Predicting growth with differential equations Look at the differential equation
$y'[x] = y[x] (2 + \sin[x]^2 - y[x])$
with $y[0] = 1$.
How do you know before you do any plotting that as $x$ advances from $0$, the plot of the solution $y[x]$ goes up near $x$'s for which
$2 + \sin[x]^2 > y[x]$
and the plot of the solution $y[x]$ goes down near $x$'s for which
$2 + \sin[x]^2 < y[x]$ ?
Why do you expect that the crests and dips of the plot of $y[x]$ are located at places where the $y[x]$ plot crosses the plot of $2 + \sin[x]^2$ ?
My answer:
The plot of $y[x]$ goes up near $x$'s for which $2+\sin[x]^2 > y[x]$ because if $2+\sin[x]^2 > y[x]$ then the expression in the parenthese will evaluate to a positive number times a positive number on the outside which results in a positve $y'[x]$ meaning it is growing. The plot of the solution $y[x]$ goes down near $x$'s for which $2+\sin[x]^2<y[x]$ because then the expression in the parentheses will evaluate to a negative number times a positive number on the outside which results in a negative $y'[x]$ meaning it is decreasing. We know that $y[x]$ cannot be a negative number because $2+\sin[x]^2$ never goes below 2. Meaning that whenever $y[x]$ crosses $2+\sin[x]^2$ it will hit a crest or dip because $2+\sin[x]^2$ is where $y[x]$ is not growing or decreasing.
I feel like I am missing the big picture and my answer is wrong here. What is the correct way to approach this problem?
 A: I would say you are taking the right approach here. Your answers to the three questions are correct, with the correct reasoning. The only sentence I don't quite agree with is

We know that $y[x]$ cannot be a negative number because $2+\sin[x]^2$ never goes below 2.

Maybe you're confusing the possible values of $y[x]$ with the possible values of its derivative $y'[x]$? 
To answer your question about possible negative values of $y[x]$: it's perfectly possible for $y[x]$ to be negative. In that case, we see that $2+\sin[x]^2 - y[x]$ is positive, so $y'[x]$ is negative. So, if $y[x]$ is negative at some point, it will have a negative slope at that point, so the function will decrease (even more) around that point.
At the initial point, for $x=0$, we have that $y[0]=1$. This means that the derivative at that point is given by
$ y'[0] = y[0] (2 + \sin(0)^2 - y[0]) = 1 \cdot(2+0-1) = 1,$
so $y[x]$ will increase around the point $x=0$. This will continue until there comes a point $x_*$ where $y[x_*] = 2 + \sin[x_*]^2$, where (as you pointed out) the derivative is zero, so the function does not grow anymore, and there will be a crest at $x_*$.
It might be insightful to study the direction field associated with this differential equation, where some possible solutions are plotted:

(plotted using DField)
