An approximate solution to a differential equation $f'(x)^2 - \omega^2 f(x)^2 = G(x)$ for small $x$

When deriving approximate solutions to differential equations, one should always build the approximation around the base solution in a proper form to ensure accuracy.

So, start with the equation in question,

$$ [f’(x)]^{2} - \omega^2 f(x)^2 = \omega^2-\alpha x^4\tag{1} $$

Since $\sinh(\omega x) $ is the exact solution to the base DE,

$$ [f’(x)]^{2} - \omega^2 f(x)^2 = \omega^2\tag{2} $$

the approximation should be constructed around $\sinh(\omega x)$ in the following product form,

$$f(x) = \sinh(\omega x)(1 +c x^4)$$

where $c$ is the approximation coefficient to be solved and it is of the order $O(\alpha)$. Then,

$$ f’(x) = \omega \cosh(\omega x) (1+cx^4)+\sinh(\omega x) \cdot 4cx^3$$

$$= \omega \cosh(\omega x) + 5c\omega x^4 + O(c^2)$$

and,

$$[f’(x)]^2= \omega^2\cosh^2(\omega x) + 10c\omega^2 x^4 + O(c^2)\tag{3}$$

Plug (3) into (1). The terms corresponding to the base DE cancels out as constructed and the following approximation coefficient can be obtained,

$$c=-\frac{\alpha}{10\omega^2}$$

Therefore, the approximation taking into account of the added term $-\alpha x^4$ in (1) is

$$f(x) = \sinh(\omega x)\left(1-\frac{\alpha}{10\omega^2}x^4\right)\tag{3}$$

The solution above should then be compared with the exact curve in your plots to judge the quality of the approximation.

Notes:

One should not expand the base solution $\sinh(\omega x)$ since it is the exact solution to the base DE. The base solution contains the essential information embedded in the base DE (2), up to all orders of $x$.

The reason that $\sinh(\omega x)$ is better than its polynomial expansion is simply because it is the exact base solution. Expanding it to just a few terms would compromise the quality of the solution, as shown by the third curve in your plots. So, it should not be surprising that $\sinh(\omega x)$ is closer to the real curve than its polynomial cutoff. (The simple expansion is only valid for small values of $x$, evident from the third curve in your plots.)

As alluded earlier, It is better to choose the product form in (3) for the approximate solution since the added term of order $x^4$ is comparable to that in (1). Furthermore, a more accurate approximation can be derived by including a higher order term as below

$$f(x) = \sinh(\omega x)\left(1-\frac{\alpha}{10\omega^2}x^4 +\frac{4\alpha}{105} x^6\right)$$