How to find a curve whose value changes for a certain percentage? I've had to find a curve $y(x)$ whose value is $N\%$ smaller than $y(x)$ for $y(x+1)$ where $x $ is an integer.I've been told that this differential equation finds the family of such curves  (this was a classic introductory example of applying differential equations for modelling mixing problems)
$$ dy=-\frac{N}{100}\, y(x)\, dx $$
This is how I believe this was obtained:
the change for some $\Delta x$ is
$$y(x+ \Delta x)-y(x) = -\frac{N}{100}\, y(x)\, \Delta x$$
after dividing the equation by $\Delta x$ and letting $\Delta x$ approach zero, we obtain the starting differential equation.My questions are:
-is this even correct?
-if it is, how can the second equation give us the correct family of curves?It assumes that, since we are multiplying percentage by $x$, that percentages are additive such that $i$ changes for $N\%$ equal one change of $iN\%$ which is incorrect.It is correct only for  $i\le1$.   
I've checked the validity of this on wolframalpha and this is either not correct (it's an approximation) or there is precision loss on wolframalpha's side.
 A: The way the problem is formulated ($x$ integer) suggests a discrete model rather than a continuous one. Thus, we may write the recurrence relationship
$$
y(x+1) = r y(x) ,
\qquad
r = 1 -\frac{N}{100} ,
$$
which is a geometric progression with common ratio $r$. The progression implies that $y(x) = r^x y(0)$ for non-negative integers $x$. The latter expression can be linked to the continuous-time exponential model by noting that $y(x) = e^{x\ln r} y(0)$.
A: The standard way of solving this kind of equation is to make an intelligent guess as to what the answer is then show it solves the equations you want. It is called guessing an ansatz.
Essentially your equation is:
$\frac{dy}{dx} = ky(x)$ for $0 < k <1$. This can be solved to give $y(x) = Ae^{kx}$ for some constant $A$. To check this equation does satisfy what you are looking for you can verify it easily enough by just doing the algebra. It is the presence of the $A$ that means you have found a family of equations.
Your method for deriving the formula does give an equation from this family but it won't be the correct one for the value of N you are using. It's hard for me to explain why here.
There are other families of equations that also give the behaviour that you asked for. Let $g(x)$ be periodic with period 1 then $Ae^{kx}g(x)$ also gives the behaviour you asked for as well as being continuous. 
