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idku

  • one year ago

Another Bernoulli DE (eq 2)

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  1. idku
    • one year ago
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    \(\large \color{black}{y'+\ln(x)y=y^{2}}\) \(\large \color{black}{y'y^{-2}+\ln(x)y^{-1}=1}\) Substitution \(\large \color{black}{v=y^{-1}}\) \(\large \color{black}{-v'=y'y^{-2}}\) DE becomes, \(\large \color{black}{-v'+v\ln(x)=1}\)

  2. idku
    • one year ago
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    Then, the integrating factor is: \(\large \color{black}{H(x)=e^{x\ln(x)-x}}\) oh, boy does that become impossible for me to do?

  3. idku
    • one year ago
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    \(\large \color{black}{-v'e^{x\ln(x)-x}+ve^{x\ln(x)-x}\ln(x)=e^{x\ln(x)-x}}\) \(\large \color{black}{ve^{x\ln(x)-x}=\int e^{x\ln(x)-x}dx}\)

  4. idku
    • one year ago
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    \(\large \color{black}{e^x=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}}\) is it possible to use this to get the integral of the right side?

  5. anonymous
    • one year ago
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    there is no elementary closed-form expression for the integral on the right, but you could use power series to write one in some given interval

  6. anonymous
    • one year ago
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    but using just that power series alone, no, that will not help

  7. idku
    • one year ago
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    well, I guess I am ought to try another example that I can chew:)

  8. idku
    • one year ago
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    \(\large \color{black}{y'+xy=y^{4}}\)

  9. idku
    • one year ago
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    ok, lets start I guess..... \(\large \color{black}{y'y^{-4}+xy^{-3}=1}\) Substitution: \(\large \color{black}{v=y^{-3}}\) \(\large \color{black}{\dfrac{-1}{3}v'=y'y^{-4}}\) \(\large \color{black}{\dfrac{-1}{3}v'+xv=1}\)

  10. idku
    • one year ago
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    \(\large \color{black}{v'-3xv=-3}\)

  11. idku
    • one year ago
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    \(\large \color{black}{v'e^{(-3/2)x^2}-3xe^{(-3/2)x^2}v=-3e^{(-3/2)x^2}}\)

  12. idku
    • one year ago
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    \(\large \color{black}{ve^{(-3/2)x^2}=\int -3e^{(-3/2)x^2}dx}\)

  13. idku
    • one year ago
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    \(\large \color{black}{\displaystyle e^x=\sum_{i=0}^{\infty}\frac{x^i}{i!}}\) \(\large \color{black}{\displaystyle e^{(-3/2)x^2}=\sum_{i=0}^{\infty}\frac{\left((-3/2)x^2\right)^i}{i!}}\) am i correct so far?

  14. idku
    • one year ago
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    \(\large \color{black}{\displaystyle e^{(-3/2)x^2}=\sum_{i=0}^{\infty}\frac{(-3/2)^ix^{2i}}{i!}}\) do I treat (-3/2)^i as a constant or I can't do that?

  15. idku
    • one year ago
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    \(\large \color{black}{\displaystyle \int e^{(-3/2)x^2}dx=\sum_{i=0}^{\infty}\frac{(-3/2)^ix^{2i+1}}{i!(2i+1)}+C}\)

  16. idku
    • one year ago
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    doesn't matter, I guess I get the method, I am just too stupid to come up with an adequate example for me.

  17. idku
    • one year ago
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    \(\large \color{black}{\displaystyle y'+y=y^n}\)

  18. idku
    • one year ago
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    \(\large \color{black}{\displaystyle y'y^{-n}+y^{-n+1}=1}\) \(\large \color{black}{\displaystyle \frac{1}{1-n}v'=y'y^{-n},~~~~~v=y^{-n+1}=1}\) \(\large \color{black}{\displaystyle \frac{1}{1-n}v'+v=1}\) \(\large \color{black}{\displaystyle v'+(1-n)v=1-n}\) \(\large \color{black}{\displaystyle v'e^{x(1-n)}+e^{x(1-n)}(1-n)v=e^{x(1-n)}(1-n)}\) \(\large \color{black}{\displaystyle ve^{x(1-n)}=\int e^{x(1-n)}(1-n)dx}\)

  19. idku
    • one year ago
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    just nvm

  20. idku
    • one year ago
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    i suck

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