16.4.1 Graphical Solutions – Direction Fields
| 16.4.1 |
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| 1. Direction Fields Of y ‘ = f (x, y) |
In this section, we’ll use the abbreviations “DE”, “DF”, and “ES” for “differential equation”, “direction field”, and “equilibrium
solution” respectively.
Approximate Solutions
Consider the DE:
looks simple and can’t be solved in terms of elementary functions. Remark that Eq. [1.2], which can be written as y‘ – y2 = x2,
is non-linear, due to the term y2
. DEs that can’t be solved in terms of elementary functions may be solved by the method of power series or by the method of special functions. Special functions are non-elementary ones that are studied in advanced
courses. The methods of power series and of special functions are presented in courses on DEs.
In this section we present a method, the method of direction fields, to construct approximate graphical solutions of DEs of the
form y ‘ = f (x, y), y ‘ = f (x), and y ‘ = f ( y
), comprising both equations that can and can’t be solved in terms of elementary functions. In the next section we’ll discuss a method to determine approximate numerical solutions. For DEs that can’t be solved in terms of elementary functions, we at this stage have to settle with their approximate solutions, which suffice for
many practical purposes.
Direction Fields (DFs)
A solution curve of a DE is the graph of a solution of the DE. To assure ourselves that the direction field (DF) of a DE indeed generates approximate graphical solutions of the DE, we consider an example DE that can be solved exactly, solve it, draw some exact-solution curves, sketch the DF of the DE, and see that the DF is consistent with the exact-solution curves and so
generates approximate graphical solutions.
Consider the DE:
y ‘ = x + y.
y = – x – 1 + Cex.
This last equation is the general solution. The graphs of some particular solutions for some values of C are sketched in
Fig. 1.1.
| Fig. 1.1 Graphs Of y = – x – 1 + Cex For Some Values Of C. |
| Direction Field Of y ‘ = x + y. |
Now consider again the DE:
y ‘ = x + y.
It says that the derivative y‘ of any solution y at any point (x, y) that’s on the graph of y equals x + y. We know that the
derivative y ‘ is the slope of the tangent of the graph of y at the point (x, y). So this equation y ‘ = x + y tells us that the slope
of the tangent at point (x, y) on the graph of y = y(x) equals the sum x + y of the x– and y
-coordinates of that point. For example, the slope of the tangent at point (1, 0) is 1 + 0 = 1, and the slope of the tangent at point (1, -3) is 1 + (-3) = -2. So
we select a number of points. Preferrably we draw grid lines that intersect the scaling points on the
x– and y -axis, and select the points of intersection of the grid lines. At each selected point, we draw a short line segment that’s centered at that point and that has slope equal to the sum of the coordinates of that point. See Fig. 1.2. Thus the line segment at each point is a short segment of the tangent line of the graph of the particular solution that passes thru that point at that point. As a consequence, each little tangent line segment represents the direction in its vicinity of the solution curve that passes thru its
midpoint. For this reason, the set of all the tangent line segments is called the direction field
(DF) or slope field of the DE. A small portion of the solution curve that passes thru or near a point is approximately parallel to the tangent line segment at that point. The DF indicates the directions in which the solution curves proceed at each point, and hence is a visualization of
the general shape of the set of all the solution curves.
We notice that the shapes of the approximate-solution curves as suggested by the DF in Fig. 1.2 are consistent with the exact-solution curves in Fig. 1.1. Now we’re assured that indeed the DF of a DE generates approximate graphical solutions of
the DE.
Consider a first-order DE of the form:
where f (x, y) is a function of the 2 variables x and y. Because the derivative y ‘ is the slope of the tangent to the graph of the
function y geometrically speaking and the rate of change of the function y algebraically speaking, the function f (not y; it’s f (x,
y), not y, that equals y ‘) in Eq. [1.3] is called the slope function or the rate function. Recall that the slope of a curve at a point
is defined to be the slope of the tangent line of the curve at that point.
Definitions 1.1 – Direction Fields
| Consider the DE: y ‘ = f (x, y). At any point (x, y), the slope of the tangent of the graph of the particular solution y = y(x) that passes thru that point equals |
Remark that we can draw a DF without knowing the general solution of the DE. DFs are a method by which we investigate the
qualitative characteristics of the solutions of the DE y ‘ = f (x, y).
They’re useful and important precisely because they can still be constructed in cases where the DE can’t be solved exactly.
Drawing A Lineal Element
As an example, to draw a lineal element with slope 3 centered at a point, we mentally choose a small distance as the unit, then we go from the point horizontally to the right for 1 unit, then go vertically up by 3 units and mark a dot there, then join the point and the dot by a line segment, and extend the line segment to the other side of the point by about the same distance
so that the point is approximately the midpoint of the line segment.
