Given the homogeneous linear ODE $$y''(x)+ P(x) y'(x) + Q(x) y(x)=0$$ where $x\in(0,\infty)$ and $P$ and $Q$ are some smooth (but not necessarily bounded) functions. I know that we can write this as a set of 2 first order ODEs in $y$ and $w=y'$, or, given a solution $y_1$, we can write a first order ODE for a second linearly independent solution $y_2$.

Now let $y_1$ and $y_2$ be two linearly independent solutions. Is it possible to write the second order ODE above as a system of two first order coupled ODEs in $y_1$ and $y_2$? If not, is it easy to show why not, or are there specific assumptions under which it can be done?

In the specific case I am considering, I do not know $y_1$ neither $y_2$, but I do know their asymptotic behaviors (i.e. leading order) as $x\rightarrow \infty$ (in my case they are some power laws $x^a$ with $a<0$, and of course $a$ is different for $y_1$ and $y_2$). I'm not sure whether this changes anything though. I thought this might be relevant since this gives a constraint on which linearly independent solutions I'm actually considering.

EDIT: what I've been trying is the following. I know that asymptotically I can write the (leading orders of the) solution as $$y = b_1 x^{a_1}+b_2 x^{a_2}$$ where $a_1,a_2<0$ and $b_1,b_2\in\mathbb{C}$. Now I wish to write the general solution as $$y=b_1(x) x^{a_1}+b_2(x) x^{a_2}$$ where $b_1(x)$ and $b_2(x)$ go to $b_1$ and $b_2$ respectively as $x\rightarrow \infty$ and where $y_1(x)=b_1(x)x^{a_1}$ and $y_2(x)=b_2(x)x^{a_2}$ are two linearly independent solutions to the ODE.

My goal is then to derive from the ODE two first order coupled ODEs in $b_1(x)$ and $b_2(x)$, or equivalently in $y_1$ and $y_2$. I tried using the fact that $b_1(x)x^{a_1}$ and $b_2(x)x^{a_2}$ are solutions on their own, but so far no luck.


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