- #1

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## y'' = \frac{\alpha}{2}y^3 - \frac{3}{2}y^2 + y - \frac{3}{x} y'##

under the following boundary conditions:

## y'(0) = 0 ##

## y(x) \rightarrow y \_ \equiv 0\ \text{as}\ x \rightarrow \infty ##

As a first step, I converted this problem into a set of coupled ODEs:

## \frac{dy}{dx} = z##

## \frac{dz}{dx} = \frac{\alpha}{2}y^3 - \frac{3}{2}y^2 + y - \frac{3}{x} z##

under the following boundary conditions:

## z(0) = 0 ##

## y(x) \rightarrow y \_ \equiv 0\ \text{as}\ x \rightarrow \infty ##

Next, my source tells me to use the shooting method to convert the BVP into an IVP, which means that I have to use two initial guesses of ## y(0) ## to be able to use the secant method to find the correct value of ## y(0) ##.

Now, my question is, according to my source, I can avoid the singularity at x = 0 using Taylor expansion as follows:

## y(r_{0}) = y_{0} + \frac{1}{16} r_{0}^{2} (2y_{0} - 3y_{0}^{2} + \alpha y_{0}^{3}) ##

I see how you can estimate ## y(r_{0}) ## where ## r_{0} ## is a tiny distance away from the origin, but I don't really see how they dervied this expression. Could anyone help me out?