### Video Transcript

In this video, weβll learn how to
use integration to find particular solutions to initial value problems involving
differential equations of the form dπ¦ by dπ₯ equals π of π₯. A differential equation is an
equation that contains derivatives or differentials. These differentials are of the form
π¦ prime equals π of π₯ or dπ¦ by dπ₯ equals some function in π₯. Weβll begin by looking at how
integrals can help us to find a general solution to these equations before
considering what the introduction of an initial value does to these solutions.

Letβs begin by recapping how
integration can help us with differential equations. Here is a simple differential
equation. dπ¦ by dπ₯ equals three π₯ squared plus four. In other words, thereβs some
function π¦ in terms of π₯ that when we differentiate it with respect to π₯, we end
up with its derivative being three π₯ squared plus four. But then, we recall that
integration and differentiation are the reverse processes of one another. In other words, if we integrate
this function, its derivative with respect to π₯, weβll end up with the general
solution to the differential equation. We obtain the original function π¦
involving some constant of integration π.

In this case then, π¦ is going to
be equal to the indefinite integral of three π₯ squared plus four with respect to
π₯. We then recall that to integrate
terms in a polynomial, we add one to the power and then divide by this new
number. Assuming that exponent is not equal
to negative one. So the integral of three π₯ squared
is three π₯ cubed divided by three. And the integral of four is four
times π₯ to the power of one divided by one. And of course, since weβre working
with an indefinite integral, we have that constant of integration π. Simplifying, and we see that π¦ is
equal to π₯ cubed plus four π₯ plus π.

This is the general solution to our
differential equation. But what about its particular
solution? In other words, how do we work out
the value of π? Well, we need an initial value. Letβs have a look at a simple
initial value problem.

Find a particular solution for the
following differential equation for which π¦ of zero equals 12. dπ¦ by dπ₯ equals
eight π₯ plus three.

Remember, we can find a general
solution for the differential equation dπ¦ by dπ₯ equals eight π₯ plus three by
first performing the reverse process for differentiation. Weβre going to integrate our
function for dπ¦ by dπ₯ with respect to π₯. So π¦ is equal to the indefinite
integral of dπ¦ by dπ₯ with respect to π₯. Or π¦ equals the integral of eight
π₯ plus three with respect to π₯. We then recall that the integral of
a polynomial term of the form ππ₯ to the πth power, where π and π are real
constants and π is not equal to negative one, is ππ₯ to the power of π plus one
over π plus one plus some constant of integration π.

Essentially, we add one to the
exponent and then divide by that new number. This means the integral of eight π₯
is eight π₯ squared over two. Then the integral of three is three
π₯. And, of course, since this is an
indefinite integral, we end up with that constant of integration π. So we obtained the general solution
of our differential equation to be π¦ equals four π₯ squared plus three π₯ plus
π.

Now, letβs refer back to the
question. Weβre told that π¦ of zero equals
12. In other words, at some point in
the equation π¦ equals π of π₯, when π₯ is equal to zero, π¦ is equal to 12. Letβs substitute these values into
our equation and see what happens. We obtain 12 equals four times zero
squared plus three times zero plus π. Well, this whole equation
simplifies to 12 equals π. And thatβs great because now we
know the value of our constant. And we can, therefore, say that π¦
equals four π₯ squared plus three π₯ plus 12.

This is known as the particular
solution. And π¦ of zero equals 12 is an
initial value. These are sometimes called initial
value problems as they allow us to find a particular solution by giving us an output
of a specific π₯-value.

Letβs now have a look at another
example.

Find the solution for the following
differential equation for π¦ of zero equals one. dπ¦ by dπ₯ minus π₯ minus π₯
squared equals zero.

In this question, we have a
differential equation and an initial value. That is, when π₯ is equal to zero,
π¦ is equal to one. Now, before we can look to solve
this differential equation, weβre going to need to perform an intermediate step. Weβre going to rearrange and make
dπ¦ by dπ₯ the subject. Now, itβs not particularly
difficult to achieve this. We add π₯ and π₯ squared to both
sides of our differential equation to find that dπ¦ by dπ₯ equals π₯ squared plus
π₯. Now, remember, we can find a
general solution for the differential equation dπ¦ by dπ₯ equals π₯ squared plus π₯
by performing the reverse process to differentiation. Weβre going to integrate with
respect to π₯. So π¦ will be equal to the integral
of dπ¦ by dπ₯ with respect to π₯ or the integral of π₯ squared plus π₯ with respect
to π₯.

