Math 250 Differential Equations
Problem Set B
February 7, 2003
From the supplemental text by Coombes et. al. you are to do problems #5,9,10,15 from problem set B (pages 59 to 67). These problems are due (via the digital drop box) Monday, February 24.

You should:

Number 5: For part (a), you'll have to "read" Mathematica's solution from DSolve to get the form (f[x,y] = c) suggested in the text. In part (b) the ContourPlot option PlotPoints -> 30 may help produce a smoother picture. For part (c) do not use ImplicitPlot, but use ContourPlot instead. You'll need to use the option Contours -> {c} for certain c-values. You can use
Graphics[{Red, PointSize[0.02], Point[{1,0}], Point[{2,2}], Point[{2,3}]}]
to create points to be combined with the ContourPlot in a Show[] command, for "marking the points" in part (d) on your plot.

Number 9: Use the terms sink, source, and node instead of stable or unstable to describe the equilibrium points. You may actually need more than 1 viewing window to accurately show the slope field for some parts. You should find the equilibrium points exactly whenever possible (i.e. use Solve instead of FindRoot if possible). It may help to start each part by plotting the right-hand-side (rhs) of each differential equation (as a function of y).

Number 10: Again, classify equilibrium points as sinks, sources, or nodes, not stable or unstable. This time a plot of the rhs of the differential equation may be necessary, before plotting the slope field. One question raised in part (b) (the one relating to the solution that lies along the "boundary") requires significant thought. Don't blow off that question! You may assume that this "boundary" solution does not go to zero as x increases (or it wouldn't lie on the boundary, it would be part of the "lower half" of the solutions). In part (c), DSolve's answer will involve odd functions, but these odd functions can be plotted and evaluated just like any other function in Mathematica. Try to plot (together) 3 or 4 solutions of each type mentioned in part (b) (those below the "boundary" solution and those above it). Try to estimate the value of c, accurate to 3 decimal places, that gives the "boundary" solution. That is, find a value of c (with 3 or more decimal places) so that the solution with y[0] = c + .0005 blows up, while the solution with y[0] = c - .0005 goes to zero, as x increases. Be warned that you won't be able to "plot" the boundary solution (roundoff error causes serious problems with such a plot), so this is about as good as we can do (that is, plot one solution just above it, and another just below it). The solution for c = 0.799 blows up to infinity at some finite value of x, say x*.  Use FindRoot to give a good estimate of this value of x*.

Number 15: Check with me early to ensure you have the right differential equation. You'll need to explain why, at time t minutes, there is 1 + t gallons of solution in the tank (for t < 9 that is). Use the constant c for the pounds of salt at t = 0 (that is S[0] = c). Note that parts (a) and (b) are to be guesses based solely on the slope field plot (you may want to increase PlotPoints in the slope field plot, to better estimate A). Part (c) has something to do with setting c = A, so that S'[0] = 0. Part (f) will likely require using the Solve command on the equation S'[t] == 0.