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anonymous

  • 5 years ago

How do I solve a differential equation if the equation becomes undefined when I plug in the initial value?

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  1. anonymous
    • 5 years ago
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    heyllo joseph hru?

  2. anonymous
    • 5 years ago
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    Show the equation :D If it's undefined at the initial value, then you probably integrated incorrectly :D

  3. anonymous
    • 5 years ago
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    dx/dy = (y-1)^2

  4. anonymous
    • 5 years ago
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    x = 1/3(y-1)^3 + C What are the initial conditions?

  5. anonymous
    • 5 years ago
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    y(0) = 1

  6. anonymous
    • 5 years ago
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    C = 0

  7. anonymous
    • 5 years ago
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    x = 1/3 (y-1)^3

  8. anonymous
    • 5 years ago
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    can you show your work in getting the solution to the differential equation?

  9. anonymous
    • 5 years ago
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    y(0) = 1 x = 0 y = 1 0 = 1/3(1-1)^3 +C 0 = 0 +C C = 0

  10. anonymous
    • 5 years ago
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    excuse me! dy/dx =...

  11. anonymous
    • 5 years ago
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    Oh:\[dx/dy = (y-1)^2\] \[\int\limits_{0}^{x}x dx = \int\limits_{0}^{y} (y-1)^2 dy\] Take the antiderivative (in this case, it works) :D \[x = (y-1)^3/3\]

  12. anonymous
    • 5 years ago
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    +C

  13. anonymous
    • 5 years ago
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    DON'T FORGET THE +C

  14. anonymous
    • 5 years ago
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    k

  15. anonymous
    • 5 years ago
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    excuse me, i made a mistake, the dy and dx are flipped from what I wrote. Is it OK?

  16. anonymous
    • 5 years ago
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    No, then the equation is completely different: \[dy/dx = (y-1)^2\] Limits using initial conditions \[\int\limits_{1}^{y} dy/(y-1)^2 = \int\limits_{0}^{x} xdx\] \[\int\limits_{1}^{y} (y-1)^{-2} dy = x\] Antidifferentiate again:\[-(y-1)^{-1} - - (0)^{-1} = x \] No solution for this diff eq.

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