Video Transcript
In this lesson, we’ll learn how to
draw slope fields, which help visualize the general solution of first-order
differential equations graphically. We’ll determine which equation a
particular slope field represents. And in reverse, we’ll work out
which slope field represents a particular differential equation. But first, let’s clarify some of
the terms we’re using.
Suppose we have an unknown function
𝑦, which is a function of 𝑥. The slope of 𝑦 d𝑦 by d𝑥 is the
derivative of 𝑦 with respect to 𝑥. A differential equation is an
equation containing the derivative of a function. A first-order differential equation
contains only the first derivative. And a differential equation defines
the slope in terms of a function 𝑓. 𝑓 can be a function of 𝑥. 𝑓 can be a function of 𝑦. Or 𝑓 can be a function of 𝑥 and
𝑦. Ideally, given the differential
equation as our starting point, we’d like to be able to solve it. This means find the antiderivative,
that is, the function 𝑦 of 𝑥. But as you probably know, for more
differential equations than not, this is not possible. There are actually very few
differential equations that can be solved exactly.
All is not lost, however, because
what we can do is plot the derivative of the function. That’s the differential equation at
various points in the 𝑥𝑦-plane. This is called a slope field or the
direction field. In this example, d𝑦 by d𝑥 is
equal to two 𝑥, which is a function of 𝑥. Each of the blue lines represents
the slope of a particular solution 𝑦 of 𝑥 at that point. So, for example, when 𝑥 is equal
to approximately 0.5, the slope of the solutions at that point is positive. In fact, we can find the slope for
any solution for 𝑥 is 0.5 by substituting 𝑥 is 0.5, which is a half, into the
differential equation. And we find d𝑦 by d𝑥 is two times
a half, which is equal to one. So that at approximately 𝑥 is
equal to 0.5, for any solution to this differential equation, the slope is one at
that point.
A particular solution to a
differential equation is defined using an initial value. So, for example, in our case, we
have three particular solutions on this graph, where solution A has an initial value
of zero, one. Solution B has an initial value of
zero, zero. And solution C has an initial value
of zero, negative two. In fact, the general solution to
this differential equation is 𝑦 of 𝑥 is equal to 𝑥 squared plus a constant of
integration.
For the particular solution A, the
constant of integration is equal to one. For solution B, the constant of
integration is equal to zero. And for solution C, the constant of
integration is equal to negative two. These are just three possible
solutions to this differential equation. There are actually an infinite
number, each defined by the constant 𝐶. And remember that the slope field
plot gives us some idea of the general behavior of the family of solutions of the
differential equation. With an initial value, we can trace
out a particular solution. And this solution follows the
direction of the slope of the line segments in the slope field.
Let’s look now at how we might
actually draw a slope field. Suppose we have the differential
equation d𝑦 by d𝑥 is equal to 𝑥 minus 𝑦, which is a function of 𝑥 and 𝑦. What this tells us is that, at any
point in the 𝑥𝑦-plane, the slope of the solution of the differential equation at
that point is equal to the 𝑥-coordinate minus the 𝑦-coordinate. And we plot a small line with this
slope at this point in the 𝑥𝑦-plane. Let’s do this for some example
points using our differential equation. If we take 𝑥 is zero and 𝑦 is one
as our first point, the slope d𝑦 by d𝑥 is equal to 𝑥 minus 𝑦, which is zero
minus one. And that’s equal to negative
one. So the slope at zero, one is
negative one.
Let’s start putting our points on
the graph so we can see how this looks in the slope field. At the point zero, one, our
solution has a slope of negative one. Let’s choose as our next point 𝑥
is zero and 𝑦 is zero. In this case, d𝑦 by d𝑥 is equal
to zero minus zero, which is equal to zero. So at this point, we have a
horizontal line which has a slope of zero. Choosing one, zero as our next
point, the slope is one minus zero, which is one. At the point one, zero therefore,
we have the positive slope of one. If we continue in this way, the
point zero, negative one has a slope of one. The point negative one, zero has a
slope of negative one. The point one, one has a slope of
zero. And the point negative one,
negative one also has a slope of zero. If we continue in this way for many
more points in the 𝑥𝑦-plane, we get the slope field. And this slope field represents the
differential equation d𝑦 by d𝑥 is equal to 𝑥 minus 𝑦.
