I have just started learning about differential equations, as a result I started to think about this question but couldn’t get anywhere. So I googled and wasn’t able to find any particularly helpful results. I am more interested in the reason or method rather than the actual answer. Also I do not know if there even is a solution to this but if there isn’t I am just as interested to hear why not.

Is there a solution to the differential equation:

f(x)=∞∑n=1f(n)(x)

**Answer**

f(x)=exp(12x) is such a function, since f(n)=2−nf(x), you have

∞∑n=1f(n)(x)=∞∑n=12−nf(x)=(2−1)f(x)=f(x)

This is the only function (up to a constant prefactor) for which ∑nf(n) and its derivatives converge uniformly (on compacta), as f′=∞∑n=1f(n+1)=f−f′ follows from this assumption. But this is the same as f−2f′=0, of which the only (real) solutions are f(x)=Cexpx2 for some C∈R.

**Attribution***Source : Link , Question Author : o.comp , Answer Author : s.harp*