Power series representation of $x$? This may not be a very good question, but I'm totally stumped.
I need to know the power series representation of $x$, or if there even is one.
I'll show you why:
I am trying to solve $y''+2xy'-y=x$ using power series.
I know that $y''=\displaystyle\sum\limits_{n=2}^\infty a_n n(n-1)x^{n-2}$, $y'=\displaystyle\sum\limits_{n=1}^\infty a_n n x^{n-1}$, and $y=\displaystyle\sum\limits_{n=0}^\infty a_n x^n$, but I don't know what the power series representation of $x$ is.
I can solve the homogeneous equation no problem by setting $y''+2xy'-y=0$, but I do not feel this is correct. 
 A: Note that $y''$ and $y$ both have a degree zero term in their series expansions, but that $2xy'$ does not. If you put your various power series together, then, you'll have $$\begin{align}y''+2xy'-y &= 2a_2-a_0+\sum_{n=3}^\infty a_nn(n-1)x^{n-2}+\sum_{n=1}^\infty 2a_nnx^n-\sum_{n=1}^\infty a_nx^n\\ &= 2a_2-a_0+\sum_{n=1}^\infty a_{n+2}(n+2)(n+1)x^n+\sum_{n=1}^\infty a_n(2n-1)x^n\\ &= 2a_2-a_0+\sum_{n=1}^\infty\bigl[a_{n+2}(n+2)(n+1)+a_n(2n-1)\bigr]x^n\\ &=\sum_{n=0}^\infty b_nx^n,\end{align}$$ where $\sum b_nx^n$ is the power series expansion of $x$. But this means that $b_n=0$ for $n\neq 1$ and $b_1=1$. Thus, what you need is: $$\begin{cases}0=2a_2-a_0 & \\1=6a_3+a_1 & \\0=a_{n+2}(n+2)(n+1)+a_n(2n-1) & \text{for all }n\geq 2\end{cases}$$
Hopefully that's enough to get you moving in the right direction for a general solution. You may even be able to get a closed (non-recursive) form for your $a_n$ in terms of $a_0$ (for even $n$) or $a_1$ (for odd $n$).
Alternately, you could use your general solution to the corresponding homogeneous equation, and use this (or some other) method to find a particular solution to $y''+2xy'-y=x$, then add the two to get your general nonhomogeneous solution.
A: $$
x=0+x+0\cdot x^2+0\cdot x^3+\dots =\sum_{n=0}^\infty a_nx^n
$$
where
$$
a_1=1,\quad a_n=0\text{ if }n\ne1.
$$
A: $$
\begin{align}
-y & =\sum_{n=0}^\infty -a_n x^n\tag{starting with 0} \\[15pt]
2xy' & = 2x\sum_{n=1}^\infty a_n n x^{n-1}\tag{starting with 1} \\
& = \sum_{n=1}^\infty 2a_n n x^n \tag{starting with 1} \\[15pt]
y'' & =\sum_{n=2}^\infty a_n n(n-1)x^{n-2}\tag{starting with 2} \\
& = \sum_{n=0}^\infty a_{n+2} (n+2)(n+1) x^n.\tag{starting with 0}
\end{align}
$$
For $n=0$, the sum of the coefficients of the $x^n$ terms is $-a_0+a_2\cdot2\cdot1$ on the left side, and $0$ on the right side.
For $n=1$, the sum of the coefficients of the $x^n$ terms is $-a_1+2a_1\cdot1 + a_3\cdot3\cdot2$ on the left side and $1$ on the right side.
For $n\ge 2$, the sum of the coefficients of the $x^n$ terms is $-a_n + 2a_n\cdot n +a_{n+2} (n+2)(n+1)$ on the left side and $0$ on the right side.
Thus one must solve the recursion $(n+2)(n+1)a_{n+2} +(n-2)a_n=0$ with the initial conditions implied by the above.
