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anonymous

  • 5 years ago

How do I solve the differential equation x^2+(x^3+8)y'=0

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  1. anonymous
    • 5 years ago
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    Can I divide each term by 1/dy to separate?

  2. anonymous
    • 5 years ago
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    This is in the form m+ny'=0

  3. anonymous
    • 5 years ago
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    see if its exact

  4. anonymous
    • 5 years ago
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    do the partial with respect to y for the m and then do the partial with respect to x in the n portion

  5. anonymous
    • 5 years ago
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    Start by isolating y' .. then integrate both sides

  6. anonymous
    • 5 years ago
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    if they are equal then it is exact

  7. anonymous
    • 5 years ago
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    But to isolate y' I would need to divide each term by y' correct?

  8. anonymous
    • 5 years ago
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    wait...is this cal 1 or differential equations and linear algebra

  9. anonymous
    • 5 years ago
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    Calc II differential equations

  10. anonymous
    • 5 years ago
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    http://1337.is/~gaulzi/tex2png/view.php?png=201103312034348813.png would look like this isolated

  11. anonymous
    • 5 years ago
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    Well you can try implicit differentiation.

  12. anonymous
    • 5 years ago
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    take the derivative of the first term with respect to x

  13. anonymous
    • 5 years ago
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    kristin wouldn't it be -x^2?

  14. anonymous
    • 5 years ago
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    sorry yes, thats correct, minor typo :)

  15. anonymous
    • 5 years ago
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    Ok no problem. Thanks

  16. anonymous
    • 5 years ago
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    I can toss up for you the answer if you want :)

  17. anonymous
    • 5 years ago
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    y' is dy/dx correct? If i were to rewrite it

  18. anonymous
    • 5 years ago
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    yeah

  19. anonymous
    • 5 years ago
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    Ok so if I were to solve y'+y=10 would I just replace y' with dy/dx and then move it all around?

  20. anonymous
    • 5 years ago
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    I'm trying to figure out the steps

  21. anonymous
    • 5 years ago
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    it can be done by the separable variables method

  22. anonymous
    • 5 years ago
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    after some modification you can get dy=-x^2/(x^3+8) dx just integrate both sides

  23. anonymous
    • 5 years ago
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    http://1337.is/~gaulzi/tex2png/view.php?png=201103312042017465.png these are the two answers you can get.. depends if you like to use ln or log

  24. anonymous
    • 5 years ago
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    I know the answer I'm trying to figure out the steps, could you show me the steps kristin? Plz.

  25. anonymous
    • 5 years ago
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    I can show it to you

  26. anonymous
    • 5 years ago
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    did you integrate like she said?

  27. anonymous
    • 5 years ago
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    I know I need to integrate each piece, I just need to know how to get each piece by itself

  28. anonymous
    • 5 years ago
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    do a u substitution

  29. anonymous
    • 5 years ago
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    u=x^3+8 du=3xdu

  30. anonymous
    • 5 years ago
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    du=3x^2 sorry

  31. anonymous
    • 5 years ago
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    you should get: integral( -1/3(1/u) dx

  32. anonymous
    • 5 years ago
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    just take u=x^3+8 --> du=3x^2dx substitute in the integral you will get \[y=-1/3\int\limits_{}^{}(1/u)du=-1/3\ln \left| u \right|\]

  33. anonymous
    • 5 years ago
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    now just substitute for u=x^3+8 \[y=-1/3\ln \left| x^3+8 \right|+c\]

  34. anonymous
    • 5 years ago
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    Oh right I got that. Thanks

  35. anonymous
    • 5 years ago
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    np

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