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Is it possible to solve the differential equation for $y(t)$ the following ODE? $$ y^{\prime \prime}(t)+ \frac{f^{\prime}(t)}{2 f(t)} y^{\prime}(t) + k^{2} y(t) = 0 $$ It can also be rewritten as $$ \frac{1}{\sqrt{f(t)}} \frac{d (\sqrt{f(t)} y^{\prime}(t))}{dt} + k^{2} y(t) = 0 $$ I tried some substituition $y(t) = e^{z(t)}$ or $e^{t z(t)}$ but I just end up with another non trivial diferential equation in $z(t)$. I am trying to find a solution that reduces to $A e^{ik t}+ B e^{-i k t}$ when $f(t)$ is independent of $t$.

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    $\begingroup$ If $f(t) = t^a$, then we get solution with Bessel functions... $$C_1t^{(1-a)/2} J_{(1-a)/2}(kt)+C_2t^{(1-a)/2} Y_{(1-a)/2}(kt)$$ $\endgroup$
    – Gerald Edgar
    Commented Dec 12 at 14:29
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    $\begingroup$ The logarithmic derivative $f'(t)/f(t)$ suggests to consider $f(t)=\exp(a_0+a_1 t+a_2 t^2+a_3 t^3)$, which is solved by the tri-confluent Heun function(HeunT[] in Mathematica). Simpler cases (some $a_k=0$) give Hermite polynomials, Bessel functions (see comment by @GeraldEdgar), or simple exponentials. $\endgroup$
    – Fred Hucht
    Commented Dec 12 at 15:13
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    $\begingroup$ @trying Please check your 2nd equation, I think it should read $f(t)$ instead of $\sqrt{f(t)}$. Else, there is a factor $1/2$ missing in the first equation. $\endgroup$
    – Fred Hucht
    Commented Dec 12 at 15:31
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    $\begingroup$ @FredHucht sorry yes, I fixed the typo. $\endgroup$
    – trying
    Commented Dec 13 at 6:57
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    $\begingroup$ @trying Ok, my comment remains correct. $\endgroup$
    – Fred Hucht
    Commented Dec 13 at 8:04

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My comment as an answer: The ODE can be simplified with the substitution $f(t)\mapsto e^{2G(t)}$ and $G'(t)=g(t)$ to read \begin{align} \label{eq:1}\tag{1} y''(t) + g(t) y'(t) + k^2 y(t) = 0 \,. \end{align} It can be seen as an harmonic oscillator with time-dependent friction term $g(t)$.

The case $g(t)\equiv 0$ gives the simplest case \begin{align} \label{eq:2}\tag{2} y(t)=A e^{ik t}+ B e^{-i k t}, \end{align} while several other choices of $g(t)$ can be integrated using Mathematica, including \begin{align} g(t) &= a && \text{exponentials} \label{eq:3a}\tag{3a}\\ g(t) &= a t && \text{Hermite polynomials} \label{eq:3b}\tag{3b}\\ g(t) &= a t^2 && \text{tri-confluent Heun functions} \label{eq:3c}\tag{3c}\\ g(t) &= a e^t && \text{Laguerre polynomials} \label{eq:3d}\tag{3d}\\ g(t) &= a t^{-1} && \text{Bessel functions }J,Y \label{eq:3e}\tag{3e}\\ &\ldots \end{align} All these cases reduce to \eqref{eq:2} for $a\to0$.

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