How to justify that the nontrivial solutions of $f'=-kf$ never vanish, without solving the ODE? Say I have a ODE: $f'(x)=-kf(x)$, and $f(0)=a$, how can I explain that if $a \not= 0$, and $k>0$ then $f(x)\neq 0$ for all $x$? (Please do not solve the equation explicitly)
 A: You can solve this one exactly using Euler's method:
$$ f(x + dx) = f(x) + f'(x) dx = f(x) - k dx f(x) = f(x) \big[ 1 - k dx \big]$$
Then let's try to estimate $f(x+1)$ when $dx = \frac{1}{n}$.  It takes $n$ steps to get from $x$ to $x+1$ so then:
$$ f(x + 1) =  f(x) \Big[ 1- \frac{k}{n} \Big]^n > 0 $$
Notice we have not actually solved the differential equation yet.
A: Assume that there is an $x$ such that $f(x)=0$. Then, by the given equation also $f'(x)=0$ which means that in this $x$ there is either a minimum, a maximum or a saddle point. Assume there is a maximum. Then $f'(x)$ has to be negative both left and right of this $x$ but this implies that $f(x)$ must be positive (I assume that $k>0$) which contradicts the local maximum. Similarly, we can contradict the local minimum. Now, a saddle point means that on the one side of $x$, $f(x)>0$ and on the other side $f(x)<0$. But, since it is a saddle point then $f'(x)$ must preserve the same sign on both sides of $x$ which is impossible. So, a contradiction, there is no such $x$ that $f(x)=0$. 

Note, that we came to this conclusion from the first equation and not from the fact that $f(0)=a\neq0$. However, the assumption $a\neq0$ is important to exclude the case that $f(x)\equiv 0$. If $f$ is constant then by the first equation $f$ must be equal to $0$, so $a\neq0$ leads to a contradiction and allows us to exclude this case.
A: If we had $f(x_*)=0$ for some $x_*$ then the initial value problem
$$y'(x)=-k\> y(x),\quad y(x_*)=0$$
would have the two different solutions $y_1(x)\equiv0$ and $y_2(x)=f(x)$.
A: Let us rewrite
$f(x)=-\frac{1}{k}f'(x)$
This means
$f'(x)=-\frac{1}{k}f''(x)$
$f(x)=\frac{1}{k^{2}}f''(x)$
and so on. In general
$f(x)=\frac{(-1)^{n}}{k^{n}}f^{(n)}(x)$
if $f(x_{1})=0$ then all derivatives vanish at $x_{1}$. This simply means that if your function exists then it is flat. The constant function is a trivial example of a flat function.
So $f(x)=a$ is the only analytic option, but this one does not have $f(x)=0$ anywhere, obviously.
(In order to finalize this we need to consider non-analytic solutions as well, but I assumed an analytic solution.)
