Video Transcript
In this video, we will learn about
the basics of differential equations. Weβll see first what a differential
equation is and how it might arise in physical problems within the areas of
mathematics and physics. Weβll introduce the terminology
associated with differential equations. And in the context of examples,
weβll see how to classify differential equations according to their type and
order.
Letβs begin then by discussing what
a differential equation actually is. Well, itβs just an equation that
contains a function and one or more of its derivatives with respect to an
independent variable. For example, the equation dπ¦ by
dπ₯ plus π¦ equals three π₯ is a differential equation. Here, π¦ is the function weβre
interested in. Itβs a function of the independent
variable π₯ and the equation includes the first derivative of π¦ with respect to
π₯.
Far more complicated differential
equations also exist. For example, the equation d two π¦
by dπ₯ squared plus four π¦ multiplied by dπ¦ by dπ₯ all squared plus three π¦
equals sin π₯. This differential equation includes
not only a first derivative dπ¦ by dπ₯ which is squared, but it also includes the
second derivative of π¦ with respect to π₯, d two π¦ by dπ₯ squared. So, we have an equation that
contains a function π¦, and its first derivative dπ¦ by dπ₯, and also its second
derivative d two π¦ by dπ₯ squared.
The first of these two examples is
known as a first order differential equation because the highest order derivative it
contains is a first derivative, the first derivative of π¦ with respect to π₯, dπ¦
by dπ₯. The second example is known as a
second order differential equation because the highest order derivative it contains
is a second derivative, d two π¦ by dπ₯ squared. More generally, differential
equations can be classified according to the highest order derivative they
contain. So, a differential equation in
which the highest order derivative is an πth derivative would be classed as an πth
order differential equation.
Examples of differential equations
arise in many physical problems. For example, simple population
growth, where the rate of increase of the population π is proportional to the
population itself. And this can be modeled by the
differential equation dπ by dπ‘ equals ππ. Here, π‘ represents time, and π is
called the constant of proportionality.
Now, itβs beyond the scope of this
video to see how we would solve a differential equation of this type. But in fact, if we had the initial
condition that the population at time zero is equal to some value π nought, then it
can be shown that the solution to this differential equation would be that the
population at time π‘ is equal to π nought multiplied by π to the power of
ππ‘. So, the population grows in an
exponential way.
Now, to solve a differential
equation in general, we need to find an expression for the dependent variable, which
in this case would be π, in terms of the independent variable, which here would be
π‘. The solution is a function such
that that function and its derivatives satisfy the differential equation. Now, generally, when we solve a
differential equation, we get not a single solution but a family of solutions, all
of which will satisfy the differential equation but will differ from one another in
the values of any constants.
If weβre also provided with
additional information, such as here we were given the value of the population at
time zero, then we can determine the values of these constants and hence obtain a
particular solution to the differential equation. This additional information is
known as a boundary condition, or sometimes an initial condition if it specifies the
value of the function when the value of the independent variable is zero.
However, solving differential
equations is not the focus of this particular video. Rather, we are focusing on
understanding and classifying them. We will, however, consider an
example of how we can confirm that a given function does indeed satisfy a
differential equation.
Is the function π¦ equals one over
two plus π₯ a solution to the differential equation π¦ prime equals negative π¦
squared?
Remember that π¦ prime is another
way of saying dπ¦ by dπ₯, the first derivative of π¦ with respect to π₯. So, weβve been given a first order
differential equation, and we want to know whether the given function π¦ is a
solution to it. That is, we need to know whether
the function π¦ satisfies this equation.
Letβs begin then by first working
out what π¦ prime, or dπ¦ by dπ₯, is equal to for this function π¦. And to do this, we can first
express π¦ in an alternative form. We can write it as two plus π₯ to
the power of negative one. We can then find this derivative
using the general power rule, which says that if we have some function π of π₯ to
the power of π, then its derivative with respect to π₯ is equal to π multiplied by
π prime of π₯ multiplied by π of π₯ to the power of π minus one.
Here, our function π of π₯ is two
plus π₯, and our power π is negative one. So, applying the general power
rule, we have π, thatβs negative one, multiplied by the derivative of two plus π₯,
which is just one, multiplied by π of π₯. Thatβs two plus π₯, to the power of
π minus one. So, thatβs the power of negative
two. We can then rewrite this as
negative one over two plus π₯ all squared. So, we know what the left-hand side
of this differential equation would be for this function π¦.
On the right-hand side, we have
negative π¦ squared. So, thatβs the original function π¦
squared and then multiply it by negative one, which is equal to negative one over
two plus π₯ all squared. To square a fraction, we can square
the numerator and square the denominator. So, we have negative one squared,
which is one, over two plus π₯ all squared.
Now, we compare our expressions for
π¦ prime and negative π¦ squared. And we see that they are both equal
to negative one over two plus π₯ all squared. And therefore, they are indeed
equal to one another. This tells us that the function π¦
equals one over two plus π₯ does satisfy the given differential equation and,
therefore, it is a solution.
Weβll now look at some further
examples to introduce the terminology needed in order to describe the wide variety
of other types of differential equations we may encounter.
Determine the order of the
differential equation d two π¦ by dπ₯ squared cubed minus π¦ triple prime to the
fourth power plus π₯ is equal to zero.
We recall first that the order of a
differential equation is the order of the highest order derivative that appears in
that equation. We can see at a glance that this
differential equation involves a second derivative, d two π¦ by dπ₯ squared. But if we look a little closer, we
see that the equation also contains π¦ triple prime, which is alternative notation
for third derivative. The highest order derivative is
three. And hence, the order of this
differential equation is three.
