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Homework Help: Solving the ODE y' = x^2 + y^2

  1. Sep 11, 2005 #1
    Hi I'm just wondering if there is a way to solve the ODE: y' = x^2 + y^2. I've skimmed through my book and I haven't found a way to do this. Any help appreciated.
     
  2. jcsd
  3. Sep 11, 2005 #2
    i suppose that y' = [tex]\frac{dy}{dx}[/tex] ?


    Is x function of y ? is y function of x ? Or both x and y function of some other argument ?

    Your are gonna need to be a bit more specific here


    marlon
     
  4. Sep 11, 2005 #3
    I think if we are provided with initial conditions Picard's iteration can come up with the required answer.
     
  5. Sep 12, 2005 #4
    I found this question while I was going through some exam papers yesterday. I'm pretty sure that in this question y is a function of x, or at least that's what the question seems to imply. I was just curious as to whether there is a standard way to solve a question like that.
     
  6. Sep 14, 2005 #5

    saltydog

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    Hey Benny, you got this? It's a particular form of the Riccati Equation right?

    [tex]y^{'}+Q(x)y+R(x)y^2=P(x)[/tex]

    so yours is:

    [tex]y^{'}-y^2=x^2[/tex]

    Thus make the substitution:

    [tex]y=\frac{u^{'}}{-u}[/tex]

    Turn the crank and get:

    [tex]u^{''}-x^2u=0[/tex]

    Then solve via power series, then take the derivative, form the quotient, then a plot. Put it all into Mathematica and back-substitute to make sure it's correct, then compare with numerical results. Or just do what you want. :smile:
     
  7. Sep 15, 2005 #6
    Thanks for your response Saltydog. I haven't seen this type of equation before and I've done only very basic DE questions with series, most of which I've forgotten by now, so it'll be a while(ie. during my 2 week break which starts next week) until I have a look at series solutions again and try to get this one out.
     
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