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Homework Help: Differential Equations: System of equations

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data
    I have the following 2 equations dx/dt = 3x+y and dy/dt = sqrt(y)
    I need to find the general solution to the system.

    2. Relevant equations

    dx/dt = 3x+y and dy/dt = sqrt(y)

    3. The attempt at a solution

    i did the y equation first using separation of variables and got
    2sqrt(y) = t+c
    so... y = (2t+A)^2 where A = 2c

    then i plugged this into the 2nd equation for y so i have dx/dt = 3x + (2t+2t)^2
    i'm not sure how to solve this. I tried using method of undetermined coefficients but i keep getting stuck. Can someone please show me step by step how to solve this? Thank you
     
  2. jcsd
  3. Feb 14, 2010 #2

    tiny-tim

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    Hi SpiffyEh! :smile:

    (have a square-root: √ :wink:)
    No, (t/2 + A). :wink:
    (You need both a complementary solution and a particular solution … I assume you've got the former?)

    Substituting u = t/2 + A may make it easier.

    If it doesn't, show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Feb 16, 2010 #3
    oops sorry completely missed the t/2...

    well when i start the method of undetermined coefficients i can ge the yh but the yp is where i have issues. Since it has to be something like the right hand side i can't figure out what to set it to. Is this the wrong method to use in this case?
     
  5. Feb 16, 2010 #4

    tiny-tim

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    Hi SpiffyEh! :wink:

    Whenever I see a polynomial on the RHS, I always try a polynomial as the particular solution. :smile:
     
  6. Feb 16, 2010 #5
    i understand that, i just don't see how it works out because of the c being in two different parts of the equation. I don't know if im just being stupid and thinking of it wrong or not.
     
  7. Feb 17, 2010 #6

    tiny-tim

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    (just got up :zzz: …)
    I'm confused … I thought you were trying to solve dx/dt = 3x + (t/2 + A)2 ? :confused:
     
  8. Feb 17, 2010 #7

    D H

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    The constant of integration from integrating the first equation has nothing to do with the constant of integration from integrating the second equation. You will have two constants of integration here. Think of them as c1 and c2, or A and B if you want.
     
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