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idku

  • one year ago

Am I solving this DE correctly?

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  1. idku
    • one year ago
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    I am just tryng some substitution problems. y' = ln(y+x)

  2. idku
    • one year ago
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    (this particular prob, I made up) z = y + x z' = y' +1 y' = z' - 1 z' - 1 = ln(z) z' = ln(z) + 1 1/ (ln(z) + 1) • (dz/dx) = 1 then I integrate both sides with respect to x, and get: Integral, 1/ (ln(z) + 1) dz= x

  3. Astrophysics
    • one year ago
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    I guess that should work, you got a separable right?

  4. Astrophysics
    • one year ago
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    Actually, I don't really know what the Ei here is, wolfram gave this http://www.wolframalpha.com/input/?i=y%27%3Dln%28y%2Bx%29 maybe @zepdrix @IrishBoy123 can help you with this one

  5. idku
    • one year ago
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    After the v sub, it is separable, but then I have an obstacle when it comes to integrating. Well, this isn't something that I so far really know how to integrate.... Wolfram gives me: \(\color{black}{\dfrac{{\rm Ei}(\log(z)+1)}{e}}\)

  6. idku
    • one year ago
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    It would be good if someone explains what this Ei really is.

  7. Astrophysics
    • one year ago
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    Oh it says "Ei(x) is the exponential integral Ei" ok still no idea haha

  8. idku
    • one year ago
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    Yes, that is my prob as well.

  9. idku
    • one year ago
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    I guess I will just take a note of this for now, and will dig into it later.... because I don't think I have a proper base knowledge for this. It seems too complex. I am better start of some normal examples on the web.... didn't mean to disturb anyone.

  10. IrishBoy123
    • one year ago
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    https://en.wikipedia.org/wiki/Exponential_integral

  11. IrishBoy123
    • one year ago
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    @SithsAndGiggles

  12. IrishBoy123
    • one year ago
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    @freckles

  13. idku
    • one year ago
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    Irishboy, it's ok.

  14. idku
    • one year ago
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    Thank you though

  15. Astrophysics
    • one year ago
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    @Zarkon

  16. freckles
    • one year ago
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    looks like: \[\int\limits_{}^{} \frac{e^x}{x} dx =Ei(x)+C \\ \int\limits \frac{1}{\ln(v)+1} dv \\ u=\ln(v)+1 \\ u-1=\ln(v) \\ e^{u-1}=v \\ e^{u-1} du=dv \\ \int\limits \frac{1}{u} e^{u-1} du =\frac{1}{e} \int\limits \frac{e^u}{u} du=\frac{1}{e} Ei(u)+C \\ =\frac{1}{e} Ei(\ln(v)+1)+C\]

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