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idku

  • one year ago

my first Bernoulli DE ... (just looked at it on lamar tutorial and will see how will I do this one)

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  1. idku
    • one year ago
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    \(\large \color{black}{y'=4y+e^xy^3}\)

  2. idku
    • one year ago
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    (this example is completely made up, not on the tutorial)

  3. amistre64
    • one year ago
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    how do you know it satisifies a Berny then?

  4. amistre64
    • one year ago
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    im thinking (couild be misremembering) that a Berny is a reduction process

  5. idku
    • one year ago
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    \(\large \color{black}{y'=4y+e^xy^{3}}\) \(\large \color{black}{y'-4y=e^xy^{3}}\) \(\large \color{black}{y'y^{-3}-4y^{-2}=e^x}\) and then I substitute, \(\large \color{black}{v=y^{-2}}\) \(\large \color{black}{v'=-2y'y^{-3}}\) ---> \(\large \color{black}{(-1/2)v'=y'y^{-3}}\) my DE becomes, \(\large \color{black}{\dfrac{-1}{2}v'-4v=e^x}\) For convenience, \(\large \color{black}{v'+8v=2e^x}\) Integrating factor: \(\large \color{black}{e^{4v^2}}\) \(\large \color{black}{v'e^{4v^2}+e^{4v^2}8v=e^{4v^2}2e^x}\) \(\large \color{black}{dv/dx(ve^{4v^2})=e^{4v^2}2e^x}\) wait I am getting stuck here a bit

  6. idku
    • one year ago
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    how would I then integrate both sides, if I have that on the RIGHT side?

  7. idku
    • one year ago
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    oh, I missed a minus on the right side (when I multiplied by -2)

  8. idku
    • one year ago
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    \(\large \color{black}{dv/dx(ve^{4v^2})=-e^{4v^2}2e^x}\)

  9. idku
    • one year ago
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    the integrating facto is wrong? is it just e^x because integrating factor is with wespect to x, not v?

  10. amistre64
    • one year ago
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    v e^(8v) = v' e^(8v) + 8v e^(8v)

  11. amistre64
    • one year ago
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    hmm, dont think thats quite right either

  12. idku
    • one year ago
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    \(\large \color{black}{v'+8v=-2e^x}\) \(\large \color{black}{v'e^{8x}+e^{8x}8v=-2e^{8x}e^x}\)

  13. idku
    • one year ago
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    like that?

  14. amistre64
    • one year ago
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    that seems better yes

  15. idku
    • one year ago
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    \(\large \color{black}{ve^{8x}=\int -2e^{9x}dx}\) \(\large \color{black}{ve^{8x}= (-2/9)e^{9x}+c}\) \(\large \color{black}{v= (-2/9)e^{x}+c/e^{8x}}\)

  16. idku
    • one year ago
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    Like this?

  17. amistre64
    • one year ago
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    so far so good

  18. idku
    • one year ago
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    \(\large \color{black}{v= (-2/9)e^{x}+C/e^{8x}}\) \(\large \color{black}{y^{-2}= (-2/9)e^{x}+C/e^{8x}}\) and then, \(\large \color{black}{y^2=\dfrac{1}{ (-2/9)e^{x}+C/e^{8x}}}\) \(\large \color{black}{y=\pm\sqrt{\dfrac{1}{ (-2/9)e^{x}+C/e^{8x}}}}\) and the rest is a matter of algebraic manipulation if I have done it correclty.

  19. amistre64
    • one year ago
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    the process seems fair

  20. idku
    • one year ago
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    Awesome, I prevailed it! Thank you amistre for your confirmation!

  21. amistre64
    • one year ago
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    youre welcome, i didnt scrutinize it, but it seemed to be in good form overall.

  22. idku
    • one year ago
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    Yes, fine for the first time. (Especially considering the fact that I am not really a "math guy")

  23. idku
    • one year ago
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    Thank you once again.

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