Series Solution to the ODE $y''+2y'+y=0$

Question

here's where I'm at, The ode we are trying to solve is, $$y''+2y'+y=0$$ I know this solution is of the form: $$\sum_{n=0}^{\infty}a_nx^n $$ And from that we get the following $$\sum_{n=0}^{\infty}(n)(a_n)x^{n-1} \quad \& \quad \sum_{n=0}^{\infty}(a_n)(n)(n-1)x^{n-2}$$ subbing them into the ODE I get: $$\sum_{n=2}^{\infty}(a_n)(n)(n-1)x^{n-2} +2\sum_{n=1}^{\infty}(a_n)(n)x^{n-1} +\sum_{n=0}^{\infty}(a_n)x^n=0$$ its a simmilar thing for the second sum. The reason the first sum starts at 2 is cause the first two terms vanish because of the factor of $n-1$ & $n$ I tried to derive a recurrance relation for the terms $a_n$ by balancing the powers and index's of each sum. I did this by introducing new variable $k=n-2$ & $m=n-1$ so the ode becomes, $$\sum_{k=0}^{\infty}(a_{k+2})(k+2)(k+1)x^{k} +2\sum_{m=0}^{\infty}(a_{m+1})(m+1)x^{m} +\sum_{n=0}^{\infty}(a_n)x^n=0$$ now since they all have the same index and same powers I think i could write this, $$\sum_{n=0}^{\infty}{[a_{k+2}(k+2)(k+1)+2a_{k+1}(k+1)+a_k)x^k]}$$ So then we derive the recurrence relation $$a_{k+2}=-\frac{(2a_{k+1}+a_k)}{(k+1)(k+2)}$$ I subbed in a few values for K and re-wrote the terms of the series in terms of $a_0$ & $a_1$ spotted a pattern which got me this recurrence relation, $$a_k=\frac{(-1)^{k+1}(ka_1+(k-1)a_0)}{k!}$$ and therefor you get a solution that looks like $$y=a_0+a_1x-x^2\frac{2a_1+a_0}{1\cdot2}+x^3\frac{3a_1+2a_0}{1\cdot2\cdot3}+x^4\frac{4a_1+3a_0}{1\cdot2\cdot3\cdot4}+...$$ So here's my problem i'm at the point now where I need to check if this really is a solution but when i differentiate my expression and sub it all in i get a few cancellations but i can't see any sort of pattern is what terms cancels with what, so i figure i must have done something wrong. Any help is appreciated!

Answer 1

HINT

\begin{align*} y^{\prime\prime} + 2y^{\prime} + y = 0 \Longleftrightarrow (y^{\prime\prime} + y^{\prime}) + (y^{\prime} + y) = 0 \Longleftrightarrow (y^{\prime} + y)^{\prime} + (y^{\prime} + y) = 0 \end{align*}