$f ' (x) = f(x - (x+1)^t + 1)$ Let $x > 0 $ and $c $ a given real $> 0.$
Let $t $ be between $0 $ and $1.$
How to find $f(x)$ or good asymptotics for $ f(x)$ such that
$$ f ' (x) = f(x - (x+1)^t + 1) $$
And $ f(1) = 1 + c$.
Also $f$ is a nonlinear function and twice differentiable for $x > 0$.
 A: I'm assuming you want asymptotics for large $x$. Changing variables $x \to (x+1)$ your equation can be written more simply as
$f'(x) =f( x- x^t +1)$
Let's write $f(x) = e^{g(x)}$.
Then:
$f'(x) = g'(x) e^{g(x)}$.
$f(x - x^t +1 ) = e^{ g(x)} e^{ g(x-x^t+1) - g(x)} \approx e^{g(x)} e^{ - (x^t-1) g'(x) } $
This approximation is accurate if $g''(t)$ is not very large. 
So this gives us $g'(x) \approx e^{ - (x^t-1) g'(x) }$.  I'm assuming that $x$ is large here, so  we should be able to work with the simpler equation
$g'(x) \approx e^{ - x^t g'(x) } $
Now if we plug in $g'(x) = x^{-t} * ( t \log x - \log t - \log \log x) $, then we get
$$ x^{-t} * ( t \log x - \log t - \log \log x)  \approx e^{ -t \log x + \log t + \log \log x}= x^t (t \log x)$$
As long as $x$ is large, $t \log x$ is much larger than $\log t$ and $\log \log x$ and the approximation is pretty close.
So exponentiating an integral of this will give you an approximate asymptotic, although sadly I don't know how good it is. This integral is not even closed form, but the leading term is going to be $x^{1-t} \log x$ times a constant. So your asymptotic should be very, very roughly $x^{ x^{1-t} /(1-t) }$.  
