<s>[INST] What Did I Do Wrong When Solving For This 2nd Order Differential Equation? (answered myself) $$
\frac{y''}{y'}+y' = f(x)
$$
I set the following to be true:
$$
y = \sum_{n=0}^{\infty} a_n x^n
$$
$$
f(x) = \sum_{n=0}^{\infty} b_nx^n
$$
Therefore:
$$
y'' = y'(f(x)-y')
$$
$$
\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(n+1)a_{n+1} x^n] \times \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n]
$$
$$
\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n \sum_{k=0}^n (k+1)a_{k+1}]
$$
Therefore
$$
(n+1)(n+2)a_{n+2} = (b_n - (n+1)a_{n+1}) \sum_{k=0}^n (k+1)a_{k+1}
$$
$$
b_n = \frac{(n+2)(n+1)a_{n+2}}{\sum_{k=1}^{n+1} ka_k} + (n+1)a_{n+1}
$$
$$
y = \sum_{n=0}^{\infty} (\frac{(n+2)(n+1)a_{n+2}}{\sum_{k=1}^{n+1} ka_{k}} + (n+1)a_{n+1})x^n
$$
Now if i set $a_n = 1$
$$
y = \sum_{n=0}^{\infty} (n+1)x^n + \sum_{n=0}^\infty (\frac{(n+2)(n+1)}{\sum_{k=1}^{n+1} k}x^n
$$
$$
\sum_{k=1}^{n+1} k = \frac{(n+1)(n+2)}{2}
$$
$$
y = \frac{1}{(x-1)^2} + \sum_{n=0}^\infty 2x^n
$$
$$
 \sum_{n=0}^\infty 2x^n = \frac{2}{1-x}
$$
Therefore
$$
f(x) = \frac{2x-1}{(x-1)^2}
$$
$$
y = \frac{1}{1-x}
$$
Therefore
$$
\frac{-2(1-x)^2}{(1-x)^3}+\frac{1}{(1-x)^2} = \frac{2x-1}{(1-x)^2}
$$
Which is true.
 [/INST] Your problem is that
you took the terms involving $b_n$
out of the summation.
Here is my take.
I'll start at
$\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(n+1)a_{n+1} x^n] \times \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n]$
but use different indices for the summations
on the right.
$\begin{array}\\
\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n 
&= \sum_{i=0}^{\infty} (i+1)a_{i+1} x^i \times \sum_{j=0}^{\infty} (b_j - (j+1)a_{j+1})x^j\\
&= \sum_{i=0}^{\infty}\sum_{j=0}^{\infty} (i+1)a_{i+1}  (b_j - (j+1)a_{j+1})x^{i+j}\\
&= \sum_{n=0}^{\infty}\sum_{j=0}^{n} (n-j+1)a_{n-j+1}  (b_j - (j+1)a_{j+1})x^{n}\\
&= \sum_{n=0}^{\infty}x^{n}\sum_{j=0}^{n} (n-j+1)a_{n-j+1}  (b_j - (j+1)a_{j+1})\\
\end{array}
$
From this,
$\begin{array}\
(n+1)(n+2)a_{n+2}
&=\sum_{j=0}^{n} (n-j+1)a_{n-j+1}  (b_j - (j+1)a_{j+1})\\
&=\sum_{j=1}^{n+1} ja_{j}  (b_{n-j+1} - (n-j+2)a_{n-j+2})\\
\end{array}
$
I'll leave it at this.
</s>