Part 1 - Analyzing equilibrium behavior

For each of the following differential equations: graph the slope field over a reasonable domain, find all equilibrium values, explain what happens to solutions that start between these values (i.e., do they head towards a particular equilibrium values, away from an equilibrium value, both?), and graph a solution or two on the slope field that demonstrates each of these different qualitative behaviors.

So, for example, if your equation were:  FormBox[RowBox[{y ', =, RowBox[{0.5, y(4 - y)}]}], TraditionalForm], your answer would look something like:

In[8]:=

<<Graphics`

Clear[y, y0, t, slopeField, solutionCurves]

RowBox[{slopeField, =, RowBox[{PlotVectorField, [, RowBox[{RowBox[{{, RowBox[{1, ,, RowBox[{0.5, y, (4 - y)}]}], }}], ,, {x, -5, 5}, ,, {y, -5, 6}, ,, AxesTrue}], ]}]}]

[Graphics:../HTMLFiles/index_5.gif]

⁃Graphics⁃

Notice, there are 2 equilibrium values:  0 and 4.  For simpler cases like this, you can probably tell what is going on by just looking at the slope field, but if the equation is an autonomous differential equation, you could also graph y vs. y' on the same axis (so here, you are graphing FormBox[RowBox[{0.5, y(4 - y)}], TraditionalForm]):

RowBox[{Plot, [, RowBox[{RowBox[{0.5, y, (4 - y)}], ,, {y, -5, 5}, ,, AxesLabel {y, y '}}], ]}]

[Graphics:../HTMLFiles/index_9.gif]

⁃Graphics⁃

The equilibrium values are where this graph crosses the horizontal axis.  Notice, that if y<0, the slopes are all negative (making y head towards -∞).  If 0<y<4, the slope is positive, so y must head upwards (towards 4).  And if y>4, the slope is negative, so y must head back down (towards 4 again).

Thus, all solutions starting below 0 head toward -∞, all solutions starting between 0 and 4 head away from 0 and towards 4, and all solutions that start above 4 head towards ∞.  So, it might make sense to choose 3 solutions to demonstrate this:  y(0) = -2, y(0) = 1, and y(0) = 5

RowBox[{solution, =, RowBox[{DSolve, [, RowBox[{RowBox[{{, RowBox[{RowBox[{y '[t], ==, RowBox[{0.5, y[t], (4 - y[t])}]}], ,, y[0] y0}], }}], ,, y[t], ,, t}], ]}]}]

Solve :: ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. More…

RowBox[{{, RowBox[{{, RowBox[{y[t], , RowBox[{RowBox[{(, RowBox[{4., , RowBox[{ᢺ ... owBox[{4., }], -, RowBox[{1., , y0}]}], )}], /, y0}], )}], ^, 1.}]}]}], )}]}]}], }}], }}]

ySol[t_] = y[t]/.solution

RowBox[{{, RowBox[{RowBox[{(, RowBox[{4., , RowBox[{, ^, RowBox[{(, RowBox[{2., , t} ... wBox[{RowBox[{4., }], -, RowBox[{1., , y0}]}], )}], /, y0}], )}], ^, 1.}]}]}], )}]}], }}]

y0 = {-2, 1, 5}

{-2, 1, 5}

solutionCurves = Plot[Evaluate[ySol[t]], {t, -5, 5}, PlotStyleBlue]

[Graphics:../HTMLFiles/index_23.gif]

⁃Graphics⁃

(Notice that you got a few solutions with a vertical asymptote.  Look back up at the slope field and see if this makes any sense.)  Now, combine these with the slope field:

Show[slopeField, solutionCurves, PlotRange {-5, 6}]

[Graphics:../HTMLFiles/index_26.gif]

⁃Graphics⁃

For each of the following differential equations, perform a similar type of analysis, showing your work in Mathematica, along with appropriate explanations to go with it.  Warning:  For some of these, you may want to use NDSolve to plot the solutions instead of DSolve.

y ' = 4y - y^3

(hint: only plot and solve for t≥0)

y ' = y(1 - ln y) (y - 4)

(Hint: use NDSolve on this one.  Also, remember that Mathematica uses Log for natural logarithm.)

Created by Mathematica  (June 4, 2004)