Should I use Taylor Expansion or the expansion of $e^x$ to express a second order differential (in this case) I've been given this equation:
$(1+x^2)\dfrac{d^2y}{dx^2} + 4x\dfrac{dy}{dx} + 2y = 0$
I've also been told that:
$y=1, \dfrac{dy}{dx} = 1$, at $x=-1$
I've been asked to find a series solution of the differential equation in ascending powers of $(x+1)$ up to and including the term in $(x+1)^4$
I've managed to do this in two different ways but they give different answers so I can only assume that one of them is wrong. However I can't see the mistake I've made in either; I'm not too sure whether it's lack of understanding or just a calculation error.
I hope you don't mind if I attach a picture of the working I've done by hand as it would take quite some time to write it all out in Latex, however I can do that if the picture's aren't clear enough.
Method 1:



Method 2:



Thank you :)
 A: Your first method appears correct. Here's a walk-through of the solution for posterity, then I'll try and address the error in the second method.
You have (for $x = -1$) that $y = 1$, $y' = 1$, and then you correctly calculate
$$
(1 + (-1)^2)\left.\frac{d^2 y}{d x^2}\right|_{x=-1} + 4(-1)\left.\frac{d y}{d x}\right|_{x=-1} + 2y(1) = 0
$$
so
$$
2\left.\frac{d^2 y}{d x^2}\right|_{x=-1} -4 + 2 = 0
$$
thus
$$
\left.\frac{d^2 y}{d x^2}\right|_{x=-1} = 1.
$$
Now, we want to differentiate to get an identity for $\left.\frac{d^3 y}{d x^3}\right|_{x=-1}$.
Note here that the derivative with respect to $x$ of $(1 + x^2)y''$ is not simply $(1 + x^2)y'''$, it is
$$
(1 + x^2)'y'' + (1+x^2)y''' = 2xy'' + (1 + x^2)y'''
$$
so after correctly differentiating the original equation, you should have
$$
2xy'' + (1 + x^2)y''' + 4y' + 4xy'' + 2y' = 0.
$$
Now you can evaluate at $x= -1$, since you know $y(-1), y'(-1), y''(-1)$ (they're all equal to $1$, coincidentally) and solve for $y'''(-1)$:
$$
-2 + 2y''' + 4 - 4 + 2 = 0
$$
so $y'''(-1) = 0$.
Now we just have to repeat the process:
$$
2y'' + 2xy''' + 2xy''' + (1 + x^2)y'''' + 4y'' + 4y'' + 4xy''' + 2y'' = 0.
$$
and plug-in what you know for $x,y,y',y'',y'''$.
The problem with the second method is basically that you're confusing $\frac{d y}{d x}$ (a function) and $\left.\frac{d y}{d x}\right|_{x=-1}$ (a value). That is, once you plug in what you know about $x,y,y'$, you are looking at an equation about specific values of $y',y''$, etc. and not a valid differential equation for $y$. (so you shouldn't try to "solve" it by guessing $y = e^{mx}$)