Now, donβt be misled by the powers
here. That is the power of three with the
second derivative and the power of four with the third derivative. The order of a differential
equation is not the highest power of the variable or any of its derivative that
appears in the equation. Itβs the order of the highest order
derivative in the equation. So, that power of three for the
first term and the power of four for the second term are entirely irrelevant in
terms of determining the differential equationβs order.
In our next example, weβll learn
the difference between linear and nonlinear differential equations.
Is the differential equation dπ¦ by
dπ₯ plus π₯ root π¦ equals π₯ squared linear?
A linear differential equation is
one which can be expressed as a linear polynomial of the unknown function, in this
case π¦ and its derivatives. What this means is that the only
powers of the unknown function and each derivative that appear in the equation are
one or zero if the equation does not contain that order derivative. And also each derivative and the
function itself are multiplied by functions of π₯ only.
So, for example, the equation two
times dπ¦ by dπ₯ plus four π₯π¦ equals three π₯ would be an example of a linear
differential equation. Because the power of both π¦ and
dπ¦ by dπ₯ is one, and theyβre each multiplied by a function of π₯ only. Whereas the equation two times dπ¦
by dπ₯ plus four π₯ over π¦ equals three π₯ is nonlinear as in the second term, the
power of π¦ is negative one. The equation four π₯ d two π¦ by
dπ₯ squared plus two π¦ dπ¦ by dπ₯ equals seven is also nonlinear as in the second
term, we see that dπ¦ by dπ₯ is multiplied by a function of π¦, not a pure function
of π₯.
More formally, we can say that a
differential equation is linear if it can be expressed in the form shown on the
screen. Each πth order derivative of π¦
and the function π¦ itself is multiplied by a polynomial in π₯ only. So, letβs consider the differential
equation that weβve been given. And we can see that it includes the
square root of π¦. Now, another way of expressing the
square root of π¦ is as π¦ to the power of one-half. And hence, this differential
equation is nonlinear, as the power of π¦ is not equal to one.
Now, notice that it isnβt the
presence of the π₯ squared term on the right-hand side that makes this differential
equation nonlinear. π₯ is the independent variable in
this equation. And it is only the powers of the
dependent variable and its derivatives, thatβs π¦ and dπ¦ by dπ₯ and so on, that
must all be equal to one in order to make the equation linear.
In our final example, weβll learn
the difference between ordinary and partial differential equations.
Which of the following
relationships is an ordinary differential equation?
Recall, first of all, that a
differential equation contains a function and one or more of its derivatives with
respect to an independent variable. If we consider the first equation,
first of all, π§ equals five π₯π¦, we see that it contains no derivatives. And therefore, this is not a
differential equation. Itβs simply an equation relating
the three variables π₯, π¦, and π§. So, we can rule out option A.
In the same way, if we consider the
final equation π¦ equals the square root of π₯ squared minus four, this isnβt a
differential equation either, as it doesnβt contain any derivatives. It just expresses the relationship
between the variables π₯ and π¦. So, weβre left with just two
possibilities, B and C. Considering the second equation, we
see that it contains an unknown variable π¦ and its derivative with respect to an
independent variable π₯. So, this is an example of a
differential equation.
But the question doesnβt just ask
us for which is a differential equation. It asks us, which is an ordinary
differential equation. So, we need to consider what this
word ordinary means in this context. The third equation does also
contain a derivative. And in fact, it is a second
derivative this time. But we see that the notation used
is slightly different. This notation represents the
partial second derivative of the variable π§ with respect to π₯.
What this means is that the
function π§ is not just a function of π₯, but also of one or more other variables,
such as π¦. The partial derivative of π§ with
respect to π₯ is the function we get if we treat each of the other variables as
constant when differentiating. And in fact, the partial second
derivative of π§ with respect to π₯ is what we get if we do this twice.
So, we return to that word ordinary
in the question. An ordinary differential equation
contains only ordinary, as opposed to partial derivatives, as the unknown function
is a function of the independent variable only. From the notation used in equation
B, we see that this contains only the function π¦ and its ordinary first
derivative. So, this is an ordinary
differential equation. Whereas option C contains a partial
derivative, and so it is known as a partial differential equation.
You may see the abbreviations O.D.E
and P.D.E used to describe ordinary and partial differential equations,
respectively. So, our answer to the question,
which of the following relationships is an ordinary differential equation, is B. A and D are not differential
equations. And C is a differential equation,
but it is a partial differential equation.
Letβs summarise then what weβve
seen in this video. Firstly, differential equations are
equations which relate a function and one or more of its derivatives. The order of a differential
equation is the order of the highest order derivative it contains. So, a differential equation in
which the highest order derivative was a third derivative would have an order of
three.
A linear differential equation can
be expressed in the form shown onscreen. The power of the function π¦ and
each of its derivatives is either one or zero. And π¦ and each of its derivatives
are multiplied by pure functions of π₯. And finally, ordinary differential
equations, or O.D.E.s, contain only ordinary derivatives, such as dπ¦ by dπ₯. Here, the function π¦ is a function
of only one independent variable π₯.
Whereas partial differential
equations, or P.D.E.s, contain partial derivatives. Here, the function π¦ is a function
of more than one independent variable. Differential equations have many
applications and can be used to model a wide range of physical phenomena.