As another example, to draw a lineal element with slope -3/2 centered at a point, we mentally choose a small distance as the unit, then we go from the point horizontally to the right by 2 units, then go vertically down by 3 units and mark a dot there, then join the point and the dot by a line segment, and extend the line segment to the other side of the point by about the
same distance so that the point is approximately the midpoint of the line segment.
For large slopes such as 100.8 or -4,766, we can use the property that the slope of a line is the trigonometric ratio tangent of
the angle that the line makes with a ray that’s parallel to and goes in the same direction as the positive x-axis. For example,
let’s say the slope is 100.8. We reach for a calculator and calculate arctan 100.8 = 89o, approximately. We draw a lineal
element that approximately makes an angle of 89o
with the ray, which isn’t drawn but simply imagined. If the angle is positive then we go in the counterclockwise direction, if the angle is negative then we go in the clockwise direction. Observe that for
small slopes, we of course can use this approach too, but using the above approach saves us time.
For fractional slopes with large numerators and/or denominators such as 193/152 or -3/85, we can first determine their
decimal forms, then proceed in one of the above approaches.
Example 1.1
Sketch the DF of the DE y ‘ = x2 + y2.
Solution
| Fig. 1.3 Calculations Of Slopes At A Number Of Points. The 1st column contains values of x and the 1st row contains values of y. A cell in any other row and column is the sum of the squares of the x-value in that row and of the y-value in that column. |
| Fig. 1.4 DF Of y ‘ = x2 + y2. |
EOS
Remark 1.1
In Example 1.1, we calculate and display in the table in Fig. 1.3 the slopes at the points (x, y) where x = -2, -1, 0, 1, and 2,
and y = -2, -1, 0, 1, and 2. Each of the 5 values of x corresponds to 5 values of y
. So there are 5 x 5 = 25 points, and thus 25 slopes to calculate. Then we draw in Fig. 1.4 little line segments at these points with these slopes. The resulting picture is the required DF. A table displaying the calculations of slopes at a number of points such as the one in Fig. 1.3 is optional unless it’s
explicitly required.
The more line segments we draw in a DF, the more detailed and the clearer the DF becomes. However it’s tedious to compute slopes and draw line segments for a large number of points by hand. Fortunately there exist computer programs that perform this task. They’re in the category of function graphers or plotters. Fig. 1.5 displays a computer-drawn DF of the DE of Example
1.1. Of course it presents a clearer picture of the DF than does Fig. 1.4.
| Fig. 1.5 Computer-Drawn DF Of y ‘ = x2 + y2. |
Example 1.2
Solution
EOS
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| 2. Graphical Solutions Of y ‘ = f (x, y) |
Example 2.1
Sketch the solution curve of the following initial-value problem without solving the initial-value problem:
Approximately determine from the solution curve its intercepts, asymptotes, end behaviors, intervals of increasing/decreasing,
intervals of concavity, and inflection points.
Note – How To Sketch
We’re asked to sketch the graph of the particular solution of the DE y ‘ = x2 + y2 that passes thru the point (1, 0) without
solving the DE and substituting x = 1 and y
= 0 to obtain the desired particular solution. For this purpose, first we draw the DF of the DE, as done in Fig. 2.1. Next we sketch the solution curve thru the point (1, 0). To do so, we start from the point (1, 0), first to one side of the tangent line segment centered at (1, 0) and approximately parallel to it, then to its other side, again approximately parallel to it. As we proceed, the curve is approximately parallel to nearby line segments. The DF and the
solution curve are sketched in Fig. 2.1.
Solution
| Approximate Solution Curve Of Initial-Value Problem: |
EOS
Remark 2.1 – On Example 2.1
The solution curve in Fig. 2.1 passes thru the points (1, 0) and (0, -1/2). This is because it’s the solution that must pass thru the point (1, 0) and because the point (0, -1/2) happens to belong to it. It passes thru the midpoints of the lineal elements at
points (1, 0) and (0, -1/2). Remember that the midpoint of the lineal element at a point (
x, y) is the point (x, y ) itself. In the picture, the solution curve doesn’t pass thru the midpoint of any other lineal element. This is because all other points considered in the picture belong to other solutions. We’re sure that the solution curve has no horizontal asymptotes. If there are no vertical asymptotes either, if we extend the length of one or both axes sufficiently, there’ll be other lineal elements such that the solution curve passes thru their midpoints. Also, whether or not there are horizontal or vertical asymptotes, if we
increase the number of points horizontally within each unit of the
x-axis and/or the number of points vertically within each unit
of the y-axis sufficiently, there’ll be other lineal elements such that the solution curve passes thru their midpoints.
