Simplest solution for differential equation Find the simplest solution:
$y' + 2y = z' + 2z$ I think proper notation is not sure, y' means first derivate of y. ($\frac{dy}{dt}+ 2y = \frac{dz}{dt} + 2z$)
$y(0)=1$
I got kind of confused, is $y=z=1$ a proper solution here? Or is disqualified because a constant is not reliant on time and something like $e^t$ is the simplest solution?
You can choose z and y however you like.
 A: $$\frac{dy}{dt}+ 2y = \frac{dz}{dt} + 2z$$
$$\frac{dy}{dt}-\frac{dz}{dt}=-2(y-z)$$
$$\frac{d(y-z)}{dt}=-2(y-z)$$
$$\frac{d(y-z)}{(y-z)}=-2dt$$
Integrating both sides, we get
$$\ln|y-z|=-2t+c$$ where $c$ is a constant of integration.
Using the given condition, we have
$$\ln|1-z(0)|=c$$
So we have that $$\ln|y-z|=\ln|1-z(0)|-2t$$ Or even better, we have that
$$\ln\left|\frac{y-z}{1-z(0)}\right|=-2t$$
Thus we have that $$y(t)=z(t)+[1-z(0)]e^{-2t}$$
That's the simplest solution possible.
A: Let $w(t)=y(t)-z(t)$.  Then, we have
$$\frac{dy(t)}{dt}+2y(t)=\frac{dz(t)}{dt}+2z(t)\implies \frac{dw(t)}{dt}+2w(t)=0$$
Hence, $w(t)=Ae^{-2t}$ for some constant $A$, from which we find that $y(t)=z(t)+Ae^{-2t}$ .  Using the initial condition, $y(0)=1$, we obtain
$$y(t)=z(t)+(1-z(0))e^{-2t} \tag 1$$
and we are done!


EDIT:
It appears that the OP would like the "simplest" functions $y(t)$ and $z(t)$ that satisfy $(1)$.  Choosing $y(t)=z(t)=1$ satisfies $(1)$ and provides the "simplest" functions that do so.

A: The answer you propose certainly can't be faulted.
There is not a unique accepted criterion of "simplest" in mathematics. Certainly we expect people to agree in many specific cases. Many items in high stakes tests are of the form "find the next term in the sequence" but what is the simplest sequence starting $1,2,4,..?$ We reasonably expect most people to say "powers of $2$ so $1,2,4,8,16,...$" But maybe the differences should be $1,2,3,4,..$ so the quadratic fit $1,2,4,7,11,...$ is simpler?
Certainly one criterion is "a polynomial if possible and, if so, one of lowest degree." That would give $1,2,4,7,11,...$
Consider the differential equation $w'(t)=-2w(t).$ The most general solution is $w=ke^{-2t}$ where $k=w(0)$. Any initial condition uniquely specifies a solution. If I denied you any initial condition and asked for the simplest answer, most people might say $w(t)=e^{-2t}$ though the criterion above prefers $w(t)=0.$ The other might qualify as the simplest non-trivial solution. All these are strictly positive or strictly negative. The general positive solution is $e^{-2t+C}$ and $C=0$ seems simplest.
Your problem could be cast as 

$y(t)$ and $z(t)$ are functions such that $y(0)=1$ and $w(t)=z(t)-y(t)$ satisfies $w'(t)=-2w(t).$ Find the simplest solution.

The must general solution is to have $y(t)$ be any function with $y(0)=1$ and $w(t)$ be any function with $w'(t)=-2w(t).$ By the reasonable criterion I suggested, $y(t)=1$ and $w(t)=0$ is indeed simplest.
BUT I might leave out $w(t)$ and say "How simple can $z(t)$ be? What about $z(t)=0?$ That forces $y(t)=e^{-2t}.$"
