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System of ODEs

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider the following system of equations
    da/dt = -kab, db/dt = kab, a(0) = a0, b(0) = b0.

    Solve these equations exactly.


    2. Relevant equations



    3. The attempt at a solution
    I added them together to get d(a+b)/dt = 0 which implies a + b = a0 + b0.
    Therefore a = a0 + b0 - b so eliminating a we get db/dt = k(a0 + b0 - b)b which is separable, but I don't know where to go from here.

    Someone please help.
     
  2. jcsd
  3. Oct 13, 2011 #2

    Mark44

    Staff: Mentor

    So db/(a0 + b0 - b)b = k
    [tex]\frac{db}{b(a_0 + b_0 - b)} = k\cdot dt[/tex]

    The left side can be integrated by using partial fraction decomposition. You could simplify the work slightly by rewriting a0 + b0 as, say, M.
     
  4. Oct 13, 2011 #3
    Thanks.

    I let M = a0 + b0 and got 1/(b(M-b)) = 1/(Mb) + 1/(M(M-b))
    and this integrates to (ln(b)-ln(M-b))/M,
    therefore we get (ln(b)-ln(M-b))/M = kt + c, so ln(b/(M-b)) = M(kt+c), so b/(M-b) = exp(M(kt+c) but what do I do now.
     
  5. Oct 13, 2011 #4

    Mark44

    Staff: Mentor

    Then b = (M - b)exp(M(kt + c))
    ==> b - bexp(M(kt + c)) = Mexp(M(kt + c))
    ==> b(1 - exp(M(kt + c))) = Mexp(M(kt + c))
    ==> b = ?

    You should be able to get rid of the constant c, since you are given that b(0) = b0.

    Finally, since a and b add up to a constant, you can solve for a.

    When you get a, by all means, check your work. Check that a(0) and b(0) turn out as expected, and then check that a'(t) = -kab, and that b'(t) = -a'(t).
     
  6. Oct 13, 2011 #5
    We get b = Mexp(M(kt+c))/(1-exp(M(kt+c))
    so b(0) = b0 = Mexp(Mc)/(1-exp(Mc)), but how do we find c?
     
  7. Oct 13, 2011 #6

    Mark44

    Staff: Mentor

    b0(1 - exp(Mc)) = Mexp(Mc)
    ==> b0 - b0exp(Mc) = Mexp(Mc)
    ==> b0 = b0exp(Mc) + Mexp(Mc) = exp(Mc)(b0 +M)
    ==> b0/(b0 +M) = exp(Mc)

    Now take the ln of both sides.
     
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