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

[askmathquestions]

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 $y'(t) = f(t)y(t)$?

 

[pasmith]

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.

 

[julian]

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.

 

[lurflurf]

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

 

[julian]

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.

 

[julian]

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.