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anonymous

  • one year ago

Find the solution for this differential equation: ((y^3)cosx - y)dx +(x+y^2)dy = 0 Helpppp

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  1. amistre64
    • one year ago
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    if its a special kind of equation, we can up one side, and then down it to the other to fill in some missing parts

  2. amistre64
    • one year ago
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    if M(x,y) dx + N(x,y) dy and My = Nx, then we have a condition is suitable

  3. anonymous
    • one year ago
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    they are not equal. the M=(3y^2)cosx - 1 and N=1

  4. amistre64
    • one year ago
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    what methods are available to you to work the problem with?

  5. anonymous
    • one year ago
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    this question needs to be solved using the integrating factor by following the instructions in the worksheet. but I think I can use homogeneous coefficient if I still can't answer it

  6. FibonacciChick666
    • one year ago
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    This will probably help you understand integrating factor a little better, check it out then let us know if it makes any more sense http://mathworld.wolfram.com/IntegratingFactor.html

  7. dan815
    • one year ago
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    |dw:1444023674912:dw|

  8. dan815
    • one year ago
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    that sub doesnt work

  9. dan815
    • one year ago
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    hmmm can u think of and substitutions to make it separable

  10. dan815
    • one year ago
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    we have to be able to get it in that dy/dx + p(x) y=g(x) form if u want to use integrating factor

  11. dan815
    • one year ago
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    http://www.wolframalpha.com/input/?i=%28%28y^3%29cosx+-+y%29dx+%2B%28x%2By^2%29dy+%3D+0

  12. dan815
    • one year ago
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    wolfram has confirm non lin equation, nothing we can do then

  13. dan815
    • one year ago
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    wolfram has all the built in bernoullis tricks so

  14. anonymous
    • one year ago
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    it can't be solve?

  15. FibonacciChick666
    • one year ago
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    hmm... did you copy the eq correctly?

  16. FibonacciChick666
    • one year ago
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    this yea? \[((y^3)cosx - y)dx +(x+y^2)dy = 0\]

  17. anonymous
    • one year ago
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    Yess

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