The solution passing through a
particular point on the slope field will follow the direction of these small line
segments. We say that this is the solution
graph with initial condition 𝑥 zero, 𝑦 zero, where in this case 𝑥 zero, 𝑦 zero
is equal to negative one, negative one. Something to notice that a lower
solution graph passing through any of the plotted line segments will have the same
slope as the segment at that point. A solution might not follow only
the plotted segments. It’s likely to weave between the
segments. Remember that the plot is just the
derivative at a selection of points. So a slope field is a way to
represent an infinite number of specific solutions. And if we know that the one we’re
looking for goes through a particular point, then following the slope lines from
that point traces out that particular solution.
Let’s look at an example.
Consider the given slope field
graph representing a differential equation. If the solution of the differential
equation contains point 𝑆, which point can also belong to the solution?
Let’s trace solution curves through
each of the points in the direction of the pattern of slopes mapped out through the
line segments. Starting at the point in question,
𝑆, we can see that the solution curve could go through the point 𝐶. Although 𝐶 doesn’t lie exactly on
the line segment, it is within the pattern of the slope field marked out by a
solution through the point 𝑆. So point 𝐶 is a contender for the
solution through 𝑆. Let’s also check the points 𝐴, 𝐵,
𝐷, and 𝐸. Following the pattern of the slope
field through the point 𝐴, the solution does not coincide with the point 𝑆. So the point 𝐴 cannot belong to
the same solution as 𝑆. Following the pattern through the
point 𝐵, again this solution does not coincide with the point 𝑆. The same applies for point 𝐷 and
also for point 𝐸, so that only point 𝐶 could belong to a solution containing the
point 𝑆.
We’ve seen how to draw a slope
field and how to trace solutions through a point. Let’s look now at an example where
we’re given the slope field and we want to find which differential equation slope
field represents.
Consider the given slope field
graph. Which of the following differential
equations is represented on the graph? A) 𝑦 prime is 𝑥 plus two over 𝑥
minus three. B) 𝑦 prime is two minus 𝑥 over 𝑥
minus three. C) 𝑦 prime is 𝑥 minus two over 𝑥
plus three. D) 𝑦 prime is 𝑥 plus two over
three minus 𝑥. Or E) 𝑦 prime is equal to 𝑥 minus
two over 𝑥 minus three.
We can see that each of the options
has a positive or negative two in the numerator and a positive or negative three in
the denominator. And if we look at our graph, the
behavior of the slope at 𝑥 is negative two and 𝑥 is positive three is quite
distinctive. At 𝑥 is negative two, the slope
field line segments are all horizontal. This means that the slope is equal
to zero. That is, 𝑦 prime is equal to zero
at 𝑥 is negative two.
So let’s try 𝑥 is negative two in
each of our possible equations and see if we get a match. At 𝑥 is negative two in our
equation A, 𝑦 prime is equal to negative two plus two over negative two plus
three. That’s equal to zero over negative
five, which is equal to zero. So equation A at the point 𝑥 is
negative two does actually match our slope field. If we look at equation B at 𝑥 is
negative two, we have 𝑦 prime is equal to two minus negative two over two minus
three. That’s equal to four over negative
five, which is not equal to zero. So equation B does not match our
slope field for 𝑥 is negative two. And we can eliminate equation
B.
Now let’s try equation C. For equation C, we have 𝑦 prime is
equal to negative two minus two over negative two plus three. That’s equal to negative four over
one, which is not equal to zero. So we can eliminate equation C. At 𝑥 is negative two in equation
D, we have 𝑦 prime is negative two plus two over three minus negative two. That’s equal to zero over five,
which is equal to zero. Equation D does match the slope
field as 𝑥 is negative two. So equation D remains a
contender. In equation E, we have 𝑦 prime is
equal to negative two minus two over negative two minus three. That gives us negative four over
negative five, which is four over five. This is not equal to zero. So it does not match the slope
field. And we can eliminate equation
E.
We’re left now with equations A and
D as possible options. Let’s now try 𝑥 is equal to
positive three in each of these equations. At 𝑥 is equal to positive three,
both equations have a denominator of zero. This means that both equations are
undefined at 𝑥 is equal to three and have a vertical slope at that point. This corresponds to the slope on
the slope field. So we’ll have to look elsewhere,
for a possible solution. Let’s choose another value, say 𝑥
is equal to zero. At 𝑥 is equal to zero, each of the
line segments in the slope field has a negative slope. So let’s see what the sign of the
slope is for each of our two remaining differential equations at 𝑥 is equal to
zero.
In equation A, 𝑦 prime is equal to
zero plus two over zero minus three, which is equal to two over negative three and
is less than zero. This corresponds to the slope at 𝑥
is equal to zero in our slope field. So equation A is still a
contender. In equation D, 𝑦 prime is equal to
zero plus two over three minus zero. That’s equal to two over three,
which is positive. And this does not correspond to the
direction of the slope at 𝑥 is equal to zero in our slope field. So we can eliminate equation D. Only equation A remains. So the slope field graph represents
the differential equation 𝑦 prime is 𝑥 plus two over 𝑥 minus three.
