Solve Diff. Eq. for g(x) w.r.t f(x) & c

[wheepep]

In the following equation:
$$g'(x) = f'\left( x + \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} \right)$$
find $g(x)$ with respect to $f(x)$ where $c$ is any constant.

 

[topsquark]

In the following equation:
find g(x) with respect to f(x) where c is any constant.

The link is broken. Can you type it out?

-Dan

 

[I like Serena]

In the following equation:
$$g'(x) = f'\left( x + \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} \right)$$
find $g(x)$ with respect to $f(x)$ where $c$ is any constant.

Hi wheepep! Welcome to MHB! ;)

I'm not sure where we want to go with this, but I can give a couple of observations.

We can observe that:
$$|f'(x)| < \sqrt{1+f'(x)^2}$$
and therefore:
$$-|c| < \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} <| c|$$From a Taylor expansion we get:
$$g'(x) = f'\left( x + \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} \right) = f'(x)+ \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }}f''(x+\theta c)$$
where $0\le \theta < 1$.If we assume that $f''$ is bounded, then we get:
$$|g'(x) - f'(x)| < |cf''(x+\theta c)| \le |c|M$$
for some upper bound $M$.
And therefore:
$$ -|c|Mx + C < g(x) - f(x) < |c|Mx + C \quad\Rightarrow\quad f(x) - |c|Mx + C < g(x) < f(x) + |c|Mx + C$$
where $C$ is an integration constant.