Lecture Notes in Differential Equations - Bruce E. Shapiro

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that since the slope of the solution y(t) at any point is dy/dt, and since
dy/dt = f, then the slope at (t, y) must be equal to f(t, y). We obtain the
direction field but dividing the plane into a fixed grid of points P i = (t i , y i )
and then then drawing a little arrow at each point P i with slope f(t i , y i ).
The lengths of all the arrows should be the same. The general principal is
illustrated by the following example.
Example 2.3. Construct the direction field of
dy
dt = t2 − y (2.11)
on the region −3 ≤ t ≤ 3, −3 ≤ y ≤ 3, with a grid spacing of 1.
First we calculate the values of the slopes at different points. The slope is
given by f(t, y) = t 2 − y. Several values are shown.
t y = −3 y = −2 y = −1 y = 0 y = 1 y = 2 y = 3
t = −3 12 11 10 9 8 7 6
t = −2 7 6 5 4 3 2 1
t = −1 4 3 2 1 0 −1 −2
t = 0 3 2 1 0 −1 −2 −3
t = 1 4 3 2 1 0 −1 −2
t = 2 7 6 5 4 3 2 1
t = 3 12 11 10 9 8 7 6
The direction field with a grid spacing of 1, as calculated in the table
above, is shown in figure 2.4 on the left. 5 At each point, a small arrow is
plotted. For example, an arrow with slope 6 is drawn centered on the point
(-3,3); an arrow with slope 1 is drawn centered on the point (-2, 3); and so
forth. (The values are taken directly with the table). A direction field of
the same differential equation but with a finer spacing is illustrated in fig
2.4 on the right. From this plot we can image what the solutions may look
like by constructing curves in our minds that are tangent to each arrow.
Usually it is easier if we omit the arrows and just draw short lines on the
grid (see figure 2.5, on the left 6 ). The one-parameter family of solutions is
illustrated on the right. 7
5 The direction field can be plotted in Mathematica using f[t , y ] := t^2 - y;
VectorPlot[{1, f[t, y]}/Norm[{1, f[t, y]}], {t, -3, 3}, {y, -3, 3}]. The normalization
ensures that all arrows have the same length.
6 In Mathematica: f[t , y ] := t^2 - y; followed by v[t , y , f ] :=
0.1*{Cos[ArcTan[f[t, y]]], Sin[ArcTan[f[t, y]]]}; L[t , y , f ] := Line[{{t,
y - v[t, y, f], {t, y} + v[t, y, f]}]; Graphics[Table[L[t, y, f], {t, -3, 3, .2}, {y, -3, 3, .2}]]
7 The analytic solution y = e −t ( e t t 2 − 2e t t + 2e t − e t 0t 0 2 + e t 0y 0 − 2e t 0 + 2e t 0t 0
)
was used to generate this plot. The solution can be found in Mathematica via
DSolve[{y’[t] == t^2 - y[t], y[t0] == y0}, y[t], t].