SECTION 1.3. QUALITATIVE TECHNIQUE: SLOPE FIELDS (Also mixing problems) January 23rd, 2007. Can the computer solve DE?. Yes! Weeeell, No. Not always. Many DE’s don’t have “analytic” solutions (i.e. you can’t always find an expression of y in terms of t .
(Also mixing problems)
January 23rd, 2007
Weeeell, No. Not always. Many DE’s don’t have “analytic” solutions (i.e. you can’t always find an expression of y in terms of t.
Maple can find analytic solutions for DE’s sometimes. Do the Maple handout.
Exercise (Maple): Solve
Find a solution with ICs T(2) = 100.
Find a solution with ICs T(2) = 20.
Finding an exact solution of a DE (i.e. what we did for separable DE’s) isn’t always possible. In fact, it isn’t even possible most of the time!
In this section, we’ll study the slope field, an important qualitative tool for analyzing DE’s. It relies on geometry to represent the families of solutions to a DE.
This tool is called qualitative because it will not give you a precise formula, but it will give you meaningful (and usually quite accurate) information about the solutions to a DE.
In fact, even if we know the exact solution to a DE, we will often do qualitative analysis as well.
You may have noticed that the solutions to the DE kind of “parallel each other.” That is, solutions that are close to each other also have similar slopes.
The slope field represents these slopes (dy/dt) on the ty-plane. For each point (y, t) we’ll compute dy/dt by plugging the values of y and t into the original DE.
Example: If y = 3, t = 2, and dy/dt = t - y, then at the point (2, 3), we have dy/dt = 2-3 = -1. So we draw a teeny line with slope -1 to indicate that any solution passing through (2, 3) has slope -1 at that point.
We are going to plot the slope field for
dy/dt = t - y.
Plot the slope field for
and plot the solution y(2) = 1.
Tell me about it:
Look at problem 33 on p. 34.
s = amount of salt (in pounds) in the water at time t
t = time in minutes
ds/dt = (rate salt is entering) - (rate salt is leaving)
ds/dt = 0.25 - (0.1)s
ds/dt = .25 - 0.1s (note that s = 2.5 is the equilibrium)
ds/(.25 - 0.1s) = dt
integrate both sides
ln(|.25 - 0.1s|)/(-.1) = t + c
s(t) = ke^(-.1t) + .25/.1 (note that s -> 2.5 when t -> infinity)
Question: why does it make sense that s=2.5 is the equilibrium?
Answer: equilibrium means “rate of salt coming in = rate of salt going out”
Now, rate of salt coming in = .25 lb per min = rate of salt going out = one-tenth of salt, which happens when salt = 2.5 lbs
Read problem 38, p. 35
You will do this lab in pairs, and each pair gets one set of parameters from Table 1.10.
What techniques can you imagine using in this lab?
What homework problem does this remind you of?