Given the following differential equation $$y'(x)=y(x)$$ It is obvious that $y$ is infinitely differentiable at any point. One would approach the problem by expanding the series of $y$ and eventually arriving at the series for $e^x$. However how could one justify that $y$ could be rewritten as its Taylor series (i.e. it is analytic). Using the theorem for uniqueness of solution one could just say that it is the solution. However in some cases one does not know whether there is a unique solution. For example $y'(x)=y(-x)$. I mean, if we want to solve this problem by findind the Taylor series for $y$, is there some way we can prove that $y$ is analytic, using only the above equation, so that replacing $y$ with its expansion is correct.

  • $\begingroup$ In that case, $y''(x)=-y'(-x)=-y(x)$ is still a well-solvable equation, then use the original equation to reduce the degrees of freedom. $\endgroup$ – LutzL Feb 5 '17 at 21:57
  • $\begingroup$ If you want to prove that solution $y'(x) = y(-x)$ is analytic then this case is no different from $y'(x) = y(x)$ at all. Suppose that classical solution exists, then solution $y(x)$ is at least $C^1$. But the derivative $y'(x)$ is $C^1$ too (because it's just a $y(-x)$, which is obviously $C^1$), so we can differentiate it and get that $y''(x)$ is $C^1$ too. The line of reasoning stays the same: we have derivatives of all orders, and thus can try to find the solution in form of Taylor series. I might be mistaken and I would be glad if someone points the error in this reasoning. $\endgroup$ – Evgeny Feb 6 '17 at 7:34
  • $\begingroup$ Yes, it is OK that $y$ is in the class $C^{\infty}$, but how do we know it is analytical? $\endgroup$ – Stoyan Apostolov Feb 6 '17 at 16:41

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