Chapter 8 · Section 8.1

8.1 Graphical Methods

Chapters 2 and 4 gave us recipes that turn special equations into formulas. Most equations, however, are not special. For $y^{'} = x^{2} + sin y$ or $y^{'} = e^{- x y}$ no recipe applies, and no formula built from familiar functions exists. Yet an engineer still needs to know how the solution behaves: does it grow, settle down, oscillate, blow up?

This chapter is about getting useful answers anyway. We start with the cheapest tool of all: a picture. The equation $y^{'} = f (x, y)$ tells us the slope of the solution curve at every point. If we draw those slopes, the solution curves almost draw themselves. No integration is needed, only evaluation of $f$ .

Line elements and direction fields

We met direction fields briefly in Section 1.2. Here we make the idea precise and learn to draw them efficiently.

Definition 8.1.1 (Line element, direction field)

Let $f (x, y)$ be defined on a region $D$ of the plane. At a point $(x_{0}, y_{0})$ of $D$ , the line element of $y^{'} = f (x, y)$ is a short line segment through $(x_{0}, y_{0})$ with slope $f (x_{0}, y_{0})$ . The collection of all line elements is the direction field (or slope field) of the equation. A picture of line elements at many points is a line element diagram.

Why is this useful? If $y = φ (x)$ is a solution whose graph passes through $(x_{0}, y_{0})$ , then $φ^{'} (x_{0}) = f (x_{0}, φ (x_{0})) = f (x_{0}, y_{0})$ . So the line element at $(x_{0}, y_{0})$ is a small piece of the tangent line to that solution curve. A dense set of line elements is a dense set of tangent pieces, and the solution curves are the curves that fit them.

The angle of a line element with the positive $x$ -axis is its inclination $α = arctan f (x_{0}, y_{0})$ , with $- 9 0^{\circ} < α < 9 0^{\circ}$ . Positive slopes rise to the right; negative slopes fall.

Example 8.1.1 (Computing line elements)

For $y^{'} = x^{2} - y$ , find the slope and the inclination of the line element at the points $(0, 0)$ , $(0, 1)$ , $(1, 0)$ , $(1, 1)$ , $(1, 2)$ , $(- 1, 0)$ , $(- 1, 2)$ , $(2, 1)$ and $(2, 0)$ .

Solution

Evaluate $f (x, y) = x^{2} - y$ . For example, at $(2, 1)$ the slope is $2^{2} - 1 = 3$ , and at $(0, 1)$ it is $0 - 1 = - 1$ .

Convert to angles. Use

α = arctan (slope)

, rounded to

0. 1^{\circ}

point	slope $x^{2} - y$	inclination
$(0, 0)$ , $(1, 1)$	$0$	$0^{\circ}$ (horizontal)
$(1, 0)$ , $(- 1, 0)$	$1$	$4 5^{\circ}$
$(0, 1)$ , $(1, 2)$ , $(- 1, 2)$	$- 1$	$- 4 5^{\circ}$
$(2, 1)$	$3$	$71. 6^{\circ}$
$(2, 0)$	$4$	$76. 0^{\circ}$

Observe. Several points share a slope. The points with slope $0$ , namely $(0, 0)$ and $(1, 1)$ , both lie on the parabola $y = x^{2}$ ; the three points with slope $- 1$ all lie on $y = x^{2} + 1$ . This is no accident, and it is the key to the faster method below.

Drawing a direction field point by point is slow: each line element costs one evaluation of $f$ and one careful angle. A computer does this in a blink, which is why direction field plotters are standard tools. By hand, we need a smarter strategy.

The method of isoclines

Example 8.1.1 suggests the trick: instead of choosing points first and computing slopes, choose a slope first and find all the points that have it.

Definition 8.1.2 (Isocline)

An isocline of $y^{'} = f (x, y)$ is a curve along which $f$ is constant. For each number $c$ , the isocline of slope $c$ is the level curve $f (x, y) = c .$ Every line element on it has the same slope $c$ .

The name means "equal inclination". Once an isocline is drawn, we can put many parallel line elements on it with a single setting of the ruler. That is the whole method.

Method in a nutshell: sketching with isoclines

Write down the isoclines $f (x, y) = c$ and identify their shape (lines, parabolas, circles, ...).
Draw a handful of them lightly, for well-spread values of $c$ (for instance $c = - 2, - 1, 0, 1, 2$ ). Always include $c = 0$ .
On each isocline, draw several short segments of slope $c$ .
Sketch smooth curves that cross every isocline with the slope of that isocline. These are approximate integral curves.

