Here's the situation: I have a function
$$e(t) = \frac{a~d(t)}{b + d(t)}$$ with first derivative $$e'(t) = \frac{a~b~d'(t)}{[b+d(t)]^2}$$
where $a$ and $b$ are constants.
For a given constant $K$ I wish to find some $d(t)$ such that $e'(t) = K$, or at least such that $\epsilon \equiv |K-e'(t)|$ is minimized.
The exact approach gives $$abd'(t) = K[b+d(t)]^2$$
But try as I might I've been unable to manipulate this algebraically to get an expression for $d(t)$ that doesn't include $d'(t)$ (which would defeat the purpose).
Is there some analytical solution I'm just not seeing? Or could anybody get me started on an optimization-based approach minimizing $\epsilon$?
I am probably missing a part of the problem; so, please, forgive me if this is off-topic.
If the equation is $$ab d'(t) = K\Big(b+d(t)\Big)^2$$ you can rewrite is as $$\frac{d'(t)}{\Big(b+d(t)\Big)^2}=\frac{K}{ab}$$ (which is moreover separable) and classical integration then leads to $$d(t)=-b \left(1+\frac{a}{a b c_1+K t}\right)=-b \left(1+\frac{a}{ c_2+K t}\right)$$ which gives for $e(t)$ a linear equation in $t$ (which was the basic assumption when saying that $e'(t)=K$).
Another approach does not require the derivatives since $e'(t)=K$ implies $e(t)=Kt+c$ and then you can solve your first equation for $d(t)$. Similarly, saying that $e'(t)$ is almost a constant could be translated as $e(t)=K t +\epsilon t^2 + c$; solve again for $d(t)$.
Is this what you are looking for ?