We have: $\frac{dx}{dt}=\frac{a}{x+b}, a>0,b>0$.
We make substitutes: $x\to\alpha h, t\to\beta \tau$.
Then the differential equation will be: $\frac{\alpha}{\beta}\frac{dx}{dt}=\frac{a}{x+b}$, We divide by $\frac{\alpha}{\beta}$ and then we have: $\frac{dx}{dt}=\frac{\beta}{\alpha}\frac{a}{x+b}$.
We choose the constants a and b as follows: $a=\alpha $ and $\beta=b$ and we get $\frac{dx}{dt}=\frac{1}{\frac{h}{b}+1}$. Now we will note $\frac{1}{b}=\theta$ and we will have $\frac{dx}{dt}=\frac{1}{\theta h +1}$.
To this problem I was asked to make both parameters disappear, which seems impossible to me.
Is there a variant? Please a hint!
Thank you !
I am not quite sure if this is what you want but if you are allowed to use shifts doing the change of variables
$$y=x+b, s=at$$
would transform
$$\dfrac{dx}{dt}=\dfrac{a}{x+b} \quad\text{to}\quad \dfrac{dy}{ds}=\dfrac{1}{y}.$$