Definitions 2.1 – Graphical Solutions And Integral Curves
| Consider the initial-value problem: To sketch the solution curve of this initial-value problem, we sketch the DF of the DE y ‘ = f (x, y), then we draw a curve thru A solution curve is the graph of a solution y, which is an antiderivative or (indefinite) integral of y‘, and thus is also called an |
Example 2.2
Sketch the DF of y ‘ = x2 + y2. Then sketch the solution curves that pass thru the points (-2, 0), (-1, 0), (0, 0), (1, 0), and (2, 0)
respectively.
Solution
| Fig. 2.2 Some Solution Curves Of y ‘ = x2 + y2. |
EOS
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| 3. Direction Fields And Graphical Solutions Of y‘ = f (x) And y‘ = f ( y) |
Differential Equations Of The Form y ‘ = f (x)
Example 3.1
Sketch the DF of the DE y ‘ = 2x – 2.
Solution
| Fig. 3.1 DF Of y ‘ = 2x – 2. |
EOS
Lineal Elements Along Any Vertical Line Are Parallel
In Example 3.1, at every point (x, y), the slope y ‘ = 2x – 2 depends only on x; it doesn’t depend on y. At any point with
x-coordinate x1, the slope is 2x1 – 2, whatever y is, so the lineal elements along the vertical line x = x1 have the same fixed
slope of 2x1 – 2, and thus are parallel. The lineal elements along any vertical line are parallel. In general, in the DF of the DE of
the form y ‘ = f (x), the lineal elements along any vertical line are parallel.
Solution Curves Obtained By Shifting A Copy Of A Known One Vertically
Now refer to Fig. 3.2. The DF is that of y ‘ = 2x – 2 of Example 3.1. Suppose a solution curve C1 that passes thru the point
(-0.5, 1) is known. Let’s shift a copy of C1 vertically up by 1.5 units and label the new curve as C2. Since the lineal elements
along any vertical line are parallel, C2 is approximately parallel to all nearby lineal elements, as each of them is parallel to a
lineal element vertically below it and to which C1 is approximately parallel. As a consequence, C2 is also a solution curve.
Similarly, the curve C3, obtained by shifting a copy of C1
vertically down by 1 unit, is also a solution curve. In general, if a
| Fig. 3.2 Solution Curves Obtained From A Known Solution Curve. |
solution curve of the equation y ‘ = f (x) is known, then every other solution curve can be obtained by just shifting a copy of the
known one vertically up or down.
Recall that the general solution of y‘ = f (x) is the general antiderivative y = F(x) + C of f (x), where F(x) is an antiderivative
of f (x) and C
is an arbitrary constant. So any 2 solution curves differ from each other by a constant. Thus if a solution curve is known, then every other solution curve can be obtained by just shifting a copy of the known one vertically up or down, as also
demonstrated by DFs.
Differential Equations Of The Form y ‘ = f ( y)
Example 3.2
Sketch the DF of the DE y ‘ = y + 1.
Solution
| Fig. 3.3 DF Of y ‘ = y + 1. |
EOS
Lineal Elements Along Any Horizontal Line Are Parallel
In Example 3.2, at every point (x, y), the slope y ‘ = y + 1 depends only on y; it doesn’t depend on x. At any point with
y-coordinate y1, the slope is y1 + 1, whatever x is, so the lineal elements along the horizontal line y = y1 have the same fixed
slope of y1 + 1, and thus are parallel. The lineal elements along any horizontal line are parallel. In general, in the DF of the DE
of the form y ‘ = f ( y), the lineal elements along any horizontal line are parallel.
Solutions Curves Obtained By Shifting A Copy Of A Known One Horizontally
Now refer to Fig. 3.4. The DF is that of y ‘ = y + 1 of Example 3.2. Suppose a solution curve C1 that passes thru the point (2,
1) is known. Let’s shift a copy of C1 horizontally left by 2.75 units and label the new curve as C2. Since the lineal elements
along any horizontal line are parallel, C2 is approximately parallel to all nearby lineal elements, as each of them is parallel to a
lineal element horizontally to the right of it and to which C1 is approximately parallel. As a consequence, C2 is also a solution
curve. Similarly, the curve C4, obtained by shifting a copy of a known solution curve C3 that passes thru the point (-1, -2)
horizontally right by 1.25 units, is also a solution curve. In general, if a solution curve of the equation y ‘ = f ( y
) is known, then arbitrarily many other solutions curves can be obtained by just shifting a copy of the known one horizontally left or right. Note
that
C4 can’t be obtained by shifting a copy of C1 horizontally. Not every solution curve can be obtained by shifting a copy of a
single known solution curve horizontally.