Remember, to integrate a polynomial
term, we add one to the exponent and then divide by that new value. So the integral of π₯ squared is π₯
cubed over three. We then see that the integral of π₯
is π₯ squared over two. And since weβre working with
indefinite integrals, we know we need that constant of integration π. So π¦ is equal to π₯ cubed over
three plus π₯ squared over two plus π. So we have the general solution to
our differential equation. But we havenβt yet used the
information that π¦ of zero equals one. Weβre going to substitute π₯ equals
zero and π¦ equals one into this general solution. When we do, we see that one equals
zero cubed over three plus zero squared over two plus π. And this whole equation simplifies
to one equals π. So we found that π equals one. And the particular solution to our
differential equation is π¦ equals π₯ cubed over three plus π₯ squared over two plus
one.

Now, itβs important to realize that
we can also apply this procedure to more complicated examples.

Find the function π, if π prime
of π‘ equals two sec π‘ times tan π‘ plus four sec π‘, when π‘ is greater than
negative π by two and less than π by two and π of negative π by three is
negative two.

In this question, we have a
differential equation. Remember, this is just an equation
involving derivatives. Here, thatβs π prime of π‘. And weβre also given an initial
value. That is, when π‘ is equal to
negative π by three, π of π‘ is equal to negative two. Now, weβre looking to work out what
the function π is. And so, we recall that integration
and differentiation are reverse processes. So our original function π of π‘
will be the integral of π prime of π‘ with respect to π‘. Thatβs the integral of two sec of
π‘ times tan π‘ plus four sec of π‘ with respect to π‘.

Now, this doesnβt look very
nice. Letβs first distribute the
parentheses of our integrand. When we do, we obtain our function
π of π‘ to be equal to the integral of two sec π‘ tan π‘ plus eight sec squared π‘
with respect to π‘. We can simplify this a little bit
by recalling that the integral of the sum of two functions is equal to the sum of
the integral of each of those functions. So π of π‘ is equal to the
integral of two sec π‘ tan π‘ with respect to π‘ plus the integral of eight sec
squared π‘ with respect to π‘.

Now, these might look really
nasty. But letβs recall some
derivatives. We know that the derivative of sec
of π‘ with respect to π‘ is tan π‘ sec of π‘. And we know that the derivative of
tan π‘ with respect to π‘ is sec squared π‘. This means the antiderivative of
tan of π‘ sec of π‘ must be sec of π‘. And the antiderivative of sec
squared π‘ must be tan π‘. And so, the integral of two sec of
π‘ tan π‘ must be two sec π‘. And, of course, itβs an indefinite
integral. So weβll add a constant of
integration π΄. Then the integral of eight sec
squared π‘ must be eight tan π‘. And weβll add a second constant of
integration π΅. Combining these two constants of
integration, we see that π of π‘ is equal to two sec π‘ plus eight tan π‘ plus
πΆ.

This is the general solution to our
differential equation. But we havenβt yet used the fact
that when π‘ is equal to negative π by three, our function is equal to negative
two. So letβs substitute these values
in. And when we do, we obtain negative
two to be equal to two sec of negative π by three plus eight tan of negative π by
three plus πΆ. If we recall that sec of negative
π by three is one over cos of negative π by three. Then two sec of negative π by
three must be two over cos of negative π by three, which gives us four. Then eight tan of negative π by
three is negative eight root three. So negative two is equal to four
minus eight root three plus πΆ.

Letβs subtract four from both sides
of this equation and add eight root three to both sides. And we see that πΆ is equal to
eight root three minus six. And weβve obtained the particular
solution to our differential equation to be π of π‘ equals two sec π‘ plus eight
tan π‘ plus eight root three minus six.

