### Video Transcript

Initial Value Problems

In this video, weβll practise
finding a specific solution to a separable differential equation when given an
initial value.

To begin, we remind ourselves of
the definition of a separable differential equation. This is any differential equation
that can be written in the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by π of
π¦. Looking at the right-hand side of
this equation, if we cannot express it as a function of only π₯ multiplied by a
function of only π¦, then it is not classed as a separable differential
equation. To find solutions to this type of
equation, we can use the following method.

Here, we divide both sides by the
function π of π¦. To make our notation easier to
follow, we can define a new function β of π¦, which is equal to one over π of
π¦. Also, a quick side note, when π of
π¦ equals zero, weβll have to divide by zero. And hence, the solution will be
lost. But we wonβt be worrying too much
about this during the video.

Okay, our next step will be to
integrate both sides of our equation with respect to π₯. Now, the left-hand side of the
equation might look a little strange to us. But one thing we might recognize is
that the chain rule can be applied to give us that dπ¦ by dπ₯, dπ₯ is equal to
dπ¦. We can use this to give us the
equation that the integral of β of π¦ with respect to π¦ is equal to the integral of
π of π₯ with respect to π₯. As an aside, let us rewind a little
bit. You will often see this method
represented in the following way. At the stage where we had β of π¦
dπ¦ by dπ₯ is equal to π of π₯, we would treat dπ¦ by dπ₯ in a similar fashion to a
fraction, allowing us to give the equivalent statement β of π¦ dπ¦ is equal to π of
π₯ dπ₯.

We would now integrate both
sides. And hopefully, you should be able
to see that we have arrived at the same stage. You should always be aware that
these steps shown on the right should be treated as a shorthand for the method that
we first showed. You often see this because it uses
fewer lines of working. But always be aware of the more
rigorous steps that are actually going on in the background. For example, in our shorthand, we
have ambiguously integrated both sides of our equation. Whereas more rigorously, we are
integrating both sides with respect to π₯. And the result just so happens to
be an integral on the left-hand side with respect to π¦.

Okay, letβs tidy things up and
continue with our method. Integrating β of π¦ with respect to
π¦ and π of π₯ with respect to π₯, we get capital π» of π¦ plus a constant, which
weβll call π one, is equal to capital πΊ of π₯ plus a constant, which weβll call π
two. And this is when capital π» and
capital πΊ are antiderivatives of a lowercase β and lowercase π, respectively. We should now pay attention to
these two π terms, which are constants of integration. Given that these are constants, we
can simplify first by subtracting π one from both sides and then simply by
redefining π two minus π one to be a new constant, which weβll call π three.

At this stage, we would have found
the general solution to our differential equation. It is likely that this will be an
implicit solution. Remember, an explicit solution is
one with π¦ on one side of the equation and some function of π₯ on the other
side. An implicit solution does not take
this form. Both forms of solution are
valid. But we often try to express our
answer in terms of an explicit solution if possible. Now, here, weβve said that the
general solution has been found because the constant π three can take any
value. And all of these will be solutions
to our differential equation.

Letβs imagine the very simple case
where π» of π¦ is just π¦ and πΊ of π₯ is just π₯. Our equation then becomes π¦ equals
π₯ plus our constant π three. Of course, this is just the
equation of a straight line. Since π three can take any value,
if we took it to be equal to zero, weβd have the straight line π¦ equals π₯. When π three is equal to minus
one, weβd have the line π¦ equals π₯ minus one. We could even take something like
π three equals π, which would give us π¦ equals π₯ plus π. All of these lines and their
corresponding equations are specific solutions to some differential equation. We found these using the general
solution to that differential equation. And weβve already shown the method
for finding a general solution to a differential equation.

Okay, so without just arbitrarily
picking numbers for our constant, how do we get from the general solution to the
specific solution? As per the title of this video,
weβll be using an initial value, sometimes called an initial condition. Letβs see an example of this using
our general solution π¦ equals π₯ plus some constant. An example of an initial value that
we might be given would be π¦ of two is equal to one. This tells us that when π₯ is equal
to two, π¦ is equal to one. It, therefore, follows that the
point two, one lies on the line of the correct specific solution that weβre looking
for.