Common mistake

An isocline is not a solution curve in general. It is a curve along which the solutions all have the same slope; solutions cross it with that slope. A solution curve follows an isocline only in the special case where the isocline itself has slope $c$ at each of its points (Example 8.1.5 shows one).

Example 8.1.2 (Parabolic isoclines)

Use isoclines to sketch the integral curves of $y^{'} = x^{2} - y$ . Then find the solution curve that is itself a parabola.

Solution

Isoclines. Setting $x^{2} - y = c$ gives $y = x^{2} - c .$ These are copies of the parabola $y = x^{2}$ , shifted down by $c$ . The isocline of slope $c = 2$ is $y = x^{2} - 2$ ; that of slope $c = - 1$ is $y = x^{2} + 1$ .
Line elements. On $y = x^{2} - 2$ draw short segments rising at $arctan 2 \approx 63. 4^{\circ}$ ; on $y = x^{2}$ draw horizontal segments; on $y = x^{2} + 1$ draw segments falling at $4 5^{\circ}$ ; and so on. Higher parabolas carry more negative slopes.
Sketch. A curve starting high on the left falls (it is above the zero isocline, where slopes are negative), crosses the parabola $y = x^{2}$ horizontally, then rises. Curves starting low rise steeply at first. Far to the right, all of them seem to crowd together along one curve (Figure 8.1.1).
A parabola solution. The picture suggests that the curves approach a parabola. Try $y = x^{2} + b x + d$ . Then $y^{'} = 2 x + b$ , while $x^{2} - y = x^{2} - (x^{2} + b x + d) = - b x - d .$ Equality for all $x$ needs $2 = - b$ and $b = - d$ , so $b = - 2$ and $d = 2$ : $y = x^{2} - 2 x + 2.$
Check against the exact solution. The equation is linear, $y^{'} + y = x^{2}$ . With the integrating factor $e^{x}$ (Section 2.3) we get $(y e^{x})^{'} = x^{2} e^{x}$ , so $y = x^{2} - 2 x + 2 + C e^{- x} .$ Since $C e^{- x} \to 0$ as $x \to \infty$ , every solution approaches the parabola, exactly as the sketch predicted.

Figure 8.1.1. Direction field of

y^{'} = x^{2} - y

. The dashed parabolas are isoclines

y = x^{2} - c

; every solution crosses the isocline labelled

c

with slope

c

. All solutions approach

y = x^{2} - 2 x + 2

(the thick curve) as

x

grows.

The zero isocline, extrema and inflection points

Two curves deserve special attention, because they tell us where solutions turn around and where they change their bending.

The zero isocline

The isocline $f (x, y) = 0$ is called the zero isocline (or nullcline). Solution curves cross it horizontally. On one side of it $f > 0$ and solutions rise; on the other side $f < 0$ and solutions fall. So a solution can have a local maximum or minimum only on the zero isocline.

The second derivative without solving

To tell a maximum from a minimum, and to locate inflection points, we need $y^{''}$ . We can get it straight from the equation. Along a solution, $y$ is a function of $x$ , so by the chain rule

y^{''} = \frac{d}{d x} f (x, y (x)) = f_{x} (x, y) + f_{y} (x, y) y^{'} = f_{x} (x, y) + f_{y} (x, y) f (x, y) .

The last expression is again a function of $x$ and $y$ only. So we know the sign of $y^{''}$ at every point of the plane, before solving anything.

Theorem 8.1.1 (Concavity from the equation)

Let $f$ , $f_{x}$ and $f_{y}$ be continuous on a region $D$ , and let $y = φ (x)$ be a solution of $y^{'} = f (x, y)$ whose graph lies in $D$ . Then at each point of the graph $φ^{''} (x) = f_{x} (x, φ (x)) + f (x, φ (x)) f_{y} (x, φ (x)) .$ In particular, the solution is concave up where $f_{x} + f f_{y} > 0$ and concave down where $f_{x} + f f_{y} < 0$ . Inflection points can occur only on the curve $f_{x} + f f_{y} = 0$ , which we call the inflection curve.

Example 8.1.3 (Turning points and inflection points)

For $y^{'} = x^{2} - y$ , find the zero isocline and the inflection curve. Decide where solutions have maxima and where they have minima.

