Problem:
Consider the following system: $$\frac{dx}{dt} = ay-bx$$
$$\frac{dy}{dt} = cx-dy,$$ where $a,b,c,d>0.$ Given positive initial conditions, I want to show that $x$ and $y$ remain positive on $[0,\infty).$
This is what I have done so far:
Let $t_1$ be the first point where $x(t_1) = 0$ and $\frac{dx}{dt}(t_1) \leq 0.$ Also let $t_2$ be the first point where $y(t_2) = 0$ and $\frac{dy}{dt}(t_2) \leq 0.$ Assume that $t_1<t_2.$ By plugging in $t_1$ in the first equation we get $0\geq \frac{dx}{dt}(t_1) = ay(t_1),$ which leads to the contradiction $y(t_1)\leq 0.$ Therefore, $t_1 =t_2.$
Help please :)
After this point I'm not sure what to do. If I can somehow show that, say, $x(t) = 0$ on $[t_1,\infty),$ then I can use the Identity Theorem to show that $x(t)=0$ for all positive reals and hence obtain a contradiction. But I'm not sure how I can show this.
Any help is highly appreciated. Thank you in advance!
Let ${x \choose y}:\mathbb{R} \to \mathbb{R}^2$ be a solution of this linear system with $x(0),y(0)>0$ and let $t_1>0$ be the first time where $x$ or $y$ is zero, w.l.o.g. $x$, say. Then $x(t_1)=0$ and $$ \forall t \in [0,t_1): ~ x(t)>0, y(t)>0. $$ Then $y(t_1) \ge 0$ and if $y(t_1)=0$ too, you get ${x(t) \choose y(t)}={0 \choose 0}$ on $\mathbb{R}$ by uniqueness of solutions of IVPs. Thus $y(t_1)>0$. Now you get a contradiction in your attempt: $$ 0 \ge x'(t_1)=ay(t_1) > 0. $$