# Nonlinear ODE by an infinite series expansion

1. Sep 6, 2009

### javicg

I have to solve the nonlinear DE y'=x²-y² by using an infinite series expansion y=$$\sum_{n=0}^{\infty} a_n x^n$$, but I've tried in vain. Maybe a change of variables would make it easier, but I don't know which one.

Thanks

Last edited: Sep 6, 2009
2. Sep 6, 2009

### loveequation

I don't think you need to change variables: just attack it directly. The trick is in writing $$y^2$$ as a double sum so that if you substitute the series into the ODE you get:

$$\sum_{n=0}^\infty n a_n x^{n-1} = x^2 - \sum_{n=0}^\infty\sum_{m=0}^\infty a_n a_m x^{m+n}$$

Next equate coefficients of like powers of $$x, i.e., x^{-1}, x^0, x^1,$$etc.
As an example, suppose you are equating coefficients of $$x^5$$, say. Then you get

$$6 a_6 = -(2a_0a_5 - 2a_1a_4 - 2a_2a_3)$$
and you can solve for $$a_6$$ from the previous coefficients you know.

You should see a pattern after a while.

Last edited: Sep 6, 2009
3. Sep 6, 2009

### Cody Palmer

I had done it the way suggested by loveequation, and you do end up with a pattern after you equate for the $$x^2$$ coefficient. I spent a couple of hours trying to find another way and was unsuccessful. The recurrence relation we obtain for the coefficients is not nice at all, but would be computable, given some initial condition, which would give us the first coefficient. The thing with infinite series solutions is that we almost never end up with something nice for the coefficients, but we do have a series that will allow us to compute approximations as close to the actual solution as we want. Good for engineering applications.

4. Sep 7, 2009

### matematikawan

Look like the equation is a special riccati equation. I had similar problem in the past and manage to get help from fellow PF.

5. Sep 7, 2009

### Cody Palmer

Okay,when I use a riccati substitution I get
$$\frac{d^2u}{dx^2} - x^2u = 0$$
But now how to solve this? If I use a series expansion on this I get a solution, but it has a rather ugly recurrence relation. Maybe there is another way to solve this particular ODE, and I am totally brain farting on it.

6. Sep 7, 2009

### matematikawan

I do not expect that the recurrence relation will be nice in this case, although the transformed equation is.

The one that I solved before
$$\frac{d^2u}{dx^2} + x^2u = 0$$
contain Bessel functions of the first and second kind as closed form.

I suspect that the equation you are solving
$$\frac{d^2u}{dx^2} - x^2u = 0$$
will have modified Bessel functions for closed form solution.

7. Sep 10, 2009

### kosovtsov

Last edited by a moderator: Apr 24, 2017