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anonymous

  • one year ago

Practice... (made up an example).

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  1. anonymous
    • one year ago
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    \(\large y'+yx=3\)

  2. anonymous
    • one year ago
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    \(\large \displaystyle \int_{}^{}x~dx~=~\frac{x^2}{2}\) (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....

  3. anonymous
    • one year ago
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    \(\displaystyle \large e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} \) \(\displaystyle \large e^{x^2/2}=\sum_{n=0}^{\infty}\frac{x^{2n}}{2^n~n!} \) \(\displaystyle \large \int e^{x^2/2}~ dx=\int \sum_{n=0}^{\infty}\frac{x^{2n}}{2^n~n!}~dx \) \(\displaystyle \large \int e^{x^2/2}~ dx=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)~2^n~n!} \) \(\displaystyle \large \int 3e^{x^2/2}~ dx=3\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)~2^n~n!} \)

  4. anonymous
    • one year ago
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    \(\displaystyle \Large y=\frac{\displaystyle 3\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)~2^n~n!}+C }{~~~~~~~~~~~e^{x^2/2}~~~~~~~~~} \)

  5. anonymous
    • one year ago
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    @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?

  6. anonymous
    • one year ago
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    @hamidic

  7. anonymous
    • one year ago
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    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.

  8. anonymous
    • one year ago
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    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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