anonymous
  • anonymous
Practice... (made up an example).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\(\large y'+yx=3\)
anonymous
  • anonymous
\(\large \displaystyle \int_{}^{}x~dx~=~\frac{x^2}{2}\) (of course, we omit the +c) So the integrating factor \(\large e^{\rm H(x)}\) is in this case \(\large e^{x^2/2}\). And now we multiply everything by this integrating factor. \(\large y'~e^{x^2/2}+y~e^{x^2/2}~x=3~e^{x^2/2}\) now the simple use of the product rule use (( which I found very clever as I read... very cool to come up with "integrating factor" in such equations as y' + p(x) y = q(x) )) \(\large \displaystyle \frac{d}{dx}\left[y~e^{x^2/2}\right]=3~e^{x^2/2}\) Integrating both sides \(\large \displaystyle y~e^{x^2/2}=\int 3~e^{x^2/2}~dx\) No closed form.... BUT....
anonymous
  • anonymous
\(\displaystyle \large e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} \) \(\displaystyle \large e^{x^2/2}=\sum_{n=0}^{\infty}\frac{x^{2n}}{2^n~n!} \) \(\displaystyle \large \int e^{x^2/2}~ dx=\int \sum_{n=0}^{\infty}\frac{x^{2n}}{2^n~n!}~dx \) \(\displaystyle \large \int e^{x^2/2}~ dx=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)~2^n~n!} \) \(\displaystyle \large \int 3e^{x^2/2}~ dx=3\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)~2^n~n!} \)

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anonymous
  • anonymous
\(\displaystyle \Large y=\frac{\displaystyle 3\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)~2^n~n!}+C }{~~~~~~~~~~~e^{x^2/2}~~~~~~~~~} \)
anonymous
  • anonymous
@radar @wio @triciaal Is this solution that I came up is super weird? Or is my example just impossible? Or did I do something totally incorrect?
anonymous
  • anonymous
@hamidic
anonymous
  • anonymous
I personally don't see any errors in your work. However, I've never taken a formal diff eq course, I just learned it on my own, so not sure if there are any special conventions about this kind of stuff. Other than that, nothing wrong with the calc itself.
anonymous
  • anonymous
I really want to say that my example is very bad, because I can't get an elementary function when I integrate e^(x^2/2), that is why y'+yx=3. If I had a teacher.... but I read just a couple hours ago in my calculus book.... tnx for checking, will see what others say if they see this.

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