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The goal is a recurrence relation for the series coefficients; maybe the

Quantile Mechanics $(93)$ to $(97)$

method, like the Frobenius method, to solve $yy’’=y’$. First, substitute $y=\sum\limits_{k=0}^\infty a_k x^{k+r}$:

$$\sum_{n=0}^\infty\sum_{m=0}^\infty(n+r+1)(n+r+2)a_ma_{n+2}x^{n+r}x^{m+r}=\sum_{k=0}^\infty(k+r+1)a_{k+1}x^{k+r}$$

Equating coefficients with Kronecker $\delta_{u,v}$:

$$(k+r+1)a_{k+1}= \sum_{n=0}^\infty\sum_{m=0}^\infty(n+r+1)(n+r+2)a_ma_{n+2}\delta_{k+r,m+n+2r}$$ and make $n$’s upper bound finite since the double sum is $0$ when $k+r\ne m+n+2r\iff k\ne m+n+r$. Following Quantile Mechanics, erase $\sum\limits_{m=0}^\infty$, $m=k-n-r$, and the upper bound of $n$ being $k-r$ is just a guess:

$$\boxed{(k+r+1)a_{k+1}=\sum_{n=0}^{k-r}(n+r+1)(n+r+2)a_{k-n-r}a_{n+2}}$$

Is this recurrence relation true and if not, what is the recurrence relation for the series coefficients of $yy’’=y’$?

Тyma Gaidash
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    Since this is an autonomous differential equation, we can use the substitution $v=y'$ to obtain $v'(y)=1/y\implies z=y'=\log y(x)+C$. That may point to why the method of Frobenius fares poorly here. On the other hand, one may have more success by working with $y$ as the independent variable by writing the ODE as $x'(y)=1/(C+\log y)$ instead. (Though in that case one should probably not expand around $y=0$.) – Semiclassical Feb 05 '23 at 00:46
  • @Semiclassical An example of the Frobenius-related method is pictured here. If one tried the same on $y’=\ln(y)+C$, it may not work. Are you sure this one does not work on $yy’’=y’$? – Тyma Gaidash Feb 05 '23 at 00:50
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    The ODE used in that case is $w''(v)=w (w')^2$, in which case the substitution $z=w'(v)$ yields $w''(v)=dz/dv=z'(w) v'(z)=w z'(w)$ and thus $z'(w)=z^2$. This solves to $z(w)=w'=-1/(w+A)$. This is still a bit ugly, but it doesn't yield the same log issues that the other ODE does. (In particular, writing the ODE as $v'(w)=-w-A$ yields $v=-w^2/2-Aw-B$. Hence $v$ is polynomial in $w$ and series reversion to recover $v$ will likely succeed.) So while a Frobenius solution may work in that case of a nonlinear equation, it doesn't guarantee it will work in others. – Semiclassical Feb 05 '23 at 01:04
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    Taking into account that, you need to inverse $$x=e^{c_1} \text{Ei}(\log (y)-c_1)+c_2$$ very little hope (in my opinion) – Claude Leibovici Feb 05 '23 at 03:47
  • @ClaudeLeibovici Yes, but the inverse error function has a fairly simple recurrence relation series, so maybe $yy’’=y’$ can at least be tried. – Тyma Gaidash Feb 05 '23 at 12:18

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