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?

$\{\begin{array}{l}\frac{dA}{dt}=-aA\\ \frac{dB}{dt}=aA-bB\\ \frac{dC}{dt}=bB\end{array}$

I have solved the first equation:

$A(t)={A}_{0}{e}^{-at}$

I want to solve this system, but I have never solved a system of ODE, can you help me?

$\{\begin{array}{l}\frac{dA}{dt}=-aA\\ \frac{dB}{dt}=aA-bB\\ \frac{dC}{dt}=bB\end{array}$

I have solved the first equation:

$A(t)={A}_{0}{e}^{-at}$

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 $({e}^{bt})$ on both sides.

EDIT:

${e}^{bt}\frac{dB}{dt}+b{e}^{bt}B=a{A}_{0}{e}^{(b-a)t}$

Now L.H.S. is equal to $\frac{d(B{e}^{bt})}{dt}$

Hence,

$\begin{array}{}\text{(k is some constant)}& B{e}^{bt}=a{A}_{0}\int {e}^{(b-a)t}dt=a{A}_{0}\frac{{e}^{(b-a)t}}{(b-a)}+k\end{array}$

which gives $B=a{A}_{0}\frac{{e}^{-at}}{(b-a)}+k{e}^{-bt}$

Now, if we add three ODE's, we have,

$\begin{array}{}\text{(}{c}_{0}\text{is some constant)}& \frac{d(A+B+C)}{dt}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A+B+C={c}_{0}\end{array}$

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

$\frac{dB}{dt}+bB=a{A}_{0}{e}^{-at}$

This is a linear differential equation in $B$ which can be solved by multiplying integrating factor $({e}^{bt})$ on both sides.

EDIT:

${e}^{bt}\frac{dB}{dt}+b{e}^{bt}B=a{A}_{0}{e}^{(b-a)t}$

Now L.H.S. is equal to $\frac{d(B{e}^{bt})}{dt}$

Hence,

$\begin{array}{}\text{(k is some constant)}& B{e}^{bt}=a{A}_{0}\int {e}^{(b-a)t}dt=a{A}_{0}\frac{{e}^{(b-a)t}}{(b-a)}+k\end{array}$

which gives $B=a{A}_{0}\frac{{e}^{-at}}{(b-a)}+k{e}^{-bt}$

Now, if we add three ODE's, we have,

$\begin{array}{}\text{(}{c}_{0}\text{is some constant)}& \frac{d(A+B+C)}{dt}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A+B+C={c}_{0}\end{array}$

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 $(3)$ to find $C$ since $(1)+(2)+(3)$ gives

$\frac{d}{dt}(A+B+C)=\frac{dA}{dt}+\frac{dB}{dt}+\frac{dC}{dt}=0\u27faA+B+C=\text{const}\u27faC=\text{const}-A-B.$

$\frac{d}{dt}(A+B+C)=\frac{dA}{dt}+\frac{dB}{dt}+\frac{dC}{dt}=0\u27faA+B+C=\text{const}\u27faC=\text{const}-A-B.$

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