Welcome back, this is Doctor Ferri. Today's lesson is on First-Order
Differential Equations. Previous lessons, we've seen that
inductors and capacitors have i-v characteristics that involve derivatives.
So, looking at our outline here, we have to be able to handle those derivatives.
And what we're going to find is that we need to be able to solve first-order
differential equations in order to be able to be able to apply the methods to
solving RC and RL circuits. The specific lesson objectives are to
examine first-order differential equations with a constant input and write
the solution and sketch the solution. Ordinary differential equations are
equations that involve functions of variables and their derivatives.
Let's look at some examples. This is a first order differential
equation that, it has a first derivative as the highest term, and that's why it's
first order. Let's look at a different one.
This one has a different forcing term in it.
So if I look at these two terms, this is s forcing term that's constant, and this
one is a forcing term that contains a sine wave.
And these are two very common forcing terms that we see in electrical circuits.
This is a second order differential equation because the highest order is
second derivative. And here, I gave the forcing term as just
a little bit more generic. In circuits, the equations that we will
be looking at involve voltages and currents, so it makes sense to write our
differential equation in terms of i or v as our variables.
Now these three differential equations. Our first order, because their highest
derivative is first derivative. In particular, we will be looking at
differential equations of this form where it's a constant forcing term.
And a circuit is a physical system. So it makes sense to look at models of
physical systems that. have first order of differential
equations. A system model typically has an input
that we'll show here, which is our forcing term to our differential
equation, and the output is the solution to the differential equation.
Where a is just a coefficient that depends on the physical properties.
Now, let's look at some common examples of other systems.
Physical symptoms that involve differential equations.
Particularly, let's look at a, a thermal system.
A heat source is the input. And the temperature at a particular
point's the output. So, when you have heat source, it takes a
little bit of time for the heat to permeate through the, the area.
And raise the temperature at a, a point. So it's modeled as a differential
equation. Typically, another example that we often
see is biological systems. Where we're looking the population growth
or decay of a system given a production rate.
Now in this course, we are interested in linear circuits, and they are solved, are
written, modeled as differential equations.
Particularly we have voltage or current as our input terms, and voltage or
current across a particular component as our output.
So, let's look at a solution to first order differential equations.
Now our approach here is not to show you how to find the solution, but just to
give you the solution and have you use it as a formula.
So given a general, generic sort of differential equation here, where you
have an initial condition, and this is given that time equals zero, and k is a
constant, and a is also a constant, and that's our coefficient right there.
The solution is given right her for time greater than or equal to zero.
Notice that we have an exponential term in there that relates to that constant
coefficient. And this K over a comes from this K over
that A. It helps for us to understand the
behavior as time gets very large. If a is greater than or equal to zero,
this exponential decays to zero. So this term k is out, and this term k is
out and we're left with K over a. I'm going to call that my steady state
value, K over a. We want to be able to graph the response.
Given the same formula for the solution to a first order differential equation,
we've already identified the steady state value.
And I'm going to call everything else in my solution as a transient.
The transients the part that the k is out.
In a time constant is something that's important to determine its a measure of
how fast the transient decays now these are two plots of solutions to first order
of each differential equations now in this first case.
The initial condition starts at 1, and then it decays, the transient decays to
steady-state value of 0. In this particular case, this solution to
this differential equation, we start out with the initial condition of 0.
And the steady-state value is 1. The time constant It's defined.
We are going to called it Tau. It's defined as a time, the exponential
transient to decay to e to the minus 1, which is 0.37 of its initial value.
Another way of saying it is 63% of it's final value.
So looking at this case over here. We want the transient de, decay to 0.37,
which is about right here, of its initial value.
And I go over and look at the time that it takes, and say that's my time
constant. So in this case the time constant is 0.5.
In this case, we want the transient to the k to you know, 63% of its final
value. So from here to here, we want it to be
about 63% of the way. So that's about 0.63 right there, and I
go over here, and this would be the time constant right there.
In both cases the time constant is 0.5, so I'm looking to reach go about 2 3rds
away, 0.63 is about 2 3rds, so about 2 3rds of the way to the final value and
over here about 2 3rds of the way to the final value.
I'd be do some sample problems dy/dt plus 10y is equal to 20.
Given an initial value I'll call it equal to one.
The solution directly from that formula where k is equal to 20 and a is equal to
10, I have y of t is equal to k over a or 2 times 1 minus e to the minus 10t plus e
to the minus 10t, for t greater than or equal to 0.
I can identify this steady state solution and that's k over a which is 2.
We can see that right here, because these are there terms will be k to 0.
And then the transient is going to be the value of time that makes this term, this
exponential term e to the minus one. Well that value of time is going to be 1
over 10, and that's always the case. The time concept is always one over this
coefficient. Next I want to plot this, versus time.
Show my scales here one, two, and I am plotting this solution y of t.
It start out, it starts out with initial condition of 1, and let me show in a
different color here. It starts out with initial condition of
1, and then it's going to reach a steady state value of 2.
Let me kind of highlight in green, what that steady state value is?
So, my solution should start out with the initial condition and exponentially
approach the steady state value. The time comes into tell us how to scale
this axis here. So, at the time constant is when I get
about 2 3rds of the way, from the initial condition, to the final value, that's
about 2 3rds right there. So, I come down here and mark that as 1
over 10 or 0.1. Then I can go ahead and just scale out
the rest of my axis in a proportional way, by time I get to 4 time constant or
four times a time constant add very very close to the final solution, so the
method again was to write down a solution, identify the steady state value
in the time constant. And then to draw your sketch from those
quantities. We're going to do one more example.
This one I want you to be a little bit more interactive in the way you do this.
So, part of this I'm going to have you do this with me.
So, dy dt plus 20y is equal to minus 2, y of zero is 2.
The solution, y of 2 is equal to minus 2 over 20, which is minus 0.1 times 1 minus
e to the minus a, or a is 20 ,so minus 20t plus y of 0 is 2e to the minus 20t
and this is for time greater than or equal to 0.
At this point I'm going to want you to pause this and I want you to find the
steady state value. The time constant, and then we'll come
back to sketch this. So hopefully, you've paused it and you're
joining me back with the, the solution. The steady state value.
Is going to be minus 0.01, which comes from K over A.
The time constant is 1 over the coefficient, which is 1 over 20.
Which is 0.05. So then if I want to sketch this, let's
call this 1, 2, this is y of t. and my steady state values minus point 1.
Let me go ahead and mark that. So minus point 1 is about right there.
And my solution, I'll draw in blue here. Let's see, I start with initial condition
of 2 and I'm going to exponentially approach my steady state value.
The time constant is 0.05. That again is the time that it's going to
take to start, to get 2 3rds of the way there.
It's about right here. The time constant is 0.05 and just
proportioning out my scale I've got 0.1. .15, and so on.
So, you should be able to work sample problems like this.
The important thing is to be able to identify the city state response, the
time constant, and to sketch it. In summary.
We've discussed how various physical phenomena are modeled by differential
equations. We've showed the solution to a generic
first-order differential equation with a constant input and initial condition.
We're using it like a formula. We've introduced a transient and
steady-state responses. And we've showed how to the sketch the
response. And plot the time constant.
In our next lesson we want to solve RC circuit equations and plot the responses.
And we're going to be doing that by using this generic method that was developed in
this lesson. Now please go to the forum to ask any
questions that you have and please work on your homework.
Thank you.