### Video Transcript

Find the solution of the
differential equation dπ¦ by dπ₯ plus nine π¦ is equal to 63 given that π¦ nought is
equal to eight.

For this type of question, the
first thing we should do is to check whether we have been given a separable
differential equation. This is a differential equation
that can be written in the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by π of
π¦. An equivalent statement to this
would be one over π of π¦ dπ¦ by dπ₯ is equal to π of π₯. We might also decide to define
another function β of π¦ to make notations slightly easier, where β of π¦ is equal
to one over π of π¦.

Now, here we should be a little bit
careful. Just because our equation has all
of the π¦ terms on the left does not mean itβs already in this form. This is because we have an addition
here. Consider the following side example
of the differential equation dπ¦ by dπ₯ plus π¦ is equal to two π₯. Although it might look like this is
a separable differential equation because we have a π¦ on the left and an π₯ on the
right. In actual fact, no matter how much
we try, we cannot express it in the correct way. This example illustrates that some
equations which look like they should be separable in fact are not. Luckily, we are not dealing with
one of these cases. However, our equation does require
a little bit of manipulation before we proceed.

We take our equation and we
subtract nine π¦ from both sides. We can then factorize the
right-hand side since both terms have a common factor of nine. We then divide both sides by seven
minus π¦. This gives us the following
equation which does indeed match the form that weβve shown here. Okay, weβre now gonna use a little
bit of a trick. We can treat dπ¦ by dπ₯ somewhat
like a fraction. Doing so allows us to reach a
different form of equation: one over seven minus π¦ dπ¦ is equal to nine dπ₯. From this form, we can then
integrate both sides of our equation. This gives us the negative of the
natural logarithm of the absolute value of seven minus π¦ is equal to nine π₯ plus
π, where π is a constant. Of course, both of our integrals
would have given us a constant of integration. However, weβve chosen to combine
both of these on the right-hand side into the constant, which weβve just called
π.

Okay, weβve now found the general
solution to our differential equation. However, this solution is in an
implicit form. An explicit solution would be one
in the form π¦ equals some function of π₯. Weβve chosen to use π’ here to
avoid confusion with this π in our definition. To make things easier to manage,
letβs work towards converting our general implicit solution to a general explicit
solution. The first thing we can do is
multiply both sides by negative one. Since we havenβt yet defined our
constant, it doesnβt matter whether we have a plus π or a minus π. So perhaps, weβll just leave it as
a plus π. Letβs clear some room to
continue.

The next thing we do is take the
exponential of both sides, which means raising π to the power of both sides of our
equation. Since this is the inverse of the
natural algorithm on the left-hand side of the equation, weβre simply left with
seven minus π¦. Next, we can simplify by
subtracting seven from both sides and then multiplying by negative one. This gives us that π¦ is equal to
negative π to the power of negative nine π₯ plus π add seven. At this stage, we might recognize
that π to the power of negative nine π₯ plus π is equal to π to the power of
negative nine π₯ multiplied by π to the power of π. Doing this gives us another way to
simplify.

Together here, we have negative π
to the power of π. But since π is currently an
undefined constant, we can redefine a new constant. Letβs say capital πΆ, which is
equal to negative π to the power of our previous lowercase π. Doing so allows us to say that π¦
is equal to capital πΆ times π to the power of negative nine π₯ plus seven. We have now found a general
implicit solution to our differential equation. At this stage, weβre ready to use
the remaining information given by the question. The question has given us that π¦
of nought is equal to eight this information is our initial value and we need to
find the particular solution to the differential equation for which this is
true.

From this statement, we know that
when π₯ is equal to zero, π¦ must be equal to eight. Another way of saying this would be
that the curve which represents our particular solution will pass through the point
zero, eight. Okay, we can substitute our known
values into the general solution to find the value of capital πΆ for our particular
solution. Since π to the power of nought is
equal to one, we have that eight is equal to πΆ times one add seven. Therefore, our constant capital πΆ
is equal to one. Substituting this value of πΆ back
into our general solution gives us the particular solution to our differential
equation, which matches the initial value given by the question. The equation that represents our
particular solution is that π¦ is equal to π to the power of negative nine π₯ plus
seven.