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.