Help solving a system of ODE

Question

I need help solving the following system of ODE. The problem is not so much getting the final solution, but getting a clear view of the process to getting there.
The system is the following: $$\begin{cases} \dfrac{dA(t)}{dt} = -aA(t) \\ \dfrac{dB(t)}{dt} = aA(t) - bB(t) \\ \dfrac{dC(t)}{dt} = bB(t)\\ \end{cases} $$

I know that $A = A(0)e^{-at}$, but I cannot figure out the expressions for B and C.
Thanks in advance!

Answer 1

This is a linear differential system of form $\dot{x}(t) = M x$, for below matrix $M$. The (matricial) solution for this ODE is $x(t, t_0, x_0) = \Phi(t, t_0) \, x_0$, such that matrix $\Phi(t, t_0)$ is called transition matrix. The transition matrix in this case is $\exp(M \, (t-t_0))$.

\begin{equation} M = \begin{bmatrix} -a&0&0 \\ a&-b&0 \\ 0&b&0 \\ \end{bmatrix} \end{equation}

There is no closed form based on exponential sum like you expect as an answer in general case. For this particular case, I performed the following Wolfram Alpha query for matrix $e^M$, let as call it N:

Therefore, the answer $x(t, t_0, x_0)$ you look for is given by $N^{t-t_0} \, x_0$

Answer 2

As you stated we have that $A = A_0 e^{-at}$. We determine $B$. The differential equation for $B$ is a linear first order differential equation of the form:

$$ \frac{dB}{dt} + P(t)B = Q(t) $$

with $P(t) = b$ and $Q(t) = aA_0e^{-at}$. The solution of such an equation can be obtained by using an integrating factor $I(t)$ such that:

$$ I(t) \left( \frac{dB}{dt} + P(t)B \right) = \left(I(t)B\right)' = I(t)Q(t) $$

such a factor can be found to be:

$$ I(t) = e^{\int P(t) dt} $$

Now we have:

$$ \left(I(t) B \right)' = I(t)Q(t)$$

Integrating both sides we get:

$$B(t) = \frac{1}{I(t)} \left[ \int I(t)Q(t) dt + C\right]$$

And so:

$$ B(t) = \frac{aA_0}{b - a} e^{-at} + \left(B_0 - \frac{aA_0}{b - a} \right) e^{-bt}$$

with $B_0 = B(t = 0)$. We can verify this is the correct solution by substituting into the first equation:

$$ \frac{dB}{dt} + P(t)B = -\frac{a^2 A_0}{b - a} e^{-at} - b\left( B_0 - \frac{aA_0}{b - a} \right) e^{-bt} + b \left( \frac{aA_0}{b - a} e^{-at} + \left(B_0 - \frac{aA_0}{b - a} \right)e^{-bt} \right)$$ $$ = \left(-\frac{a}{b - a} + \frac{b}{b - a} \right) a A_0 e^{-at} $$ $$ = a A_0 e^{-at} = Q(t) $$