y” + (1-x) y= 0 (Part I)

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In this post we will solve the following basic second-order equation with corresponding initial condition:

$\displaystyle y'' + (1-x)y = 0, \quad y(0) = 1, \quad y'(0) = 0$

We rewrite the equation (inserting $\epsilon$ ):

$\displaystyle y'' + (1-\epsilon x)y = 0$

We expand $y_{\epsilon}(x)$ as:

$\displaystyle y_{\epsilon}(x) = \sum_{n=0}^{\infty} a_n(x) \epsilon^n$

The differential equation becomes:

$\displaystyle y_{\epsilon}(x)'' + (1-\epsilon x)y_{\epsilon}(x) = 0$

$\displaystyle \sum_{n=0}^{\infty} a_n''(x) \epsilon^n + (1-\epsilon x) \sum_{n=0}^{\infty} a_n(x) \epsilon^n = 0$

$\displaystyle \sum_{n=0}^{\infty} a_n''(x) \epsilon^n + \sum_{n=0}^{\infty} a_n(x) \epsilon^n - x \sum_{n=0}^{\infty} a_n(x) \epsilon^{n+1} = 0$

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