Solution

Zero isocline. $x^{2} - y = 0$ , that is, the parabola $y = x^{2}$ . Above it $y^{'} < 0$ (solutions fall); below it $y^{'} > 0$ (solutions rise).
Second derivative. Here $f_{x} = 2 x$ and $f_{y} = - 1$ , so $y^{''} = f_{x} + f f_{y} = 2 x + (x^{2} - y) (- 1) = y - x^{2} + 2 x .$
Maxima or minima? On the zero isocline $y = x^{2}$ , so $y^{''} = x^{2} - x^{2} + 2 x = 2 x$ . A solution crossing $y = x^{2}$ at some $x > 0$ has $y^{'} = 0$ and $y^{''} > 0$ there: a local minimum. A crossing at $x < 0$ gives $y^{''} < 0$ : a local maximum.
Inflection curve. $y^{''} = 0$ means $y = x^{2} - 2 x$ . Solutions are concave up above this parabola and concave down below it.
Where the two curves meet. $x^{2} = x^{2} - 2 x$ only at $x = 0$ , the origin. The solution through the origin is horizontal and inflecting there: it has a horizontal inflection point, not an extremum. (From the formula of Example 8.1.2 this is the member $C = - 2$ ; you can check $y^{'} (0) = y^{''} (0) = 0$ and $y^{'''} (0) = 2 \neq = 0$ .)
A finer count, from the formula. For $y = x^{2} - 2 x + 2 + C e^{- x}$ we have $y^{'} = 0$ exactly when $(2 x - 2) e^{x} = C$ . The left side has its smallest value $- 2$ at $x = 0$ . So members with $C > 0$ have one minimum (at some $x > 1$ ), members with $- 2 < C < 0$ have a maximum at some $x < 0$ and a minimum at some $0 < x < 1$ , and members with $C < - 2$ never turn. The picture alone already told us the qualitative story; the formula confirms it.

Figure 8.1.2. Solutions of

y^{'} = x^{2} - y

for

C = - 3, - 2, - 1, 1

. Circles: extrema, all on the zero isocline

y = x^{2}

(maxima for

x < 0

, minima for

x > 0

). Squares: inflection points, all on

y = x^{2} - 2 x

. The two curves meet at the origin.

Now explore on your own. The plotter below draws the direction field of $y^{'} = x^{2} - y$ ; click anywhere to trace the solution through that point. Then type another right-hand side, such as $f = y - x^{2}$ or $f = (y - 1) (3 - y)$ , and compare with the sketches you predict from the isoclines.

Two families with simple isoclines

Autonomous equations: horizontal isoclines

When $f$ does not depend on $x$ , the equation $y^{'} = g (y)$ is called autonomous. Its isoclines $g (y) = c$ are horizontal lines. The zero isocline consists of the horizontal lines $y = y^{*}$ with $g (y^{*}) = 0$ . On such a line, the line elements are horizontal and lie along the line itself, so each root of $g$ gives a constant solution $y \equiv y^{*}$ , called an equilibrium solution. By Theorem 8.1.1 with $f_{x} = 0$ , the inflection curve is where $g^{'} (y) g (y) = 0$ .

Example 8.1.4 (An autonomous equation)

Sketch the integral curves of $y^{'} = (y - 1) (3 - y)$ . Identify the equilibrium solutions and the level at which non-constant solutions inflect.

Solution

Equilibria. $(y - 1) (3 - y) = 0$ gives $y = 1$ and $y = 3$ . Both constant functions are solutions.
Sign of the slope. Test one value in each band: $y = 0 y = 2 y = 4 : (- 1) (3) = - 3 < 0, : (1) (1) = 1 > 0, : (3) (- 1) = - 3 < 0.$ So solutions between the equilibria rise, and solutions outside them fall.
Isoclines. $(y - 1) (3 - y) = c$ is a quadratic in $y$ with discriminant $4 - 4 c$ . For $c < 1$ it gives two horizontal lines placed symmetrically about $y = 2$ ; for $c = 1$ only the line $y = 2$ . The largest possible slope is $1$ , reached on $y = 2$ .
Concavity. With $g (y) = (y - 1) (3 - y) = - y^{2} + 4 y - 3$ we have $g^{'} (y) = 4 - 2 y$ , so $y^{''} = g^{'} (y) g (y) = (4 - 2 y) (y - 1) (3 - y) .$ Between the equilibria, $y^{''}$ changes sign at $y = 2$ : rising solutions are concave up below $y = 2$ and concave down above it. Every such solution has an S-shape with its inflection point on the line $y = 2$ .
Long-term behaviour. Solutions starting between $1$ and $3$ climb towards $y = 3$ ; solutions above $3$ fall towards it. So $y = 3$ attracts its neighbours (it is stable). Solutions near $y = 1$ move away from it (it is unstable); those below $1$ fall without bound.
Optional check. Separating variables and using partial fractions gives $\frac{y - 1}{3 - y} = K e^{2 x}$ , that is, $y = \frac{1 + 3 K e ^{2 x}}{1 + K e ^{2 x}} .$ For $K > 0$ this increases from $1$ (as $x \to - \infty$ ) to $3$ (as $x \to \infty$ ), in agreement with the sketch.

