sure if you are giving me a hint or just making a comment. I am sure that they both will, but I am trying to eliminate all 'x' or all 'y' from one equation so that I can solve for it. Just like with a system of regular equations (i.e. not DEs).

    Just not sure how to approach it.
     
  5. Sep 29, 2010 #4

    Mark44

    Staff: Mentor

    I'm giving you a hint. I was able to find solutions for x and y by "guessing" that each was some multiple of e2t. The solutions I found worked, but it's possible that I didn't get the complete solution.

    Alternatively, and more in line with what you're trying, you could solve for y' in the second equation, integrate to get y, and then substitute for y and y' in the first equation. Then you would have a single 2nd order, linear equation in one dependent variable.
     
  6. Sep 29, 2010 #5
    One approach is to treat the system in terms of the differential operators and then just use simple elimination keeping in mind the terms are operators. So writing it as:

    [tex]
    \begin{aligned}
    (D-1)x&+(D+1)y&=e^{2t}\\
    D^2 x&+Dy&=3e^{2t}
    \end{aligned}
    [/tex]

    Now, to solve for y, I can operate on the first with the D^2 operator and on the second with the (D-1) operator:

    [tex]
    \begin{aligned}
    D(D-1)x&+D(D+1)y&=De^{2x} \\
    D^2(D+1)x&+D(D+1)y&=(D+1)3e^{2x}
    \end{aligned}
    [/tex]

    Notice now I can subtract the second from the first and obtain:

    [tex]D(D-1)x-D^2(D+1)x=De^{2x}-(D+1)3e^{2x}[/tex]

    [tex](D^2-D-D^3-D^2)x=2e^{2x}-(6e^2x}+3e^{2x})[/tex]

    [tex](D^3+D)x=e^{2t}[/tex]

    Solve that with the set of [itex](a_1,a_2,a_3)[/itex] arbitrary constants. Do the same for y with [itex](b_1,b_2)[/itex], back-substitute into the DE, equate coefficients to determine the relationship between the a and b coefficients.

    See "Differential Equations" by Rainville and Bedient
     
  7. Sep 30, 2010 #6
    Hi again Mark44 :smile: I still don't see it. The second equation is

    x'' + y' = 3e2t or y' = 3e2t - x''

    I don't see how I can "integrate to obtain y." y' is in terms of 't' and 'x.'

    As I recall, some refer to this as the annihilator method right? This was the approach I was going for, but I just did not have the proper foresight.
     
  8. Sep 30, 2010 #7

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This equation can be directly integrated to

    y = (3/2) e2t - x' + c
     
  9. Sep 30, 2010 #8
    OH! I still have to wrap my head around the fact that these are differential equations. As such, I did not have to 'solve' the second equation completely for y interms of t. I see now that x' in the above can be directly substituted into the 1st equation leaving only x in terms of t.

    Thanks guys for your patience. :smile:

    Casey
     
  10. Sep 30, 2010 #9
    Nicer way. Thank you. I didn't know that either. :)
     
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