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RolyPoly

\[y' = x^3 (1-y)\]

  • one year ago
  • one year ago

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  1. cnknd
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    this is easily separable...

    • one year ago
  2. RolyPoly
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    \[y' = x^3(1-y)\]\[\frac{dy}{1-y} = x^3 dx\]\[-\ln |1-y| = \frac{1}{4}x^4+C\]So far so good?

    • one year ago
  3. RolyPoly
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    lol I know, the whole chapter I'm doing now is separable.

    • one year ago
  4. cnknd
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    ugh... check ur right hand side

    • one year ago
  5. RolyPoly
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    Right hand side? \[x^3 dx\]?

    • one year ago
  6. hartnn
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    i think its fine here u can isolate y easily.

    • one year ago
  7. RolyPoly
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    If it's fine, then I continue.. y(0) = 3 \[-\ln |1-3| = \frac{1}{4}(0)^4+C\]\[-\ln |2| = C\] So far so good?

    • one year ago
  8. RolyPoly
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    Hmm.. -ln2

    • one year ago
  9. cnknd
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    ok x^2 does not integrate to 1/4*x^4

    • one year ago
  10. cnknd
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    but C would still be -ln2

    • one year ago
  11. RolyPoly
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    It's x^3!

    • one year ago
  12. hartnn
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    continue....

    • one year ago
  13. RolyPoly
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    So, \[-\ln |1-y| = \frac{1}{4}(x)^4-\ln 2\]\[\ln2-\ln |1-y| = \frac{1}{4}(x)^4\]\[\ln\frac{2}{1-y} = \frac{1}{4}(x)^4\]\[\frac{2}{1-y} = e^{\frac{x^4}{4}}\]\[2= ({1-y} )e^{\frac{x^4}{4}}\]\[y = \frac{e^{\frac{x^4}{4}}-2}{\frace^{\frac{x^4}{4}}}{

    • one year ago
  14. RolyPoly
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    Sorry, please ignore the last line. OS just freezed my page.

    • one year ago
  15. RolyPoly
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    But that doesn't look good at al..

    • one year ago
  16. hartnn
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    better to do this \( -\ln2+\ln |1-y| = -\frac{1}{4}(x)^4\)

    • one year ago
  17. RolyPoly
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    \[-\ln2+\ln |1-y| = -\frac{1}{4}x^4\]\[\frac{1-y}{2}=e^{\frac{x^4}{4}}\]\[1-y=2e^{\frac{x^4}{4}}\]\[y=1-2e^{\frac{x^4}{4}}\]

    • one year ago
  18. hartnn
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    e^{-...}

    • one year ago
  19. RolyPoly
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    Oh!! \[e^{-\frac{x^4}{4}}\]

    • one year ago
  20. hartnn
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    everything else is correct

    • one year ago
  21. RolyPoly
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    Got it :)

    • one year ago
  22. RolyPoly
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    Thanks!!

    • one year ago
  23. cnknd
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    \[\ln \left| 1-y \right| = -\frac{ 1 }{ 3 }x ^{3} + \ln2\] \[\left| 1-y \right| = 2e^{-\frac{ x ^{3} }{ 3 }}\] then you have to solve the absolute value part which gives u a piecewise function (this is a pain in the retrice

    • one year ago
  24. cnknd
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    lol i like how they censor inappropriate words

    • one year ago
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