### Video Transcript

Consider the given slope field
graph. Which of the following differential
equations is represented on the graph? A) π¦ prime is π₯ plus two over π₯
minus three. B) π¦ prime is two minus π₯ over π₯
minus three. C) π¦ prime is π₯ minus two over π₯
plus three. D) π¦ prime is π₯ plus two over
three minus π₯. Or E) π¦ prime is equal to π₯ minus
two over π₯ 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 π₯ is negative two and π₯ is positive three is quite
distinctive. At π₯ is negative two, the slope
field line segments are all horizontal. This means that the slope is equal
to zero. That is, π¦ prime is equal to zero
at π₯ is negative two.

So letβs try π₯ is negative two in
each of our possible equations and see if we get a match. At π₯ is negative two in our
equation A, π¦ prime is equal to negative two plus two over negative two plus
three. Thatβs equal to zero over negative
five, which is equal to zero. So equation A at the point π₯ is
negative two does actually match our slope field. If we look at equation B at π₯ is
negative two, we have π¦ prime is equal to two minus negative two over two minus
three. Thatβs equal to four over negative
five, which is not equal to zero. So equation B does not match our
slope field for π₯ is negative two. And we can eliminate equation
B.

Now letβs try equation C. For equation C, we have π¦ prime is
equal to negative two minus two over negative two plus three. Thatβs equal to negative four over
one, which is not equal to zero. So we can eliminate equation C. At π₯ is negative two in equation
D, we have π¦ prime is negative two plus two over three minus negative two. Thatβs equal to zero over five,
which is equal to zero. Equation D does match the slope
field as π₯ is negative two. So equation D remains a
contender. In equation E, we have π¦ 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.

Weβre left now with equations A and
D as possible options. Letβs now try π₯ is equal to
positive three in each of these equations. At π₯ is equal to positive three,
both equations have a denominator of zero. This means that both equations are
undefined at π₯ is equal to three and have a vertical slope at that point. This corresponds to the slope on
the slope field. So weβll have to look elsewhere,
for a possible solution. Letβs choose another value, say π₯
is equal to zero. At π₯ is equal to zero, each of the
line segments in the slope field has a negative slope. So letβs see what the sign of the
slope is for each of our two remaining differential equations at π₯ is equal to
zero.

In equation A, π¦ 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 π₯
is equal to zero in our slope field. So equation A is still a
contender. In equation D, π¦ prime is equal to
zero plus two over three minus zero. Thatβs equal to two over three,
which is positive. And this does not correspond to the
direction of the slope at π₯ 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 π¦ prime is π₯ plus two over π₯ minus three.

In this example, we were given a
slope field. And we had to fit the appropriate
differential equation to the graph.