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Need help with a complex differential equation!

  1. Sep 19, 2013 #1
    Hi guys!

    I am really stuck at quite a complicated (from my point of view) differential equation. I would really appreciate any hints or suggestions on how to tackle and solve it if it is possible, thanks!


    $$(\frac{x^{n-1}-y^{n-1}}{n-1})-(\frac{x^{n-2}-y^{n-2}}{n-2})p=\frac{1}{2}y^{n-3}(p-y)^2y'(x)$$

    p and n - parameters

    where $$n≥3,\ p<x<1$$
     
  2. jcsd
  3. Sep 19, 2013 #2

    Simon Bridge

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    Have you tried:
    1. putting the LHS over a common denominator?
    2. putting n=3,4,5 ... writing it out and seeing if you recognize the form of the equation?
     
  4. Sep 19, 2013 #3
    Yes, and the equation is more or less solvable when n=3, but starting from n=4 there is no more difference of squares which allowed for simplification... Say, for n=5 we have:
    $$\frac{x^3(3x-4p)+4py^3-3y^4}{y'(x)}-6y^2(p-y)^2=0$$

    or the general form of the same equation:

    $$\frac{x^{n-2}((n-2)x-(n-1)p)+(n-1)py^{n-2}-(n-2)y^{n-1}}{y'(x)}-\frac{(n-2)(n-1)}{2}y^{n-3}(p-y)^2=0$$

    But it still does not shed any light for me on possible ways to find a solution...
     
  5. Sep 20, 2013 #4

    Simon Bridge

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    Usually you'd have to explore a bit more before light dawns.
    What level are you doing this at?

    I am a bit puzzled as to why you keep putting the y' in the denominator.

    The general form seems to be something like f(x)+g(y)=h(y)y' right?
    Or s(y)y' = p(x)/q(y)+1 ... both nearly things you know strategies for perhaps?

    What happens if you change variables:

    ##z=y^{n-2}## then ##z'=(n-2)y^{n-3}y'##

    ... looks close to something useful.

    Also
    - why is it important that p<x<1 ? (may be a clue - can you exploit this restriction?)
    - is n an integer?
    - is p > 0 too? (i.e. can x be negative?)
     
    Last edited: Sep 20, 2013
  6. Sep 20, 2013 #5
    I am sorry but I did not understand the question about "what level am I doing it at", could you, please, specify?

    As for the general form you are right, f(x)+g(y)=h(y)y' is the one! I do not know much about differential equations, could you, please, refer me to a method, which deals with such forms of equations?

    Again, changing variables seems promising, but the squared term (p-y) distorts the picture...

    For the questions:
    -This restriction applies from the set up this equation was derived, I do not think it can be a clue in a mathematical solution, but I cannot completely reject such possibility...
    - Yes, n is an integer.
    - Yes, p is strictly positive and less than x, y is also strictly positive.
     
  7. Sep 20, 2013 #6

    Simon Bridge

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    Education level.

    Is this for a course at some sort of school or university or college or something?

    There is no one method, just a bunch of strategies.
    IF it were y'=f(x)/g(y) you'd have no problem though right?

    Yes - you'll probably have to look for another kind of substitution.

    Can you find one where z'=h(y)y' ? Would that help?
    Do you see the kind of thinking that is needed now?

    It's always a clue.

    What is the setup?
     
  8. Sep 20, 2013 #7
    Thank you so much, Simon, for all your suggestions and hints! I really appreciate it, but unfortunately I cannot say more about the problem beyond its mathematical formulation...
     
  9. Sep 21, 2013 #8

    Simon Bridge

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    In that case I cannot help you.
    Good luck.
     
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