Figure 8.1.3.

y^{'} = (y - 1) (3 - y)

. The isoclines are horizontal, so the field looks the same at every

x

and each solution is a horizontal shift of another. Solutions approach the stable equilibrium

y = 3

and leave the unstable one

y = 1

; S-shaped curves inflect on

y = 2

Why the field does not change with x

For an autonomous equation, if $y = φ (x)$ is a solution then so is $y = φ (x - a)$ for every constant $a$ : shifting a curve sideways does not change its slopes at given heights. That is why the curves in Figure 8.1.3 are copies of one another. Population models, chemical reactions and cooling laws are often autonomous, so this picture appears again and again in applications.

Isoclines through the origin

When $f$ depends only on the ratio $y / x$ , the isoclines $f = c$ become $y / x = constant$ : straight lines through the origin. Equations of this type were solved in Section 2.2 with $y = vx$ ; graphically, they are just as pleasant.

Example 8.1.5 (An isocline that is a solution)

Sketch the integral curves of $y^{'} = 1 + \frac{2 y}{x}$ for $x \neq = 0$ . Which isocline is itself an integral curve?

Solution

Isoclines. $1 + 2 y / x = c$ gives $y = \frac{c - 1}{2} x,$ a line through the origin (the origin itself excluded, since $f$ is undefined at $x = 0$ ). For instance, $c = 0$ gives $y = - x /2$ , and $c = 5$ gives $y = 2 x$ .
When is a line a solution? The line $y = m x$ has slope $m$ . It lies on the isocline with $c = 1 + 2 m$ . It is an integral curve exactly when its own slope matches the slope of the line elements on it: $m = 1 + 2 m ⟹ m = - 1.$ So $y = - x$ (the isocline $c = - 1$ ) is an integral curve on each half-line $x > 0$ and $x < 0$ . Check: $y^{'} = - 1$ and $1 + 2 (- x) / x = - 1$ . $✓$
The other curves. The equation is linear: $y^{'} - \frac{2}{x} y = 1$ . With the integrating factor $x^{- 2}$ we get $(y / x^{2})^{'} = 1/ x^{2}$ , so $y = - x + C x^{2} .$ Each of these parabolas has slope $- 1$ at the origin: they all touch the line $y = - x$ there, as Figure 8.1.4 shows.
Comment. Infinitely many integral curves approach the origin. This does not contradict the uniqueness theorem of Section 1.3, because $f$ is not even defined on the line $x = 0$ .

Figure 8.1.4. For

y^{'} = 1 + 2 y / x

the isoclines (dashed) are lines through the origin. Only the isocline

c = - 1

, the line

y = - x

, is also an integral curve. The parabolas

y = - x + C x^{2}

are all tangent to it at the origin.

What a picture can and cannot tell us

Graphical methods work for any first-order equation $y^{'} = f (x, y)$ , linear or not, as long as we can evaluate $f$ . In a few minutes they show the qualitative behaviour: equilibria and their stability, turning points, concavity, how solutions behave far away. For many engineering questions ("does the temperature settle?", "can the level overshoot?") that is already the answer.

But a sketch has clear limits:

It gives no formula, and no reliable numbers. Reading $y (1.5)$ off a hand sketch may be wrong in the first decimal.
Its accuracy is unknown. Small drawing errors can accumulate along a long curve.
A careful sketch by hand is slow, and it becomes misleading near points where $f$ is undefined or very large.

The rest of the chapter replaces the picture by computations whose accuracy we can control: power series (Section 8.2), successive approximations (Section 8.3), and numerical step-by-step methods (Sections 8.4 to 8.8). Keep the direction field in mind, though. It is the best way to check whether a computed answer is believable.

