When to Taylor expand in a differential equation

Question

I've come across a differential equation:

$\dfrac{d\theta}{dx} = f(\theta)$,

whose analytic solution is very complicated, but in this situation it is valid to Taylor expand functions around $\theta = a$. I observe that solving

$\dfrac{d\theta}{dx} = f(a) + f'(a)(\theta - a) + O((\theta - a)^2)$,

where

$f'(\theta) \equiv \dfrac{df}{d\theta}$,

whilst truncating terms of higher order than $1$, should yield

$\theta(x) = A e^{f'(a)x} + a - \dfrac{f(a)}{f'(a)}$.

Yet, if instead one divides through by $f$ and then expands similarily:

$\dfrac{1}{f(\theta)} \dfrac{d\theta}{dx} = 1$

$= \lgroup\dfrac{1}{f(a)} - \dfrac{f'(a)}{f(a)^2} (\theta - a) + O((\theta - a)^2) \rgroup \dfrac{d\theta}{dx}$,

a solution of different nature is obtained:

$\theta(x) = \dfrac{f(a)}{f'(a)} + a \pm \sqrt{\lgroup \dfrac{f(a)}{f'(a)} + a \rgroup^2 - \dfrac{2 f(a)^2}{f'(a)} (x + c)}$.

My question is how can these two seemingly different solutions be consolidated? And which, if either, is more valid?

Edit: For those wondering why I expressed the differential equation in this particular form, $\theta(x)$ is a physical variable dependent on its position in space (and so one would intuitively view its behaviour in this way), and $\dfrac{d\theta}{dx}$ was obtained from:

$\dfrac{d^{2}\theta}{dx^2} = g(\theta)$.

Answer 1

The right thing to do is expand the whole differential equation around $x=x_0$, $\theta = \theta(x_0)$. If $\theta(x) = a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + \ldots$ and $f(y) = c_0 + c_1 (y-a_0) + c_2 (y-a_0)^2 + \ldots$ then the DE says

$$ a_1 + 2 a_2 (x - x_0) + 3 a_3 (x - x_0)^2 + \ldots = c_0 + c_1 (a_1 (x - x_0) + a_2 (x - x_0)^2 + \ldots) + c_2 a_1^2 (x - x_0)^2 + \ldots $$

so that

$$ \eqalign{a_1 &= c_0 \cr a_2 &= c_1 a_1/2 = c_0 c_1/2 \cr a_3 &= (c_1 a_2 + c_2 a_1^2)/3 = c_0 c_1^2/6 + c_0^2 c_2/3\cr}$$ etc.

Answer 2

In the first place you should write your differential equation in the form $${dx\over d\theta}={1\over f(\theta)}\ ,$$ which immediately gives solutions of the form $$x(\theta)=G(\theta)+C\ .$$ Concerning your "expansion", you cannot hope that truncating it after one or two terms would produce something that is, identically in $x$, equal to the correct solution. Such an expansion up to $O(x^k)$ for some finite $k$ can only give the first few terms of the Taylor expansion of the desired solution $x\mapsto\theta(x)$.