| Fig. 3.4 Solution Curves Obtained From A Known Solution Curve. |
The Differential Equations y ‘ = f (x) And y ‘ = f ( y)
| For the DE of the form y ‘ = f (x), in its DF, the lineal elements along any vertical line are parallel, and if a solution curve is For the DE of the form y ‘ = f ( y), in its DF, the lineal elements along any horizontal line are parallel, and if a solution curve is known, we can obtain arbitrarily many other solution curves by shifting a copy of the known one horizontally right or left. Note that in general not every solution curve can be obtained by shifting a copy of a single known solution curve horizontally. autonomous. |
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Equilibrium Solutions
For convenience a copy of the picture in Fig. 3.4 is re-produced in Fig. 4.1 with the horizontal line y = – 1 becoming solid and
colored dark blue. The DE is y‘ = y + 1.
In Fig. 4.1, the constant function y = -1 is a solution of the DE y‘ = y + 1 (check: y = -1 for all x, y‘ = 0, y + 1 = (-1) + 1
= 0, so y‘ = y + 1). The DE y‘ = y + 1 has infinitely many solutions. So the quantity y = g(x) such that y‘ = y
| Fig. 4.1 The equilibrium solution of y‘ = y + 1 is y = -1. |
many possibilities for its formula g(x).
For convenience, let’s think of x as time. Let’s consider a particular value of x, for example the initial time x = 0. There, if the
value of y is -0.75 then y behaves in the way indicated by the curve C1; if the value of y is -1 then y behaves in the way
indicated by the horizontal line y = -1; if the value of y is -1.75 then y behaves in the way indicated by the curve C4; if the
value of y is 3.25 then y behaves in the way indicated by the curve C2; etc. The overall behavior of any individual y on its
domain depends on its value at a particular time, for example at the initial time x = 0, which is reasonable and believable.
The solution y = -1 is constant. If the quantity y takes on the constant solution y = -1, then of course it’s constant, it doesn’t
change as x changes, it’s in an equilibrium state. Consequently the constant solution y = -1 is called an equilibrium solution
of the DE y‘ = y + 1. If at x = 0 the quantity y takes on the value -1, then it’s an equilibrium solution. Note that of course the
solution “curves” of the equilibrium solutions are horizontal lines.
Finding Equilibrium Solutions
As done above, equilibrium solutions can be found from DFs. If a DE can be solved, it may be possible to obtain its equilibrium
solutions, if they exist, from the general solution. Let’s solve y‘ = y + 1:
y‘ = y + 1,
y‘ – y = 1,
y‘e–x – ye–x = e–x,
( ye–x)’ = e–x,
ye–x = – e–x + C, C = arbitrary constant; this C isn’t related to the curve names Ci‘s in Fig. 4.1,
y = – 1 + Cex,
y = Cex – 1.
We see that we can obtain the equilibrium solution of y‘ = y + 1 from its DF and its general solution. For this particular DE, it’s
possible too to obtain its equilibrium solutions directly from its equation without solving it, as follows. If y = k is an equilibrium
solution, then y‘ = 0, consequently it must be that y + 1 = 0 also, hence k + 1 = 0, it follows that k = -1; we conclude that y =
-1 is the equilibrium solution. In generall, to find equilibrium solutions of y‘ = f( y), we note that if y = k, where k is a constant,
is an equilibrium solution, then y‘ = 0, consequently it must be that f(k) = 0 also, where we substitute y = k. Therefore we
solve the equation f(k) = 0 for k. If it can be solved and if a constant solution exists, say k = k1, then y = k1 is an equilibrium
solution of y‘ = f( y). If there are no constant solutions then there are no equilibrium solutions.
Example 4.1
Find the equilibrium solutions of the following DEs if they exist.
a) y‘ = y2 + 1.
b) y‘ = x + y.
c) y‘ = y3 – 17y2 – 60y.
Solution
a) If y = k is an equilibrium solution then y‘ = 0, then k2 + 1= 0, which is false (in the realm of real numbers). So there are
no equilibrium solutions.
b) If y = k is an equilibrium solution then y‘ = 0, then x + k = 0, then k = –x, which is false because –x isn’t a constant. So
there are no equilibrium solutions.
c) If y = k is an equilibrium solution then y‘ = 0, then k3 – 17k2 – 60k = 0, then k(k2 – 17k – 60) = 0, then k(k + 3)(k – 20) =
0, then k = 0, -3, or 20. So the equilibrium solutions are y = 0, y = -3, and y = 20.
