Let $x \in \mathbb{R}$ be a variable and $c\in\mathbb{R}$ a parameter. Also, let $f(x,c)$ be a function dependent on $x$ and $c$. Furthermore, define a Differential Equation which is solved by the function $y(x)$:

$$\left(\frac{d}{dx}\right)^2 y(x) + f(x,c)y(x)=0$$

Suppose, the general Differential Equation is too hard to solve, but for practical reasons only the solution with $c\approx 0$ is required. Naively I would expand $f(x,c)$ around $c=0$ and then compute the approximate Differential Equation. To the order $O(c^1)$ for example I would have to solve the following (where a prime denotes a derivative in respect to $c$):

$$\left(\frac{d}{dx}\right)^2 y(x) + \big(f(x,0)+f'(x,0)~c + O(c^2)\big)y(x)=0$$

If $f(x,0)$ and $f'(x,0)$ turn out to be simple enough and if we neglect the $O(c^2)$ terms, this approximated equation might turn out to be solvable.

Does this give a mathematically valid approximation for the solution $y(x)$? Maybe there are some subtleties which I did not consider? For instance, I have the feeling that the solution $y(x)$ should also be considered as a function of $c$ (as in $y(x,c)$) and be somehow involved in the expansion. It would be nice if someone knowledgeable could shed some light onto this.

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    $\begingroup$ It would depend on the function $f$. $\endgroup$ – Avi Steiner Mar 3 '13 at 16:36
  • $\begingroup$ @Kagaratsch If you stay around 0 for c, your solution method is correct. $\endgroup$ – AnilB Mar 14 '13 at 23:36

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