Csdrhrt

I have two questions about ordinary differential equations:

1. Consider the first-order linear differential equation $x'(t)=a(t)x(t)+b(t)$, with initial condition $x(t_0)=x_0$. Suppose that $x_0>0$. By continuity, we know that there is a neighborhood of $t_0$, say $(t_0-delta,t_0+delta)$, where $x(t)>0$. Is it possible to know the maximum interval where $x(t)>0$ ?




2. Consider the Riccati differential equation $x'(t)=a x(t)^2+b x(t)+c$ (here $a$, $b$ and $c$ are constants), with initial condition $x(t_0)=x_0$. Applying the Fröbenius method (search of a power series solution), one can find a formal solution $x(t)=sum_{n=0}^infty x_n (t-t_0)^n$, where the coefficients $x_n$ satisfy a difference equation. Is is possible to establish the convergence of this series on a maximum interval? I was not able to do so, due to the difficulty arisen from the term $x(t)^2$.

I do not know if these two questions belong to well-studied topics in the literature or not. In the former case, a reference would be great.

1. Assuming $a(t)$ and $b(t)$ are continuous, solutions are globally defined. The maximal interval where $x(t)>0$ is $(c,d)$ where $d$ is the least root $> t_0$ of $x(t)=0$ (or $+infty$ if there is no such root), and similarly $c$ is the greatest root $<t_0$ or $-infty$. Of course the first-order linear DE has an explicit solution in terms of integrals, but
   in general those integrals can't be expressed in closed form, nor can we express the roots in closed form.




2. Your Riccati equation's solution is of the form
   $$ x(t) = A tan(B (t - t_0) + C) + D$$
   for appropriate $A,B,C, D$ (which I'll leave it to you to find). This has singularities when $B (t-t_0) + C= (2k+1)pi/2$ for integer $k$. The radius of convergence of the power series solution is the distance from $t_0$ to the closest singularity in the complex plane.

1) Asking about the positivity of $x$ is the same as asking about the positivity of $y(t)=e^{-A(t)}x(t)$ where $A'=a$. Then $$y'(t)=e^{-A(t)}b(t).$$ Any upper bound for that expression will give some lower bound for the position of a root.

2) You can transform the Riccati equation into a linear ODE of second order by setting $y=-frac{u'}{au}$ with $$u''-frac bau'+cu=0$$ where the recursion formula for the power series coefficients will be easier to derive. The roots of $u$ are the poles of $y$, so that bounds on the roots are also bounds on the poles.

Alternatively take $r_1,r_2$ to be the roots of $ar^2+br+c$ and consider $$u=frac{y-r_1}{y-r_2}implies u'=-a(r_2-r_1)u$$
which allows a direct computation of the solution.