EOS
Remarks 4.1 – On Example 4.1
a) For part c, suppose we use the DF of the DE to find the equilibrium solutions. If the range of the y-axis is from b1 to b2 with
b1 < -3 and 0 < b2 < 20, then we would miss the equilibrium solution y = 20. So we should use the DE itself whenever
possible to find its equilibrium solutions without solving it.
b) For part d, suppose we use the DF of the DE to find the equilibrium solutions. If the y-coordinates of the points where we
draw lineal elements include the numbers with values m + 0.5 where m is any integer, then the equilibrium solutions are
visible, as illustrated in Fig. 4.2. Otherwise, they aren’t, as illustrated in Fig. 4.3, and likely we may be misled to declare that
there are no equilibrium solutions. So, again, we should use the DE itself whenever possible to find its equilibrium solutions
without solving it.
| Fig. 4.2 The equilibrium solutions are visible. |
| Fig. 4.3 The equilibrium solutions aren’t visible. |
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| 5. Matching Differential Equations With Direction Fields |
Example 5.1 – Matching DEs With DFs
Match the DE with its DF labelled a thru d in Fig. 5.1. Give reasons for your answer.
1) y‘ = x(3 – y). [1]
2) y‘ = sin x cos y. [2]
3) y‘ = x + y – 2. [3]
4) y‘ = 3 – y. [4]
| Fig. 5.1 DFs To Match DEs. |
Solution
1) b. The slope of [1] at the origin is (0)(3 – 0) = 0. Only b and d have slope 0 at the origin. The slope of [1] at (1, 1) is (1)(3 –
1) = 2. The slope of b at (1, 1) appears to be 2, while the slope of d at (1, 1) is clearly strictly between 0 and 1. So [1]
matches b.
2) d. The slope of [2] at the origin is sin 0 cos 0 = (0)(1) = 0. Only b and d have slope 0 at the origin. Now b has already been
found to match [1]. So [2] must match d.
3) a. The slope of [3] at the origin is 0 + 0 – 2 = -2 < 0. Only a has a negative slope at the origin. So [3] must match a.
4) c. Equation [4] is different from any other equation. Now a, b, and d have already been matched. So [4] has to match c.
EOS
Remark 5.1 – On Example 5.1
For part 4. The equations y‘ = sin2 x and y‘ = 1 – cos2 x, although looking different, are actually the same, since sin2 x + cos2 x
= 1, and thus sin2 x = 1 – cos2 x, for all real x.
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1. Sketch the DF of the DE y ‘ = x2 – y2. Find equilibrium solutions if they exist.
Solution
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2. Sketch the solution curve of the following initial-value problem without solving the given DE:
Approximately determine from the solution curve its intercepts, asymptotes, end behaviors, intervals of increasing/decreasing,
intervals of concavity, and inflection points.
Solution
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3.
b) Determine all the equilibrium solutions. Are they visible in the given DF? Why or why not?
Solution
Note
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4. Match the DEs with the DFs labelled A thru D. Give reasons for your answer.
a) y‘ = 2x + y – 1.
b) y‘ = (2 – x) y.
c) y‘ = ex/10ey/3.
d) y‘ = 2 – x.
Solution
a) C. The slope of a at the origin is 2(0) + 0 – 1 = -1 < 0. Only C has a negative slope at the origin. Thus a must match C.
b) B. The slope of b at the origin is (2 – 0)(0) = 0. Only B has slope 0 at the origin. Thus b must match B.
c) D. The slope of c at the origin is e0/10e0/3 = 1 > 0. Only A and D have a positive slope at the origin. There the slope of D
appears to be 1 while that of A clearly is > 1. Thus c must match D.
d) A. The equation in d is different from any other equation. Now B, C, and D have already been matched. Thus d must match
A.
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5. The DF of the DE:
together with the solution curve that passes thru the origin (0, 0) is shown. Obtain from this solution curve the solution curve
that passes thru the point (0, 2) and the solution curve that passes thru the point (0, -1). Explain how you obtain these
solution curves.
Solution
The given DE is of the form y ‘ = f (x
). So the lineal elements on any vertical line are parallel. The solution curve that passes thru the point (0, 2) is obtained by shifting a copy of the given solution curve vertically up by 2 units. The solution curve that
passes thru the point (0, -1) is obtained by shifting a copy of the given solution curve vertically down by 1 unit.
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