TITLE: Initial Value Problems and integral curves
QUESTION [1 upvotes]: I have a question regarding my introductory differential equations course. I'm lost regarding this question:
"Consider the following Initial Value Problem: $dy/dt = f(y)$, with initial condition $y(0) = y_0$, where $f(y)$ and $f'(y)$ are continuous for all $y$ (implies a unique solution). Suppose that $f(y_0) \neq 0$ (cannot equal $0$). Then it can be shown that the solution to the above Initial Value Problem is monotonic (strictly increasing or decreasing). Provide a careful argument that justifies this claim, based on sketches of integral curves."
Any ideas on how to go about this? I'm not quite sure what $f(y_0)$ represents or if the variable $t$ is involved (if it isn't obvious by now it's my first DE course). Any help/guidance on this question would be greatly appreciated. Thanks everyone

REPLY [0 votes]: I can show the argument based not on sketches, however, you can draw them too for more clarity.
Assume, without loss of generality, that $f(y_0) \equiv f(y(0)) > 0$. It means that in the initial moment of time $t = 0$ the derivative is strictly positive:
$$
\frac{dy(0)}{dt} = f(y(0)) > 0,
$$
which is equivalent to the fact that the solution $y(t)$ must be strictly increasing near $t = 0$.
Assume now that the solution $y(t)$ non-monotone, in general. Therefore, the strict increase will change somewhere to the non-strict increase (or decrease), i.e. there exists $t_0$ such that
$$
\frac{dy(t_0)}{dt} = f(y(t_0)) = 0.
$$
Note that $y(t_0) = C > 0$.
Now we consider the Initial Value Problem 
$$
\frac{dy}{dt} = f(y) \quad \mbox{with the initial condition} ~ y(t_0) = C.
$$
But the constant function $y(t) = C$ becomes a solution of this problem, since it satisfies the equation and the initial data. Moreover, this solution is unique. Therefore, this solution is becomes a solution of you first Initial Value Problem.
But it contradicts the assumption
$$
\frac{dy(0)}{dt} = f(y(0)) > 0.
$$
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