Taylor series of infinitely differentiable function Suppose that $f$ is infinitely differentiable on an open interval that contains 0. Suppose that $f'(x)=2xf(x)$ and $f(0)=1$. Express $f$ as a Taylor series around 0. What is the radius of
convergence of this Taylor series? [Hint: Think of an exponential function.]
 A: A possibility, to get you started: (there are others, this one generalizes to other differential equations)
Write $f$ as a power series, with radius of convergence $R\in[0,\infty]$:
$$
f(x) = \sum_{n=0}^\infty a_n x^n, \qquad x\in(-R,R) \tag{1}
$$
and note that the assumption gives already $a_0 = f(0)=1$.
Now,from there  we get the expression of the power series for $f'$:
$$
f'(x) = \sum_{n=1}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^{n}, \qquad x\in(-R,R) \tag{2}
$$
as well as that of $x\mapsto 2xf(x)$:
$$
2xf(x) = \sum_{n=0}^\infty 2a_n x^{n+1} = \sum_{n=1}^\infty 2a_{n-1} x^{n}, \qquad x\in(-R,R) \tag{3}
$$
But, since $f'(x)=2x f(x)$ for all $x$, if $R>0$ then we know from (2) and (3) that the coefficients (by unicity of Taylor series) must satisfy
$$
2a_{k-1} = (k+1)a_{k+1}, \qquad k\geq 1 \tag{4}
$$
along with the $a_0=1$ from before.
Compute the coefficients using (4), then check that indeed the corresponding radius of convergence will be positive.
A: If your aim is to solve indirectly using series:  let's try to find the $n$th derivative using the given relation$$f'(x)=2xf(x)$$
$$f'(0)=0$$
$$f''(x)=2xf'(x)+2f(x)\implies f''(0)=2$$
$$f'''(x)=2xf''(x)+4f'(x)\implies f'''(0)=0$$
$$f^{(4)}(x)=2xf'''(x)+6f''(x)\implies f^{(4)}(0)=12$$
$$f^{(5)}(x)=2xf^{(4)}(x)+8f'''(x)\implies f^{(5)}(0)=0$$
It seems like:
$$f^{(n)}(x)=2xf^{(n-1)}(x)+2(n-1)f^{(n-2)}(x)$$
This checks out for $n+1$. Then
$$f^{(n)}(0)=2(n-1)f^{(n-2)}(0)$$
If $n$ odd: $$f^{(2k+1)}(0)=2(2k)f^{(2k-1)}(0)=2(2k)\times\cdots\times f'(0)=0$$
If $n$ even  $$f^{(2k)}(0)=2(2k-1)f^{(2k-2)}=2(2k-1)2(2k-3)\times\cdots\times 2\times3\times 2\times 1 \times f(0)$$$$\boxed{\color{blue}{f^{(2k)}(0)=2^{k}\times (2k-1)!!}}$$
So that finally
\begin{equation}
f(x)=\sum_k f^{(2k)}(0)\underbrace{\frac{x^{2k}}{(2k)!}}_{\substack{\text{product of integers} \\
 \text{from $1$ to $2k$}}}=\sum_k 2^{k}\times \underbrace{(2k-1)!!}_{\text{product of odd integers}}\frac{x^{2k}}{\underbrace{(2k)!!}_{\substack{\text{product of}\\\text{integers of form}\\\text{2r}}}(2k-1)!!}\end{equation}
$$\color{blue}{f(x)=\sum_k\frac{x^{2k}}{k!}}$$
A: Hint...$\frac{f'(x)}{f(x)}=2x\Rightarrow \ln(f(x))=?$
