CHAPTER 08.05: HIGHER ORDER AND COUPLED DIFF Eq Back: Part 1 of 2
In this segment we are going to talk about higher
order and coupled order ODE's. So we have higher or coupled ODE's and what we
want to be able to figure out is that how do we solve higher order and coupled
ODE's by using our numerical methods. So this is just the background information
which I am going to show you. Examples of how to solve higher order and coupled
ordinary differential equations are given in a seperate segment. So let's see
that how do we go about doing that. The first thing which I need to talk to you
about is that all the methods which we have talked about, like Euler's Method,
Runge Kutta Second Order Method, Runge Kutta Fourth Order Method, we are only
able to solve differential equations which are of this particular form, which
means that we are only able to solve differential equations which have, which
are first order and they can be written in this particular form which we have
and some initial condition is given. Surely, the initial value is not
necessarily supposed to be zero. It can be any number which is given to you. So,
but since we have this as a first order ordinary differential equation form does
that mean that we can, we are only, then relegated to solve only first order
ordinary differential? What happens if we have coupled ordinary differential
equations? What happens when we have higher order ordinary differential
equations? How do we use our methods of Runge Kutta Second Order Method, Fourth
Order Method, and Euler's Method to be able to solve differential equations
which are of that form? So in order to do that what I am going to do is I am
going to take an example and show you how you can reduce these coupled ordinary
differential equations. In this case, it is a first order ordinary differential
equations so you can use your Runge
Kutta Methods and Euler's Method to be able to do that. So let's do that here.
Let's suppose somebody gives me coupled differential equations like this:
2(d2y/dx2) + 3(dy/dx) - 5(dz/dx). So you are already finding out that the reason
why you are writing coupled ordinary differential equations is because you
already have two dependent variables, y and z in this particular ordinary
differential equation. And so you would need, and that's why you would need
coupled ordinary differential equations. 7z = e^-x So that's the first ordinary
differential equation given to us and we are given another simultaneous ordinary
differential equation. 2(d2z/dx2) - 29(dy/dx) + 31 (dz/dx) - 37(y) + 41z = sin(x)
So these are the two second order ordinary differential equations which are
given to us but they are coupled because you are already finding out that you
have an instance of y and z there and then you have an instance of y and z in
the second one so they have to be solved simultaneously, that's why there are
called coupled ordinary differential equations. And then you have certain
initial conditions. You need an initial condition of y. You need the initial
condition on the first deriviative of y. Then you will need an initial condition
on z. And then you will need an initial condition on the derivative of z. And
now what we need to do is be able to solve this set of coupled ordinary
differential equations. Now, in order to be able to do that by using the
numerical methods which we have been talking about such as Euler and Runge Kutta
Methods we need to reduce these coupled ordinary differential equations into
some set of first order ordinary differential equations because we said that we
can only use first order differential equations to, for the numerical methods.
So one of the things which I am going to do is I'm going to let dy/dx be equal
to, let's suppose, u. So if I let dy/dx equal to u what's going to happen is
that when I take this dy/dx = u and substitute it in here, I will get du/dx, so
that's how I'm reducing this to a first order ordinary differential equation.
And also, wherever there is an instance of dy/dx, for example, I am going to put
u there so that I will be able to represent it as a function of dependent
variables and independent variables like I need it for the numerical methods. So
I am going to dy/dx=u, so that will become my first ordinary differential
equation. In fact, dy/dx=u becomes our first ordinary differential equation
itself. Then, I will also assume, I will also say let dz/dx be equal to w. And
this becomes my second ordinary differential equation and the reason why I am
doing that is because once I substitute dz/dx=w, so for example, in this
particular equation when I substitute dz/dx=w I get dw/dx so I get the first
derivative of w with respect to x here. And wherever there is a dz/dx, for
example here, I will substitute a w there. So, you can see that this is a
mechanism of being able to reduce my second order coupled ordinary differential
equations into first order differential equations. So let me go ahead and do the
substitution and see what happens. So let's suppose if I do the substitution,
these two substitutions, let me go ahead and do them in the differential
equations which are given to me. |