In this segment, we'll talk about higher order ordinary differential equations.
So we'll just talk about a little bit of the background with how we tackle higher
order differential equations so that we can use numerical methods to solve them.
Differential equations which we saw, they are of this particular form, whether we are using Euler's
method, Runge-Kutta 2nd order method, or make it a fourth order method of any of the methods
which you have learned so far. So they are all of the form of the first order ordinary differential
equation, and hence we got to figure out that if we have higher order ordinary differential
equations, like second, third, fourth, fifth order, how do we go about solving them? So let's
review. When we had a differential equation like this, let's suppose we had two dy by dx plus 3y
is equal to 5x and we are given, let's suppose, the initial condition at 0 to be 7. So what we
did was, in order to be able to solve by Euler's method or by Runge-Kutta method, we rewrote this
differential equation in the form which was needed to be able to do so. So we simply took 3y to the
right side divided by 2, and we got y(0) = 7, and this becomes our function f(x,y). So let's go and
see that how we do that for higher order ordinary differential equations to get a similar format.
So when we have a format like this one, here is a second order differential equation,
here is a third order differential equation, we cannot simply say that hey we can rewrite
it as the first order ordinary differential equations as I showed you in the previous slide.
So what do we do in this particular case? How do we use the first order ordinary
differential equation format to be able to write equations which are of second order,
third order, and so on and so forth? So let's go and look at that.
So what we're going to do is we're going to go do this to an example. We will take this differential
equation which we have here, it's a second order ordinary differential equation, and we'll rewrite
it in terms of two simultaneous first order ordinary differential equations so that we can
use the methods which we have learned for Euler's method and Runge-Kutta 2nd order method on first
order ordinary differential equations to do so. So, let's say, what we're going to do is we're
going to take this dy by dt and we're going to introduce another variable called, let's suppose,
z. So z is an introduced dependent variable, and we are so I'm going to say "say." So what that
means that this is not a variable which is there, but we are introducing it. So if we do that,
what's going to happen is that this differential equation right here, we can write as 5 dz by dt.
Why? Because we can say the second derivative of y with respect to t is now dz by dt.
So that's the case, then we can write 13 and so dy over dt, we're going to put z
plus 17y we're going to keep it like that 23 e to the power minus t. And now what we
can do is we can say hey hello dz by dt can be rewritten as 23 e to the power minus t.
Let's take 13z to the right side, let's take 17y to the right side, and divide it by 5. And the
question is that hey if we are going to rewrite it like this, we got to also have a corresponding
initial condition on it. The initial condition will not will be on z. A common misconception
about students that they think that it should be or something else, but it should be on the
dependent variable which you have here. So we have say whatever z_0. z_0 is dy by dt at 0, and that
is equal to 11. So that becomes our first order ordinary differential equation. However, what
we have to think about now is that hey we have this differential equation which is given to us,
but this is not the one uh this is not the only one which we have to solve. We have
reduced the second order differential equation to a first order one, but it does not come by
itself because what you're going to see here is that you have z as well as y as the dependent
variables introduced, so we should have another differential equation to carry on with this.
So rewriting what we have is now dz by dt is equal to 23 e to the power minus t
minus 17y minus 13z divided by 5, and we have the condition that z_0
is 11. Now keep in mind, this is now some function that's, suppose we called f1,
of not only t, which is the independent variable, but also y and z. And, as I mentioned that we have
only one differential equation, we need another one. And the other one is nothing but the dy by
dt is equal to z. That is your next differential equation first order differential equation, and
that's also a function of t, y, and z. Although y and t are not in the expression right here, but
they're still a function of t, y, and z. And this is subject to initial condition on this variable,
the dependent variable y, and that we know that y_0 is equal to 7. So what we have is that this
is the first second order differential first first order differential equation, and this is
the second first order differential equation, and these are now simultaneous. These are simultaneous
2 first order ODEs, which we can now tackle with Euler's method, or Runge-Kutta 2nd order method,
or Runge-Kutta 4th order method to solve. So that is the background
of how we uh how we take care of higher order differential equations in order to
be solved them by the numerical methods we know. And that's the end of this segment.