System of ODEs

squenshl

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.

Mark44

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.

squenshl

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.

Mark44

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).

squenshl

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?

Mark44

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.