Series solution to nonlinear differential equation

[teroenza]

Homework Statement

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

Homework 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}

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.

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)

 

[haruspex]

Homework Statement

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

Homework 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}

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.

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)

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

 

[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.

 

[haruspex]

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