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idku

  • one year ago

let me try another Bernoulli DE (example 3)

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  1. idku
    • one year ago
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    \(\large y'+e^{-x}y=y^7\)

  2. idku
    • one year ago
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    \(\large y'y^{-7}+e^{-x}y^{-6}=1\) \(\large\dfrac{-1}{6}v'+e^{-x}v=1\) \(\large v'-6e^{-x}v=-6\) \(\large v'e^{e^{6x}}-e^{e^{6x}}6e^{-x}v=-6e^{e^{6x}}\)

  3. idku
    • one year ago
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    \(\large ve^{e^{6x}}=\int -6e^{e^{6x}}dx\) well, I guess I am not yet able to integrate this, but \(\large y^{-6}=\dfrac{\int -6e^{e^{6x}}dx}{e^{e^{6x}}}\) \(\large y^{}=\dfrac{\left(e^{e^{6x}}\right)^6}{\left(\int -6e^{e^{6x}}dx\right)^6}\)

  4. idku
    • one year ago
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    Can someone give me a better example?

  5. idku
    • one year ago
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    \(\large y'-(1/x)y=y^4\)

  6. Empty
    • one year ago
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    Two things! Slight mistake on exponents here: \(\large y^{}=\dfrac{\left(e^{e^{6x}}\right)^{1/6}}{\left(\int -6e^{e^{6x}}dx\right)^{1/6}}\) Also, if you really needed this integral, you could use the taylor series: \[\ e^{e^{6x}}=\sum_{n=0}^\infty \frac{e^{6nx}}{n!}\] Then integrate term by term, until you have a reasonable enough approximation for whatever application you want to use it for, cause like let's face it in the real world 10 digits of accuracy is not bad at all haha

  7. idku
    • one year ago
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    yes those are 1/6's

  8. idku
    • one year ago
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    I didn't know about that taylor series. Something to work on in my little math world.

  9. idku
    • one year ago
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    making a sandbox -:(

  10. idku
    • one year ago
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    I will do that taylor series at some other time, I want to digest the Bernoulli technique first.

  11. Empty
    • one year ago
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    Oh you probably already know this power series! \[e^y=\sum_{n=0}^\infty \frac{y^n}{n!}\] \[y=e^{6x}\]

  12. idku
    • one year ago
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    oh just a sub, I am so ...don';t want to violate the site's policy.

  13. Empty
    • one year ago
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    Haha hey at least it turned out being easier than you thought rather than it being harder than you thought! xD

  14. idku
    • one year ago
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    I should have figured that (after having taken calc 2)

  15. idku
    • one year ago
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    thanks.

  16. idku
    • one year ago
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    I will proceed to my next example I started to do then for now.....

  17. idku
    • one year ago
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    \(\large y'y^{-4}-(1/x)y^{-3}=1\) \(\large v'/(-3)-(1/x)v=1\) \(\large v'+(3/x)v=3\) \(\large v'e^{3\ln x}+e^{3\ln x}(3/x)v=e^{3\ln x}\) \(\large ve^{3\ln x}=e^{3\ln x}\) \(\large ve^{3\ln x}=\int x^3 dx \) \(\large vx^3=(1/4)x^4+C \) \(\large v=\dfrac{\frac{1}{4}x^4+C}{x^3}\) \(\large y^{-3}=\dfrac{\frac{1}{4}x^4+C}{x^3}\) \(\large y=\dfrac{(\frac{1}{4}x^4+C)^{3}}{x^9}\)

  18. idku
    • one year ago
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    I left out the integral in line 5

  19. idku
    • one year ago
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    but I fixed it in line 6

  20. Empty
    • one year ago
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    Are you verifying your solutions by plugging them back into the original differential equation?

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