Can this system of ODE's be solved?

It's my first time dealing with a system of ODE's. I don't know if it is even possible to find solutions. I tried to solve it but i didn't make it.

Here is my system : $\forall t\in\mathbb R,\,a,b,c\in]0,1],$ $$ \left\{ \begin{aligned} \dot x(t)&=-ax(t)y(t)-bx(t) \\ \dot y(t)&=ax(t)y(t)-cy(t) \\ \dot z(t)&=bx(t)+cy(t) \end{aligned} \right. $$

Is it solvable ? And if it is, how to solve it ? Thanks for the help in advance !

Edit : I find something with $y$ : $$\dot y = -\frac{1}{a}\left(\frac{\ddot x}{x} - \left(\frac{\dot x}{x}\right)^2\right)$$

If $b = c$ then $\dot x + \dot y = -b(x+y)$. Let $\alpha\in\mathbb R,\,x+y = \alpha e^{-bt}$.

And let $f = y-x$, we have $\dot y- \dot x = 2axy - c(y-x) = \frac{a}{2}((x+y)^2-(x-y)^2) -c(y-x)$

So : $$\dot f +cf= \frac{a}{2}(\alpha e^{-2bt}-f^2)$$

Solution 1:

You can deduce from the first two DE: $$\dfrac {\dot x}{ \dot y}=\dfrac {axy+bx}{-axy+cy}$$ $$\dfrac {(-ax+c)dx}{ x}=\dfrac {(ay+b)dy}{y}$$ $$-ax+c \ln x=ay+b\ln y +K$$ Then you can deduce $z(t)$ once you solve this system.