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

1. Sep 11, 2005

### Benny

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. Sep 11, 2005

### marlon

i suppose that y' = $$\frac{dy}{dx}$$ ?

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

3. Sep 11, 2005

### heman

I think if we are provided with initial conditions Picard's iteration can come up with the required answer.

4. Sep 12, 2005

### Benny

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.

5. Sep 14, 2005

### saltydog

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

$$y^{'}+Q(x)y+R(x)y^2=P(x)$$

so yours is:

$$y^{'}-y^2=x^2$$

Thus make the substitution:

$$y=\frac{u^{'}}{-u}$$

Turn the crank and get:

$$u^{''}-x^2u=0$$

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.

6. Sep 15, 2005

### Benny

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.