# Simple system of Linear ODEs

I have this simple system of ODEs for rates $$r_1,r_2$$ and $$a(t), b(t)$$:

$$a' = r_2b - r_1a$$ $$b' = -r_2b + r_1a$$

I am trying to solve for $$a(t), b(t)$$, but I am not sure what I am doing wrong. I first start with differentiating the first ODE:

$$a'' = r_2b'-r_1a'$$

$$\; \; \; \;$$ $$= r_2[-r_2b + r_1a] - r_1a'$$

$$\; \; \; \;$$ $$= r_2[(-a'-r_1a) + r_1a] - r_1a'$$

$$\; \; \; \;$$ $$= -r_2a' - r_1a'$$

$$\; \; \; \;$$ $$= a'(-r_2-r_1)$$

$$\rightarrow a(t) = a_0e^{(-r_2-r_1)t}$$

Following the same logic I also get:

$$\rightarrow b(t) = b_0e^{(-r_2-r_1)t}$$

However I am confused as to why both $$a$$ and $$b$$ have the same decay rates. I might be wrong, but I would have assumed that the solutions would have been

$$a(t) = a_0e^{(r_2-r_1)t}$$

$$b(t) = b_0e^{(r_1-r_2)t}$$

Where did I go wrong?

• Don't you think your last edit removed important information from the question? – rafa11111 Dec 10 '18 at 19:45

## 2 Answers

Hint:

Notice that $$a' = -b'$$

With this in mind, look again in your calculations step by step.

You will find a mistake and you will answer your own question

The conclusion from $$a''=-ra',$$ $$r=r_1+r_2$$, is not $$a(t)=a_0e^{-rt}.$$ As a second order linear differential equation it has two basis solutions and two integration constants.

Instead, you should get $$a(t)=a_0+c(e^{-rt}-1).$$ Insert into $$b=(a'+r_1a)/r_2$$ to get $$b(t)=\frac{r_1(a_0-c)-r_2ce^{-rt}}{r_2}=b_0-c(e^{-rt}-1)\implies b_0=\frac{r_1a_0-rc}{r_2}$$ so that $$c=\frac{r_1a_0-r_2b_0}{r}.$$