### Video Transcript

Consider the differential equation
π¦ is equal to one over π₯ dπ¦ by dπ₯ minus three. Part i) Draw a slope field at the
four indicated points in the given figure. Part ii) Let π¦ is equal to π of
π₯ be a particular solution of the differential equation. Write an equation of the tangent to
the graph of π at the point zero, two. Part iii) Find the approximate
value of π of 0.02 using the equation of the tangent line in part two. Part iv) Find π¦ is equal to π of
π₯ with π of zero is equal to two.

For the first part of this
question, weβre required to draw a slope field at the four points in the figure. These points are negative one, one;
negative one, two; zero, two; and one, one. In order to draw the slope field
for each of these points, we need to find the value of dπ¦ by dπ₯ at these
points. Letβs start by writing our
differential equation in terms of dπ¦ by dπ₯. We obtain that dπ¦ by dπ₯ is equal
to π₯ multiplied by π¦ plus three. We start by substituting in π₯
equals negative one and π¦ equals one. We obtain that dπ¦ by dπ₯ is equal
to negative four. We can write this value into a
table.

Next, we can use π₯ is equal to
negative one, π¦ is equal to two. We obtain that dπ¦ by dπ₯ is equal
to negative five. Next, we use π₯ is equal to zero,
π¦ is equal to two, giving dπ¦ by dπ₯ is equal to zero. Finally, we use π₯ is equal to one,
π¦ is equal to one, giving us the result that dπ¦ by dπ₯ is equal to four. Now, we have found the values of
dπ¦ by dπ₯ at these four points. In order to draw the slope field at
these four points, we simply draw a line with the gradient of dπ¦ by dπ₯ at the
points.

The slope field at negative one,
one will have a gradient of negative four. At negative one, two, it will have
a gradient of negative five. At zero, two, its gradient will be
zero. And at one, one, its gradient will
be four. So this completes the first part of
the question.

In part two, weβre told that π is
a particular solution to the differential equation. And weβre required to find the
equation of the tangent to the graph at the point zero, two. Looking at our slope field, we can
see that at the point zero, two, dπ¦ by dπ₯ is equal to zero. Now, our tangent will be a straight
line. So it will be of the form π¦ is
equal to ππ₯ plus π, where π is the gradient of the tangent and π is the π¦-axis
intercept of the tangent. We have just shown that at the
point zero, two, the gradient dπ¦ by dπ₯ is equal to zero. Therefore, we know that π will be
zero in our equation. So our tangent will be of the form
π¦ is equal to π.

Now, weβre finding the tangent of
the graph at the point zero, two. Zero, two is on the π¦-axis. Therefore, zero, two will be the
π¦-axis intercept of our tangent. At this point, the value of π¦ is
two. Therefore, we can conclude that the
equation of our tangent is π¦ is equal to two.

In part three, we need to find the
approximate value of π of 0.02 using the equation of the tangent line in part
two. Now, our tangent in part two was
tangential at the point zero, two. Here, the π₯-value is zero. In part three, weβre asked to
approximate the value of π of 0.02. Here, the π₯-value is 0.02. Since 0.02 is very close to zero,
this means that weβre able to use the tangent from part two in order to estimate the
value of π of 0.02.

In order to do this, we simply find
the π¦-value on our tangent when π₯ is equal to 0.02. Now, the equation of our tangent is
π¦ is equal to two. And itβs not depending on π₯. This means that π¦ hold the value
of two regardless of the π₯-value. Therefore, we can use this tangent
to approximate the value of π of 0.02 as two.

For part four of the question, we
need to find π¦ is equal to π of π₯ with π of nought equal to two. In this part of the question, we
need to find a particular solution. Therefore, weβre going to need to
solve our differential equation. Now, our differential equation is
of the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by β of π¦, where in our case
π of π₯ is equal to π₯ and β of π¦ is equal to π¦ plus three. What this means is we can use the
separation of variables method in order to solve this differential equation.

We can write our differential
equation as dπ¦ over π¦ plus three is equal to π₯ dπ₯. And now, we can integrate. We have that the integral of one
over π¦ plus three with respect to π¦ is equal to the integral of π₯ with respect to
π₯. Weβre able to integrate π₯ with
respect to π₯ quite easily. We simply increase the power of π₯
by one and divide by the new power. This gives us one-half π₯
squared. Next, we need to integrate one over
π¦ plus three with respect to π¦.

Now, we know that the integral of
one over π¦ with respect to π¦ is equal to the natural logarithm of the absolute
value of π¦. Now, we could use a π’ substitution
here in order to show that our integral of one over π¦ plus three with respect to π¦
is equal to the natural logarithm of the absolute value of π¦ plus three. However, this is fairly
straightforward. And so, we can just use this result
here. We obtain that the natural
logarithm of the absolute value of π¦ plus three is equal to one-half π₯
squared.

Now, you may be wondering where the
constants of integration for our indefinite integrals have gone. And thatβs because we havenβt added
them in yet. We add a constant to each side of
the equation: plus π on the left and plus π on the right. Now, we can move this constant both
to the right-hand side. And we obtain a constant on the
right of π minus π. However, since π minus π is also
a constant, we can relabel this as π. Now that we have an equation for π¦
in terms of π₯, letβs make π¦ the subject of this equation.

We can write both sides of the
equation as powers of π. Now that the natural logarithm in
the left-hand side of the equation is the inverse function of π, therefore these
two functions cancel out with one another. Therefore, weβre left with the
absolute value of π¦ plus three is equal to π to the power of a half π₯ squared
plus π. Since we know that π is positive
and it is simply being raised to a power, we know that the right-hand side of our
equation is always positive. Therefore, we can lose the absolute
value signs here, since the right-hand side of our equation is always going to be
greater than zero.

Next, we can split up the
right-hand side of our equation. We write it as π to the power of
one-half π₯ squared times π to the power of π. Now, π to the power of π is
simply another constant since both π and π are constants. Therefore, we can rewrite this as
another constant π. Now, we simply subtract three from
both sides of the equation. And weβre left with π¦ is equal to
ππ to the power of one-half π₯ squared. Now, we have nearly found the
particular solution here. All that remains is to find the
value of our constant π.

Weβll be using information given in
the question. And that is that π of nought is
equal to two. π of nought is equal to two tells
us that at π₯ equals zero, π¦ will be equal to two. Therefore, we substitute π₯ equal
to zero and π¦ equal to two into our equation. We obtain that two is equal to π
multiplied by π to the power of zero. π to the power of zero is simply
one. Therefore, we find that π is equal
to two. From here, we reach our solution
which is that π¦ is equal two π to the power of one-half π₯ squared. Weβve now found solutions to all
four parts of this question.

Part one, we found the slope field
for the four points in the figure. For part two, we found that π¦ is
equal to two. For part three, we found that π of
0.02 is approximately equal to two. And for part four, we found that π¦
is equal to two π to the power of one-half π₯ squared.