In this example, we were given a
slope field. And we had to fit the appropriate
differential equation to the graph.
In our next example, we start with
a differential equation and need to find which of the given graphs matches the
differential equation.
Which of the following is the slope
field of the differential equation 𝑦 prime is equal to two 𝑥 plus three 𝑦 minus
five.
The way we’re going to approach
this is by a process of elimination. First, we’ll find a point on each
graph, where 𝑦 prime, the slope, is equal to zero. That’s a flat line segment. If our differential equation does
not equal to zero at such a point, then we can eliminate the associated graph. If we start with graph A, d𝑦 by
d𝑥 is equal to zero at the point one, negative one. So the slope in graph A is equal to
zero at the point one, negative one. Now, if we substitute 𝑥 is equal
to one and 𝑦 is negative one in our differential equation, we have 𝑦 prime is
equal to two times one plus three times negative one minus five. That’s equal to two minus three
minus five, which is negative six. This is nonzero. So straight away, we can eliminate
graph A.
Now let’s look at graph B. Graph B has a slope of zero when 𝑥
is negative one and 𝑦 is negative one. If we try this in our equation, we
have 𝑦 prime is equal to two times negative one plus three times negative one minus
five. That’s equal to negative two minus
three minus five, which is equal to negative 10. This is nonzero, so we can now
eliminate graph B. Now let’s look at graph C. One of the points in graph C, where
d𝑦 by d𝑥 is equal to zero, is the point three, two. With 𝑥 is three and 𝑦 is two in
our equation, we have 𝑦 prime is two times three plus three times two minus
five. That’s equal to seven, which is
nonzero. And so we can eliminate equation
C. Remember, we’re trying to find a
graph with the point where the slope is equal to zero and where that point makes our
differential equation also equal to zero.
Let’s look at graph D. The slope is equal to zero in graph
D at the point one, one. Substituting 𝑥 equal to one and 𝑦
equal to one into our differential equation, we have 𝑦 prime is equal to two times
one plus three times one minus five. That’s equal to two plus three
minus five, which is equal to zero. This matches with the slope on the
graph at the point one, one. So graph D is a contender. Now, let’s look at graph E. We can see that, in graph E, at the
point one, one, we have a slope of zero. We know already from graph D that
the point one, one satisfies our differential equation. So let’s try another point on graph
E which has a slope of zero. The slope is equal to zero in graph
E at the point four, minus one. So let’s try this in our
differential equation. We have 𝑦 prime is two times four
plus three times negative one minus five. That’s eight minus three minus
five, which is equal to zero. So far then, graph E also matches
our differential equation.
We now have two contenders, graph D
and graph E. So let’s look again at graph D. In graph D, the slope is equal to
zero at the point one, one. And this matches with our
differential equation. The slope is also equal to zero at
the point three, negative two. So let’s try this point in our
differential equation. We have 𝑦 prime is two times three
plus three times negative two minus five. That gives us six minus six minus
five. And that’s equal to negative five,
which is not zero. So we can now eliminate graph
D. The only graph out of the five that
could possibly match our differential equation now is graph E.
Let’s just try another couple of
points in graph E with slope of zero and see if they match our differential
equation. The slope on the graph of zero at
the point four, negative one and in fact 𝑦 prime in the differential equation is
also equal to zero. The slope on the graph is equal to
zero at the point negative two, three. And 𝑦 prime is also equal to zero
at the point negative two, three. Out of our five graphs then, we can
say that graph E is the only graph that matches the differential equation 𝑦 prime
is two 𝑥 plus three 𝑦 minus five.
Let’s just summarize now what we’ve
seen. For a first-order differential
equation of the form d𝑦 by d𝑥 is equal to 𝑓 of 𝑥, 𝑦 or d𝑦 by d𝑥 is equal to
𝑓 of 𝑥, or d𝑦 by d𝑥 is equal to 𝑓 of 𝑦, we can visualize a general solution
without actually solving for the function 𝑦. We can do this by drawing a slope
field. To do this, we draw our line
segments, each with the slope d𝑦 by d𝑥 at various points 𝑥, 𝑦 in the
𝑥𝑦-plane. From this, we can get an idea of
the pattern of the general solution of the differential equation. And with an initial condition,
which is a starting point 𝑥 zero, 𝑦 zero, we can trace a particular solution
through the slope field following the direction of the slope of the line
segments.