Now, the real power of this process
is in contextual examples. Letβs have a look at one.

The acceleration, π meters per
square second, of a particle at time π‘ seconds is given by the equation π equals
negative three sin of 4π‘ for π‘ is greater than or equal to zero. The initial velocity of the
particle is three meters per second and the initial displacement of the particle is
negative two meters. Find an equation for the
displacement, π , of the particle at time π‘ seconds.

Letβs begin by recalling the
relationship between acceleration, velocity, and displacement. Letβs say we have a function for
displacement, π , in terms of π‘. The velocity of an object is its
rate of change of displacement with respect to time. And, of course, when we consider
rate of change, weβre actually thinking about derivatives. So we can say that π£ is equal to
dπ by dπ‘. Similarly, acceleration is defined
as rate of change of velocity with respect to time. So acceleration is dπ£ by dπ‘. And since π£ itself is the first
derivative of π with respect to time, we can say that acceleration could also be
known as the second derivative of displacement with respect to time.

Now, in this question, weβve been
given a function for acceleration in terms of time. To be able to find a function for
displacement, weβre going to need to do the reverse process for differentiation. Weβre going to integrate our
function for acceleration with respect to time. That will give us a function for
velocity. Then, weβre going to integrate once
more to find our function for displacement. So letβs begin by integrating our
function for acceleration. We know that π£ must be equal to
the integral of negative three sin of four π‘ with respect to π‘. Well, remember, we can take out any
constant factors and focus on integrating the function with respect to π‘. So we can say that the integral of
negative three sin of four π‘ with respect to π‘ is negative three times the
integral of sin of four π‘ with respect to π‘.

Then, we recall the standard result
for the integral of sin of ππ₯ with respect π₯. Itβs negative one over π cos
ππ₯. And so, this means the integral of
sin of four π‘ must be negative one-quarter cos of four π‘. And, of course, we need that
constant of integration here. Iβve called that π΄. Distributing our parentheses, we
therefore see that π£ is equal to three-quarters cos of four π‘ plus π΅. And the reason our constant of
integration has become π΅ is because weβve changed it. Weβve multiplied it by negative
three.

Weβre told that the initial
velocity of the particle is three meters per second. In other words, when π‘ is equal to
zero, π£ must be equal to three. So letβs substitute these values
into our function for velocity. That gives us three equals
three-quarters cos of zero plus π΅. And since cos of zero is one, we
see that three is equal to three-quarters plus π΅. To solve for π΅, we subtract
three-quarters from both sides of the equation. And we find that π΅ is equal to
nine-quarters. And so, weβve found that velocity
is equal to three-quarters cos of four π‘ plus nine-quarters.

Our next step is to integrate our
function for velocity to find a general function for displacement. Thatβs the indefinite integral of
three-quarters cos of four π‘ plus nine-quarters with respect to π‘. This time, we recall that the
integral of cos of ππ₯ is one over π sin of ππ₯. And so, the integral of
three-quarters cos of four π‘ must be three-quarters times a quarter sin of four
π‘. Then the integral of nine-quarters
is nine-quarters π‘. So we have a general function for
displacement. Itβs three-quarters times
one-quarter sin of four π‘ plus nine-quarters π‘ plus, of course, that constant of
integration, π. This simplifies to three sixteenths
sin of four π‘ plus nine-quarters of π‘ plus π.

Thereβs one piece of information
weβve not yet used. The initial displacement of the
particle is negative two meters. So when π‘ is equal to zero, not
only is the velocity three meters per second, but π is equal to negative two. Letβs substitute these values into
our equation. Thatβs negative two equals three
sixteenths sin of zero plus nine-quarters of zero plus π. And sin of zero is zero, and
nine-quarters of zero is zero. So we see that negative two is
equal to π. And so, our very final step is to
replace π with negative two in our equation for the displacement. And we find the equation with the
displacement π of the particle at time π‘ seconds is three sixteenths sin of four
π‘ plus nine-quarters π‘ minus two.

In this video, weβve seen that we
can use integration to find general solutions of differential equations. We also saw that we can use initial
values to evaluate the constant. And this is called the particular
solution to the differential equation.