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2y"+y'=0 has a solution in the form c1+c2(e^-x/2)

Through power series, I got enter image description here

The official solution does not include my factored x for the Maclaurin of (e^-x/2). How did I go wrong? Or do I incorporate the extra x differently?

Anne Bauval
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timetravel_0
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  • https://math.stackexchange.com/help/notation – Anne Bauval May 10 '23 at 06:06
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    @SassatelliGiulio they are applying https://en.wikipedia.org/wiki/Frobenius_method but hiding the crucial steps. – Anne Bauval May 10 '23 at 06:24
  • I love the crucial steps. – timetravel_0 May 10 '23 at 06:37
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    Well write them please (not in comments; rather inside your post). – Anne Bauval May 10 '23 at 06:42
  • I've posted an answer to your question. Here are some tips. Your mistake was in the Taylor series for $e^x$. You should always check what the limits of a sum are before applying a standard summation formula. Also, the power of each $x$ term should be exactly equal to the factorial argument in the denominator, i.e., try converting $\frac{x^n}{n+1}$ terms into $\frac{x^{n+1}}{n+1}$ terms before applying the formula. – Amogh May 10 '23 at 07:34
  • Of course! the OP did not deign to write the main part, to derive the 3rd line from the 2nd one (find the recurrence relation on the $a_n$'s induced by the differential equation). + mistaken penultimate line, as noted below by Amogh. – Anne Bauval May 10 '23 at 07:52

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The second to last line should be: $$y=a_0+a_1x\left(1-\frac{x}{4}+\frac{x^2}{24}-\frac{x^3}{192}+\frac{x^4}{1920}\cdots\right)$$ Using summation notation: $$y=a_0+a_1x\sum_{n=0}^\infty \frac{x^n(-1)^n}{2^n(n+1)!}$$ Multiplying the $x$ into the summation: $$y=a_0+a_1\sum_{n=0}^\infty \frac{x^{n+1}(-1)^n}{2^n(n+1)!}$$ $$y=a_0+a_1\sum_{n=0}^\infty \frac{2}{-1}\frac{x^{n+1}(-1)^{n+1}}{2^{n+1}(n+1)!}$$ $$y=a_0-2a_1\sum_{n=0}^\infty \frac{\left(-\frac{x}{2}\right)^{n+1}}{(n+1)!}$$ Since the summation starts at $n+1=1$ and not $n+1=0$: $$y=a_0-2a_1\left(e^{-\frac{x}{2}}-\frac{\left(-\frac{x}{2}\right)^0}{0!}\right)$$ $$y=a_0+2a_1-2a_1e^{-\frac{x}{2}}$$ Since $a_0+2a_1$ and $-2a_1$ are just arbitrary constants: $$y=c_1+c_2e^{-\frac{x}{2}}$$

Amogh
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    We should not answer "poor quality posts" like this one: no effort to improve the question by adding the requested details. If the poster had accepted to do that, they probably would have found their mistake. Please don't do for them the work they are able to. This is not the role of this site. – Anne Bauval May 10 '23 at 07:32
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    @AnneBauval Alright, I'll make sure to follow that next time. – Amogh May 10 '23 at 07:36
  • You can do it also this time: you can delete this answer. – Anne Bauval May 10 '23 at 07:38
  • Thank you bro @Amogh. Don't mind the rude people, your help is always welcome :) – timetravel_0 May 10 '23 at 08:41