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Find (if it exists) the solution for the differential equation.

  1. Mar 3, 2012 #1
    Find (if it exists) the solution for the differential equation:
    dy/dx = -2xtan(y)
    given the initial value y(0) = /4

    My steps in tackling this, by using the theorem of existence and unicity I would take domain of f(x,y) and the domain of the partial derivative, then figure out the common interval and see whether the initial value is within this interval to find whether a unique solution exists. However, my dilemma. I understand how to find the domain for say dy/dx = √x because its just x in the function, so y would be all real values and x would be ≥ 0. But in my above example there is x AND y in the function, how do I find the domain?

    Once I figure out whether it has unique solution or not I think I can find solution.
  2. jcsd
  3. Mar 3, 2012 #2


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    Hi Lengalicious! :smile:

    find the solution first … separate the variables! :wink:
  4. Mar 3, 2012 #3
    But how can I find the solution if I don't know if one exists? :s
  5. Mar 3, 2012 #4


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    Once you do separation of variables, if you're still not convinced that you know there's a solution, just take the solution you got from separation of variables and plug it into the differential equation to make sure it really works
  6. Mar 3, 2012 #5


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    how can you find the silver lining if you don't know if one exists? :smile:
  7. Mar 3, 2012 #6
    ok so after i seperate variables and integrate i get ln(sin(y)) = -x^2 + C , how do I solve for y? Also was my integration correct?
  8. Mar 3, 2012 #7
    yes it is correct.

    Try by setting sin(y) = A and solve for A. Then substitute back and solve for y.
  9. Mar 3, 2012 #8
    Ok and once I've done that do I insert the intial values for the particular solution and that is the final answer?
  10. Mar 3, 2012 #9
    Yes you insert your initial values and find the C. Then your done:tongue2:
  11. Mar 3, 2012 #10
    Ok thank you very much
  12. Mar 3, 2012 #11
    Just to double check when substituting A you get ln(A)=-x2+C

    I still not sure how to solve for x because the C is there, without the c it is just A = e-x2, but with the C there what do I do, or do i ignore the C ?
  13. Mar 3, 2012 #12
    Well just do the same with the C there, what difference does it make?
  14. Mar 3, 2012 #13
    So ln(A)=-x^2+c
    Last edited: Mar 3, 2012
  15. Mar 4, 2012 #14
    I assume you mean y=arcsin(e^(c-x^2)) ?
    Now to find the C, plug in the values in this equation
    ln(sin(y)) = -x^2 + C
    as it is way easier to solve for C
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