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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.
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