Exercises

Find the isoclines of $y^{'} = 3 x - y$ . Which isocline is also an integral curve? Sketch a few solutions.
Answer
Isoclines: parallel lines $y = 3 x - c$ of slope $3$ . The isocline $c = 3$ , the line $y = 3 x - 3$ , is an integral curve. (Exact: $y = 3 x - 3 + C e^{- x}$ ; all solutions approach this line.)
For $y^{'} = y - x^{2}$ , find the zero isocline, the inflection curve, and the one solution that is a parabola. Do the other solutions approach it as $x \to \infty$ ?
Answer
Zero isocline $y = x^{2}$ ; $y^{''} = y - x^{2} - 2 x$ , so the inflection curve is $y = x^{2} + 2 x$ ; parabola solution $y = x^{2} + 2 x + 2$ . No: $y = x^{2} + 2 x + 2 + C e^{x}$ , so the others move away from it.
Describe the isoclines of $y^{'} = x^{2} + 4 y^{2}$ . Where is the slope zero? What can you say about every solution?
Answer
For $c > 0$ the isoclines are the ellipses $x^{2} + 4 y^{2} = c$ ; the slope is $0$ only at the origin. Since $y^{'} \geq 0$ everywhere, every solution is nondecreasing, and solutions get steeper the farther they are from the origin.
For the autonomous equation $y^{'} = y (2 - y)$ , find the equilibrium solutions, the bands where solutions increase or decrease, and the level where they inflect.
Answer
Equilibria $y = 0$ (unstable) and $y = 2$ (stable). Increasing for $0 < y < 2$ , decreasing for $y < 0$ and $y > 2$ . Rising solutions inflect on $y = 1$ .
For $y^{'} = cos y$ , find all equilibrium solutions and the inflection levels. What happens to the solution with $y (0) = 0$ ?
Answer
Equilibria $y = \frac{π}{2} + kπ$ ( $k$ an integer). $y^{''} = - sin y cos y$ , so non-constant solutions inflect where $sin y = 0$ , that is on $y = kπ$ . The solution with $y (0) = 0$ increases and approaches $π /2$ as $x \to \infty$ (and $- π /2$ as $x \to - \infty$ ), inflecting at the origin.
Find the isoclines of $y^{'} = - \frac{2 y}{x}$ ( $x \neq = 0$ ). Which of them are integral curves? Verify that $y = C / x^{2}$ are solutions and sketch them.
Answer
Lines $y = - \frac{c}{2} x$ through the origin. A line $y = m x$ is an integral curve only if $m = - 2 m$ , i.e. $m = 0$ : the $x$ -axis ( $y \equiv 0$ ). $y = C / x^{2}$ gives $y^{'} = - 2 C / x^{3} = - 2 y / x$ .
Show that the isocline of slope $1$ of $y^{'} = x - y + 1$ is an integral curve, and explain from the direction field why all solutions approach it as $x \to \infty$ .
Answer
Isocline $c = 1$ : $y = x$ , which has slope $1$ , so it is a solution. Above it $y^{'} < 1$ and below it $y^{'} > 1$ , so curves are pushed towards it. Exact: $y = x + C e^{- x}$ .
For $y^{'} = x^{2} + y$ , find the zero isocline and the inflection curve.
Answer
Zero isocline $y = - x^{2}$ ; $y^{''} = 2 x + x^{2} + y$ , so the inflection curve is $y = - x^{2} - 2 x$ .
Describe the isoclines of $y^{'} = 4 - x^{2} - y^{2}$ . In which region do solutions rise, and where do they fall? Where is the slope largest?
Answer
Circles $x^{2} + y^{2} = 4 - c$ centred at the origin ( $c < 4$ ). Solutions rise inside the circle of radius $2$ (the zero isocline) and fall outside it. The largest slope, $4$ , occurs at the origin.
Find the slope and inclination (to $0. 1^{\circ}$ ) of the line elements of $y^{'} = x y - 1$ at $(2, 1)$ , $(0, 3)$ and $(1, - 1)$ .
Answer
Slopes $1$ , $- 1$ , $- 2$ ; inclinations $4 5^{\circ}$ , $- 4 5^{\circ}$ , $- 63. 4^{\circ}$ .
Explain why two different integral curves of $y^{'} = x^{2} - y$ can never cross each other.
Answer
$f = x^{2} - y$ and $f_{y} = - 1$ are continuous everywhere, so by the existence–uniqueness theorem (Section 1.3) exactly one solution passes through each point. Two crossing curves would give two solutions through the crossing point.
For $y^{'} = (y - 1) (3 - y)$ , find the solution with $y (0) = 2$ (use the formula of Example 8.1.4). Find its limits as $x \to \pm \infty$ and its inflection point.
Answer
$K = 1$ : $y = \frac{1 + 3 e ^{2 x}}{1 + e ^{2 x}}$ . Limits $3$ (as $x \to \infty$ ) and $1$ (as $x \to - \infty$ ); inflection point $(0, 2)$ .