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Consider the following ODE with parameters $\alpha,\beta,\gamma \in \mathbb R$

$$f'(t)= \begin{pmatrix} \alpha-\beta t & \gamma t \\ \gamma t & -(\alpha-\beta t) \end{pmatrix} f(t).$$

This ODE is non-autonomous and the matrix also does not commute with its derivatives, so diagonalization is not going to bail us out either.

However, since the coefficients are first order polynomials. Can we still derive an explicit solution?

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Let us reduce it to a scalar linear ODE. In general, $$f'=\left(\begin{array}{cc}A&B\\ C&D\end{array}\right)f$$ is equivalent to $$w''+pw'+qw=0,$$ where $w$ is the first component of $f$, and $p=-(B'/B+A+D),$ and
$q=-A'+AB'/B+AD-BC$. So, if my computation is correct, we obtain $$p=-1/t,\quad q=-(\gamma^2+\beta^2)t^2+2\alpha\beta t-\alpha^2+\alpha/t.$$ Further, on can kill $p$ by setting $w=y\sqrt{t}$ and obtain $$y''+\left(-(\beta^2+\gamma^2)t^2+2\alpha\beta t-\alpha^2+\frac{\alpha}{t}-\frac{3}{4t^2}\right)y=0.$$ Edit. One can show that $0$ is an apparent singularity, so this equation can be reduced to the Weber equation (parabolic cylinder) $u''=(t^2+c)u$, but a simpler way to do this reduction is indicated in the answer of Michael Renardy. There are countably many $c=c_n$ for which this equation has an elementary solution.

For generic $\alpha,\beta,\gamma$, the solutions are not elementary functions..

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  • $\begingroup$ I am puzzled by your last comment. When beta and gamma are zero, then the original ODE completely decouples, no? In fact, it even suffices to just have gamma equal to zero. $\endgroup$ – Martinique Apr 12 at 13:56
  • $\begingroup$ I removed the last sentence, which was incorrect, but everything else seems to be correct. Including the conclusion. $\endgroup$ – Alexandre Eremenko Apr 12 at 15:05
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Diagonalize the part of the matrix that is proportional to $t$. You get a system of the form $$x'=\lambda t x+\tilde\alpha x+\tilde\beta y, y'=-\lambda t y+\tilde\gamma x+\tilde\delta y.$$ Now use the first equation to eliminate $y$. You end up with a single equation of the form $$x''+(at+b)x'+(ct^2+dt+e)x=0.$$ This equation has just one irregular singular point at infinity and can be reduced to the parabolic cylinder equation.

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  • $\begingroup$ There are countably many explicit polynomials $F_n$ such that whenever the relation $F_n(\alpha,\beta,\gamma)=0$ holds for some $n$, there is an elementary solution (eigenfunction of harmonic oscillator). $\endgroup$ – Alexandre Eremenko Apr 12 at 23:32

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