First order differential equation concept issue Find differential equation  $y^{\prime} = f(t, y)$ satisfied by $y(t) = 4 \, e^{2 t} + 3$.
Solution:
Compute derivative of $y$,
$y^{\prime} = 8 \, e^{2 t}$
Write right hand side above, in terms of the original function $y$, that is,
$y = 4 \, e^{2 t} +3$ ----> $y - 3 = 4 \, e^{2 t}$ ----> $2(y-3)= 8 \, e^{2 t}$
Get a differential equation satisfied by $y$, namely
$y^{\prime} = 2y - 6$
So my issue with that last answer. How is this a solution? Does it mean that if you somehow take an integral of $2y - 6$ you should end up with the original $y(t) = 4 \, e^{2 t} + 3$ ???
It seems that there two different derivatives of $y(t)$
one is:
$y^{\prime} = 8 \, e^{2 t}$
the other is:
$y^{\prime} = 2y - 6$
and I don't get it, can someone explain?

Also a bit offtopic, but the way $y^{\prime} = f(t, y)$ is written kinda bugs me.
Shouldn't it be written like $y^{\prime} = f(t, y(t))$ to show that the function $f$ contains t as an independent variable and the function y(t) which contains variable t as an input to itself (dependent variable t) ??? That's kinda an essential information, so surprised it's omitted in the writings.
 A: Here's a more detailed version of Matthew Leingang's answer. Let's look at an analogous problem where the unknown is a number instead of a function.

Find a polynomial equation $p(y) = 0$, with integer coefficients, satisfied by $y = 1 + \sqrt{3}$.

Solution
Compute the square of $y$,
$$y^2 = 4 + 2\sqrt{3}.$$
Write the right-hand side above in terms of the original number $y$. That is,
$$y^2 = 2y + 2.$$
Rearrange to get a polynomial equation satisfied by $y$, namely
$$y^2 - 2y - 2 = 0.$$
Issues
How is this a solution?
This is a solution because, if you plug $y = 1 + \sqrt{3}$ into the polynomial $p(x) = x^2 - 2x - 2$, you get zero. In other words, you found a polynomial $p$ that makes the equation $p(y) = 0$ true.
Does it mean that if you somehow take a square root of $2y + 2$, you should end up with the original $y = 1 + \sqrt{3}$?
Yes! This is a surprising fact, one we might not have known before solving the problem. You can check it with a calculator.
It seems there are two different squares of $y$. One is $y^2 = 4 + 2\sqrt{3}$. The other is $y^2 = 2y + 2$.
For most numbers $x$, the numbers $x^2$ and $2x + 2$ would indeed be different. The fact that $y^2$ and $2y + 2$ are the same is a special property of the number $y = 1 + \sqrt{3}$.
A: What you have are two differential equations. In one, you have $y'  = 8e^{2t}$, so you have the derivative of $y(t)$ in terms of the value of $t$. In the second, you have $y' = 2y - 6$, so you have the derivative in terms of the value of $y$. It happens that $y(t) = 4e^{2t} + 3$ is a solution for both. Now it's not true that the two DEs are the same - in fact, the solution set for the first is $y = 4e^{2t} + C$, where $C$ is some constant, while the solution set for the second is $y = Ae^{2t} + 3$, where again $A$ is some constant value. However, you can see that the point where those two solution sets coincide is when $A = 4$ and $C = 3$.
This is not some crazy thing - there are always multiple ways to represent things, and there are plenty of equations that happen to have coincident solutions - consider the equations $y = 4x + 3$ and $2y + x = 0$; there are a wide number of $x$ and $y$ values that satisfy the first (for example, $(0, 3)$ or $(1000, 4003)$) and the same for the second, and as it happens there's one pair that satisfies both - $(-\frac{2}{3}, \frac{1}{3})$.
As for the fact that you want to write $y = f(t, y(t))$, that's kind of fine, but as you go further into the depths of mathematics you'll find that as long as something is generally understood in its context, it will be simplified as much as possible so you don't have to keep writing it out. You've probably already done so in some fashion without even realising it. And when things start getting more complicated (n-th order differential equations in m variables, where some of those variables could technically relate to each other) you don't want to have to be writing every single thing out every time.
EDIT: Also, the two different DEs to some extent represent different ideas of what's going on. In the first, it's saying that the thing "pushing" the value of $y$ is an exponential force over time. In the second, it's saying that the force on $y$ is related to how big $y$ is - this will become more important when you start looking at DEs representing physical systems, e.g. harmonic motion with a driving force (which might be something like $y'' = -k_1y + e^{-k_2 t} y^2$) or, say, falling under gravity with drag ($y'' = -k_1 y^{-2} - k_2 y'^2$ or thereabouts). 
A: To your first point about “how is this a solution?”—this exercise is not a usual one.  Usually we are given a differential equation and asked to find a solution.  In this case you are given a function and asked to find an equation it is the solution for.
Here is a version of the same exercise from algebraic equations: 

Find a polynomial $P(x)$ with rational coefficients such that $P(x) = 0$ is satisfied by $a = \sqrt{2}$.

Solution: if $a =\sqrt{2}$, then $a^2 = 2$, so $a^2-2 = 0$, so $P(x) = x^2-2$ works.  
So is $x^2-2=0$ a “solution”?  Not in the usual sense.  Only to this exercise.
