Addison Trujillo

Answered

2022-07-01

Set of ODE: how can I solve it?
I want to solve this system, but I have never solved a system of ODE, can you help me?
$\left\{\begin{array}{l}\frac{dA}{dt}=-aA\\ \frac{dB}{dt}=aA-bB\\ \frac{dC}{dt}=bB\end{array}$
I have solved the first equation:
$A\left(t\right)={A}_{0}{e}^{-at}$

Answer & Explanation

Danika Rojas

Expert

2022-07-02Added 9 answers

After putting solution of first equation in 2nd, we get,
$\frac{dB}{dt}+bB=a{A}_{0}{e}^{-at}$
This is a linear differential equation in $B$ which can be solved by multiplying integrating factor $\left({e}^{bt}\right)$ on both sides.
EDIT:
${e}^{bt}\frac{dB}{dt}+b{e}^{bt}B=a{A}_{0}{e}^{\left(b-a\right)t}$
Now L.H.S. is equal to $\frac{d\left(B{e}^{bt}\right)}{dt}$
Hence,
$\begin{array}{}\text{(k is some constant)}& B{e}^{bt}=a{A}_{0}\int {e}^{\left(b-a\right)t}dt=a{A}_{0}\frac{{e}^{\left(b-a\right)t}}{\left(b-a\right)}+k\end{array}$
which gives $B=a{A}_{0}\frac{{e}^{-at}}{\left(b-a\right)}+k{e}^{-bt}$
Now, if we add three ODE's, we have,

Now put $A$,$B$ to get $C$.

antennense

Expert

2022-07-03Added 7 answers

If you solve the first two equations as Avatar suggested you don't really need to substitute into $\left(3\right)$ to find $C$ since $\left(1\right)+\left(2\right)+\left(3\right)$ gives
$\frac{d}{dt}\left(A+B+C\right)=\frac{dA}{dt}+\frac{dB}{dt}+\frac{dC}{dt}=0⟺A+B+C=\text{const}⟺C=\text{const}-A-B.$

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