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Need Help with Heterogeneous Differential Equation

  1. Apr 19, 2013 #1
    Hi Everyone!

    I am really stuck at one differential equation:

    [itex](p-y)^{2}y'(x)+(x-y)(x+y-2p)=0[/itex]

    where [itex]p<x<1; \ 0<y<p; \ p - parameter, 0<p<1[/itex]

    I have a suspicion that it does not have a solution, but is there a way to prove it mathematically in this case? I would appreciate any hints on tackling this equation. Thank you!
     
  2. jcsd
  3. Apr 20, 2013 #2
    Hi !

    The solutions of the ODE cannot be expressed on the form y(x), but on a parametric form :
     

    Attached Files:

  4. Apr 20, 2013 #3
    Thanks, JJacquelin!

    I am not strong in differential mathematics and I am not sure what parametric solution stands for, so sorry for a naive question, but is it possible to somehow plot the parametric form solution or at least verify that if it could have been expressed on a y(x) form its slope would be downward?
     
  5. Apr 20, 2013 #4
    There is no need to solve the equation and no need to know explicitly y(x) to answer the question. Just show that y'(x) < 0 from the equation.
     
  6. Apr 20, 2013 #5
    Do you mean the initial equation, where we are given that basically y<p<x and therefore the slope is evidently negative, or do we know for sure the signs of [itex]X(\theta) \ \ and \ \ \theta [/itex]? In this case from the parametric solution I believe it would be a more formal proof...
     
  7. Apr 20, 2013 #6
    Yes, only the initial equation with all the specified conditions. Forget the analytic resolution.
     
  8. Apr 20, 2013 #7
    Ok, thanks again! One last naive question, could you, please, tell me if the analytical solution also shows the downward slope of the function, because I am not sure I completely understand it...
     
  9. Apr 21, 2013 #8
    Forget the analytical solution. It should be a thousand times more complicated than using directly the differential equation with the associated conditions in order to prove that y'(x)<0.
     
  10. Apr 21, 2013 #9
    Yes, this I understand, but I wonder does the analytical way you solved this equation allows to come to some kind of explicit solution that would map x to y, so that when, say, p=0,5 or any other number withing given range, we would always be able to identify the value of function y, given the value of argument x?
     
  11. Apr 21, 2013 #10
    What you expect suppose to express analytically the inverse (reciprocal) function of X([itex]\theta[/itex]) which is not possible in the general case. If it was possible, no need to a prametric form of solution : I would have written y(x) instead of.
     
  12. Apr 21, 2013 #11
    Ok, thanks again for all your answers!
     
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