Question

Differential equation: Sketch direction field:

y'= y-y^2

Answer 1

\[y'=y-y^2=y(1-y)\]\[\frac{y'}{y(1-y)}=1\]\[\frac{y'}{y}+\frac{y'}{1-y}=1\]\[(\ln{y}-\ln{(1-y)})'=1\]\[\ln\frac{y}{1-y}=x+C\]\[\frac{y}{1-y}=e^{x+C}\]\[\frac{1-y}{y}=e^{-x-C}\]\[y^{-1}=1+e^{-x-C}\]\[y=\frac{1}{1+e^{-x-C}}=\frac{e^x}{e^x+C_1}\]

Answer 2

Sketching out a field on this website would probably be a bit tricky, but the idea behind the problem is to essentially pick a bunch of points and find the slope at those times.
\[y \prime = y - y^2\]is your slope, so pick points close to the origin. Say...(0, 0)
Plugging that into your equation will give you \[y \prime=0 - 0^2 = 0\] Since the slope is 0, you would draw a horizontal line at the origin.
Try with (0, 1):
\[y \prime = 1 - 1^2 = 0\] The slope is still 0, so draw a horizontal line one notch above the origin.
Try with (-1, 0):
\[y \prime = 0 - 0^2 = 0\]Since you don't have an x value in your equation for the slope, changing the x (or horizontal) value won't affect it. It's not necessary to note that fact, but it'll make drawing the slope field a little less tedious.
Try with (1, 2):
\[y \prime = 2 - 2^2 = -2\] Now you actually have a nonzero slope! To draw this one, think of how you would draw the slope of a regular line. Slope is rise over run, so -2 would go down 2 notches and one to the right.
It's pretty tedious, but assuming your teacher isn't being cruel you shouldn't have to repeat this too many times. It goes by faster if you can recognize a pattern while drawing it out.

Answer 3

Here's a very badly drawn depiction. Accuracy as to what the slope is exactly usually isn't too much of an issue, as long as you can tell the slope changes. Once you start getting things like slope = 8 and above (or negative and below), it'll pretty much just look like vertical lines.