Video Transcript
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.
