So I have a question that says find the radius of convergence after I have found the power series solution of a given differential equation. I know to find the radius of convergence you take
$$ p=\lim_{n \rightarrow \infty} \left\lvert\frac{C_n}{C_{n+1}}\right\rvert $$
but, I don't understand where $C_n$ and $C_{n+1}$ come from. Could someone please explain how I would find $C_{n}$ and $C_{n+1}$?
Best Answer
An elementary (and not particularly smart) approach could be:
Let's say the problem is $$\begin{cases}y'(x)=f(x,y(x))\\ y(0)=y_0\end{cases}$$
Since $C_n=\dfrac{y^{(n)}(0)}{n!}$, you "only" need to compute $y^{(n)}(0)$.
You know that $y'(0)=f(0,y_0)$.
Deriving in $x$ the first equation you get $$y''(x)=\frac{\partial f}{\partial x}(x,y(x))+y'(x)\frac{\partial f}{\partial y}(x,y(x))$$
Whence $y''(0)=\dfrac{\partial f}{\partial x}(0,y_0)+y'(0)\dfrac{\partial f}{\partial y}(0,y_0)$. Notice that you have already calculated $y'(0)$ the step before.
Keep deriving $$y^{(3)}(x)=\\= \dfrac{\partial^2 f}{\partial x^2}(x,y(x))+2y'(x)\dfrac{\partial^2 f}{\partial x\partial y}(x,y(x))+y''(x)\dfrac{\partial f}{\partial y}(x,y(x))+(y'(x))^2\dfrac{\partial^2 f}{\partial y^2}(x,y(x)) $$
Again, you can evaluate everything in $x=0,\ y=y_0$ and get $y^{(3)}(0)$.
The formulas rapidly worsen the more you derive, but perhaps the specific instance of the problem simplifies the calculations.