Taylor expansion of a function where denominator is x - at point $x_0 = 0 $

Question

I'm trying to solve Problem 2, b for equation (2). Using integrating factor, I have solved the equation (2) as $$ y = \frac{ce^{-x^2}}{x}$$

Now, I need to get $y(0), y'(0), y''(0) $ and $y'''(0)$ to expand this equation in the first four terms of taylor series. However, $y(0) $ has denominator of $x$. How should I solve this problem?

Answer 1

You can't take a Maclaurin expansion precisely because none exists unless $y=0$. One can, however, get a Laurent expansion as follows:

$$y=\frac Cxe^{-x^2}=\frac Cx\sum_{n=0}^\infty\frac{(-1)^n}{n!}x^{2n}=\frac Cx-Cx+\frac C2x^3-\frac C6x^5+\dots$$