# Series solution to nonlinear differential equation

1. Sep 25, 2014

### teroenza

1. The problem statement, all variables and given/known data

By truncating the differential equation below at n=12, derive the form of the solution, obtaining expressions for all the ancoefficients in terms of the parameter $\lambda$.

2. Relevant equations
The ODE is:
$$\frac{\mathrm{d^2}\phi }{\mathrm{d} x^2} = \frac{\phi^{3/2}}{x^{1/2}}$$

I am told that the solutions have the form:

$$\phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{3/2}$$

3. The attempt at a solution
The way I have solved series equations before was to equate the coefficients of like powers of x. In this case, however, I have those x underneath the root, and I don't know how to go about doing this. I have the terms, via Mathematica. But I don't know how to get relationships between the coefficients.

Code (Text):
p[x_] := 1 + \[Lambda]*x + Sum[Subscript[a, n]*x^(n/2), {n, 3, 12}];
pdd[x_] :=
Sum[Subscript[a, n]*(n^2 - 2*n)/4 x^((n - 4)/2), {n, 3, 12}];

LHS = pdd[x] - x^(-1/2)*(p[x])^(3/2)

Last edited by a moderator: Sep 25, 2014
2. Sep 26, 2014

### haruspex

Do you mean $$\phi = 1+ \lambda x + \sum_{n=2}^{12}a_{n} x^{n}$$?

3. Sep 26, 2014

### teroenza

No, but I did make an error in the above. It should be:

$$\phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{n/2}$$

But I can't figure out how to edit my original post.

4. Sep 26, 2014

### haruspex

That suggests to me the substitution x = u2. Haven't tried it, though.