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?



Solution to simple non-autonomous ODE
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..