Algebraically, we can find the
equation of the correct specific solution by substituting in the known values into
the general solution. When π₯ equals two, π¦ equals
one. Of course, a simple rearrangement
gives us that the constant π three is equal to negative one. Substituting this back into our
general solution, we get the equation π¦ equals π₯ minus one. A visual representation of what
weβve done here is that we looked at the point two, one. And we found the equation of the
line for which it lies on. This line is the specific solution
which matches the initial value we were given.

Now that we understand our method,
letβs take a look at an example.

Find the equation of the curve
which passes through the point negative eight, one given that the gradient of the
tangent at any point is equal to two times the square of the π¦-coordinate.

Letβs first interpret the
information given by this question. Weβre told about the gradient to
the tangent of a curve at any point. This gradient can be represented by
dπ¦ by dπ₯. Weβre told that this is equal to
two times the square of the π¦-coordinate. So in other words, is equal to two
π¦ squared. Here, we recognize that weβve been
given a differential equation. Not only that, weβve been given a
separable differential equation. This is one that can be expressed
in the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by π of π¦. In our case, weβll be taking π of
π₯ to be equal to the constant two and π of π¦ to be equal to π¦ squared.

Given an equation of this form, we
can find the general solution in the following way. We first divide both sides of our
equation by π of π¦. In our case, this is π¦
squared. This gives us that one over π¦
squared dπ¦ by dπ₯ is equal to two. Next, we can integrate both sides
of our equation with respect to π₯. Now, the chain rule tells us that
dπ¦ by dπ₯ dπ₯ is equal to dπ¦. This gives us that the integral of
one over π¦ squared with respect to π¦ is equal to the integral of two with respect
to π₯. As a side note, you often see
methods where before the integration, dπ¦ by dπ₯ is treated somewhat like a
fraction. Both sides are multiplied by
dπ₯.

And the equivalent statement is
obtained: one over π¦ squared dπ¦ is equal to two dπ₯. Both sides of the equation are then
integrated. And we reach the same state as we
did before, but with less working. Always remember that this is just a
shorthand for the more rigorous steps that weβve shown here. Returning to our method, one over
π¦ squared can be expressed as π¦ to the power of negative two. You might be more comfortable
performing the integration in this form to get negative one over π¦ plus the
constant of integration π one is equal to two π₯ plus some other constant of
integration, π two.

Usually, we donβt worry about
putting a constant on both sides of this equation because we just define a new
constant, which in this case would be π two minus π one. And this could just go on one side
of the equation. Now, at this stage, we have found
the general solution to our differential equation. And itβs in implicit form, which
means itβs not in the form where π¦ is equal to some function of π₯. We found the general solution for
all lines where the gradient of the tangent at any point is equal to two times the
square of the π¦-coordinate.

Now, our next thought might be to
convert our implicit solution into an explicit solution. But as it turns out, this isnβt
necessary. This is because the general
solution is not our final destination. Rather, we are seeking a particular
solution. We want the line which passes
through the point negative eight, one. This information is an initial
value. And weβll be finding the particular
solution which matches in the following way. We know that if negative eight, one
lies on the line, when π₯ is equal to negative eight, π¦ must be equal to one. We can substitute these known
values into the equation for our general solution. We then perform some
simplification. And we find that our constant π
three is equal to 15.

We can now substitute this value
into our general solution which weβll turn it into a particular solution. Negative one over π¦ is equal to
two π₯ plus 15. Now, if we wanted, we could convert
this to an explicit solution first by multiplying by negative one and then raising
both sides to the power of negative one. To recap, we used the information
given in the question to form a separable differential equation. The particular solution to this
which passes through the point negative eight, one is π¦ equals negative one over
two π₯ plus 15. And this is the answer to our
question.

Letβs now take a look at another
example to practise this form of question.

Find the equation of the curve that
passes through the point zero, negative one given dπ¦ by dπ₯ is equal to negative
six π₯ minus four divided by four π¦ plus 13.

