Show without solving the ODE that an equality is an invariant after initial condition

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I am studying invariants for systems of ODEs. For example, I have proved that if $x'(t)=f(x(t))$ such that $x\, (t_0) > c$, and if $f(k)>0$ for all $k \geq c$, then the derivative will keep being greater than $0$, and therefore $x(t)$ for all $t\geq t_0$. I proved this by contradiction showing that if $t_1$ is the infimum of all those points not satisfying the conclusion of the statement, then $t_1$ would have to satisfy it and, by continuity, its neighbourhood also has to do it. However, I have not been able to generalise this argument for equalities yet.

Consider for instance the system of ODEs $$\begin{cases} x_1'(t) = A\cdot x_1^2(t)+B\cdot x_1(t),\\ x_2'(t) = A\cdot x_1(t)\cdot x_2(t)+B\cdot x_2(t). \end{cases}$$ Show that if $x_1(t_0) + x_2(t_0)=0$, then $x_1(t) + x_2(t)=0$ for all $t\geq t_0$.

Notice that $x_1'(t) + x_2'(t) = (A\cdot x_1(t)+B)\cdot(x_1(t)+x_2(t))$ which at least for $t_0$ we know it to be equal to $0$. I have tried to combine Grönwall's (in)equality, the mean value theorem and intermediate value theorem in various unfruitful arguments. Hopefully, the math-StackExchange community can provide some insights into solving this particular problem. Then I can try to generalise said insight.

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Take any solution of the equation $$ x_1'(t) = A x_1^2(t)+B x_1(t). $$ Then, as you wrote, $$ y'(t) = C(t)y(t), $$ where $$y(t)=x_1(t)+x_2(t)\quad\text{and}\quad C(t)=Ax_1(t)+B.$$ Being a linear equation, if $y(t_0)=0$, then $y(t)=0$ for all $t$.