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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!

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    $\begingroup$ Do you know the characteristic polynomial of a differential equation? Also, try $$y(t) = C \cdot e^t$$ for some constant $C \in \mathbb{R}$. $\endgroup$ – Viktor Glombik Feb 10 at 16:45
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    $\begingroup$ This method solves the equation easily. $\endgroup$ – Git Gud Feb 10 at 16:46
  • $\begingroup$ Hi, thanks for the comment, i know that this solves the equation nicely i was just trying to do it this way as an exercise in finding series solutions since im pretty new to doing them $\endgroup$ – Questlove Feb 10 at 16:53
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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*}

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  • $\begingroup$ Thanks for the answer, just wanted to ask if this reduces the problem to solving $y'+y=0$? Since any solution of that would also be a solution to the original ODE. $\endgroup$ – Questlove Feb 10 at 22:32
  • $\begingroup$ I mean without loosing any solutions $\endgroup$ – Questlove Feb 10 at 22:47
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    $\begingroup$ The idea it to make the substitution $w = y^{\prime} + y$ in order to obtain the solution $w_{p} = ke^{-x}$, whose Maclaurin expansion is given by $$e^{-x} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \ldots$$ Then you can solve the equation $y^{\prime} + y = w_{p}$ through the method of series so that you get the desired solution. $\endgroup$ – APC89 Feb 10 at 22:49
  • $\begingroup$ Thanks that makes alot of sense $\endgroup$ – Questlove Feb 10 at 22:52

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