For this question, we first noticed
that we have been given a separable differential equation. This is an equation that can be
written in the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by π of π¦. Solving this separable differential
equation will give us a general solution, which will include a constant π. Now, the question has also given us
an initial value, which is the fact that the curve should pass through the point
zero, negative one. Weβll be able to use this
information to narrow down the general solution that we find into a specific
solution. For now, letβs just work on the
general solution.

For now, weβll put aside our
initial value and just work towards the general solution. To find this, weβll first express
our differential equation in the following form, where we have some function of π¦
dπ¦ by dπ₯ is equal to π of π₯. For our equation, we can do this by
multiplying both sides by four π¦ plus 13. This gives us that four π¦ plus 13
dπ¦ by dπ₯ is equal to negative six π₯ minus four. The next thing we can do is to
treat dπ¦ by dπ₯ somewhat like a fraction. If we pretend weβre multiplying
both sides by dπ₯, we obtain an equivalent statement four π¦ plus 13 dπ¦ is equal to
negative six π₯ minus four dπ₯.

The next step we take is to then
integrate both sides of our equation, giving us an integral with respect to π¦ on
the left-hand side and an integral with respect to π₯ on the right-hand side. Now, you should always remember
that what we have done here is essentially a trick. dπ¦ by dπ₯ is not actually a
fraction and cannot always be treated like one. However, weβve done so here to save
us from working since it gives us an equivalent answer. Going back to our method, we now
perform our integrations. Doing so gives us the
following.

Now, itβs worth noting that both of
our integrals would actually have given us a constant. However, we can simply combine
these two constants into one term. Instead of π one and π two, we
simply have a π, which is equal to π one minus π two. Again, this leads to less
unnecessary work for us. After some small simplifications,
we have arrived at the general solution to our differential equation. This means that we have found an
equation which describes all the lines for which dπ¦ by dπ₯ is equal to negative six
π₯ minus four divided by four π¦ plus 13.

However, we need to find which one
of these lines passes through the point that weβve been given: zero, negative
one. This initial value that we have
been given will allow us to find that particular solution. We know that if the line passes
through the point zero, negative one that when π₯ is equal to zero, π¦ must be equal
to negative one. We can, therefore, take our general
solution, substitute in π¦ equals negative one and π₯ equals zero, and solve this to
find the value of our constant π. With a little bit of working, we
find that the value of π here is negative 11. We can now substitute this value of
π back into our general solution to find the specific solution. This is the specific solution to
the differential equation which passes through the point zero, negative one, which
is our initial value.

Now, you might notice that the
solution that we have been given is implicit, which means itβs not in the form π¦
equals some function of only π₯. Usually, itβs a good idea to try
and express your solution in explicit form. However, in this case, you should
not worry. It is not actually possible to give
the answer to this problem in the form of a single explicit equation since we have a
multiple powers of π¦ on the left-hand side. Again, this is no cause for
concern, since the implicit form is a perfectly valid solution. And we have indeed answered our
question.

Let us now take a look at one final
example of an initial value problem.

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.

Okay, with our final example done,
letβs finish off the video by going through some key points. A separable differential equation
can be expressed in the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by π of
π¦. There are also other ways in which
this can be expressed, which may be more helpful depending on the problem. The general solution to a separable
differential equation can be found using integration techniques. Within this method, you often see
dπ¦ by dπ₯ treated somewhat like a fraction. But remember, this is just a trick
which saves your working. The general solution to a
differential equation will include a constant π.

An initial value, for example, π¦
of π is equal to π, can be used to find a particular solution to the differential
equation. This can be done by substituting in
π₯ equals π and π¦ equals π into your general solution and then solving to find
the constant π. Putting this value of π back into
your general solution will give rise to a particular solution, specifically one that
matches the initial condition that you used.

As a final side note, solutions can
either be explicit or implicit. Explicit solutions are of the form
π¦ equals some function of π₯. Implicit solutions are not of this
form. Itβs usually a good idea to try and
give an explicit solution to a problem. But sometimes this is
impossible. So donβt worry too much if you
cannot do this.