# Can we solve a non-autonomous diffeq via Taylor series?

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In summary, the conversation discusses the possibility of finding a Taylor series solution for a non-autonomous differential equation. It is mentioned that in principle, it is possible to find a Taylor series solution for the non-autonomous case, but it may not always exist, as shown in the example where the function is not analytic.

I've occasionally seen examples where autonomous ODE are solved via a power series.

I'm wondering: can you also find a Taylor series solution for a non-autonomous case, like ##y'(t) = f(t)y(t)##?

In principle, yes. You end up with$$(n+1)a_{n+1} = \sum_{k=0}^n \frac{f^{(n-k)}(0)}{(n-k)!}a_k$$ where the right hand side only involves $a_k$ which are already known.

topsquark
pasmith said:
In principle, yes. You end up with$$(n+1)a_{n+1} = \sum_{k=0}^n \frac{f^{(n-k)}(0)}{(n-k)!}a_k$$ where the right hand side only involves $a_k$ which are already known.
What if

$$f (t) = \left\{ \begin{matrix} e^{ -\frac{1}{t^2} } & t > 0 \\ 0 & t < 0 \end{matrix} \right.$$

Wouldn't all the ##a_n##'s be zero aside from ##a_0##? And the Taylor series solution give ##y(t)= y(0)##?

The general solution to the differential equation ##y'(t) = f(t) y (t)## is

$$y(t) = y(0) e^{\int_0^t f (t') dt'}$$

For the example above, the solution would be

$$y (t) = y(0) e^{\int_0^t e^{ -\frac{1}{t^{'2}} } dt'}$$

which isn't a constant.

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topsquark
That function is not analytic. Its not equal to its Taylor series so you cannot substitute the series for the function.

PeroK
lurflurf said:
That function is not analytic. Its not equal to its Taylor series so you cannot substitute the series for the function.
I was illustrating that when ##f(t)## is not analytic there may not exist a Taylor series solution to the differential equation ##y'(t) = f(t) y(t)##. The OP was asking if we can solve the differential equation via a Taylor series. And the answer is not always.

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I was using the fact that the Taylor series of the function

$$f (t) = \left\{ \begin{matrix} e^{ -\frac{1}{t^2} } & t > 0 \\ 0 & t < 0 \end{matrix} \right.$$

at the origin converges everywhere to the zero function in order to prove that a Taylor series solution of the differential equation at the origin converges everywhere to the constant function. Meaning that a Taylor series solution fails to solve the differential equation for this choice of ##f(t)##. Meaning that the differential equation cant always be solved via a Taylor series.

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