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Solving the ODE y' = x^2 + y^2

  • Thread starter Benny
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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.
 

Answers and Replies

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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
 
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I think if we are provided with initial conditions Picard's iteration can come up with the required answer.
 
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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.
 
saltydog
Science Advisor
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Benny said:
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.
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:
 
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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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