Welcome back, this is Dr. 
Ferri. 
Today's lesson will be on second order 
differential equations. 
In previous lessons we've looked at 
solutions to first order differential 
equations and we've applied that to the 
analysis of RC and RL circuits. 
So looking at our module on reactive 
circuits here. 
that covers, that first order analysis 
covers up through RC and RL circuits. 
So now we want to introduce second-order 
differential equations, so that we can 
examine more complicated circuits, 
particularly ones that involve RLC 
circuits. 
They include both inductors and 
capacitors, as well as resistors. 
Our lesson objectives are to examine 
second-order differential equations with 
a constant input. 
We want to be able to determine the 
steady-state solution and determine the 
type of transient response that we can 
get out of it. 
And then we want to see what the actual 
response looks like if we plot it and 
recognize the characteristics of that. 
So as we talked about before with first 
order differential equations, And 
ordinary Differential equation is just an 
equation with functions and variables of 
their derivatives. 
Looking at this equation right here, 
we've got a second derivative, so that 
makes it a second order differential 
equation and were just assuming that 
we've got constant coefficients here. 
And that our forcing term here is just a 
constant. 
Many physical systems are governed by 
differential equations like this second 
order differential equations. 
Where the forcing term is the input to 
our system. 
And then the output is the solution to 
our differential equation. 
Vibratory system and RLC circuit are 
examples of this. 
In a vibratory system our forcing term, 
our actual physical force that we apply 
at the system is the input. 
And then the solution to the differential 
equation is a position. 
In our RLC circuit, our input is a 
voltage, or current source, and our 
output is a voltage or current across a 
particular element. 
So let's look at a generic second-order 
differential equation. 
And I say generic so that we can apply it 
to Rlc circuits, vibratory systems or 
anything that's governed by second order 
differential equations. 
And in, in the system that were looking 
at, were assuming that these are constant 
but unknown parameters, a1 and a2 and k 
is also constant. 
We assume that we have known values for 
initial conditions, so y of 0 is known, 
That's time equals zero and the first 
derivative at time equals zero is also 
known. 
Our solution is made up of a steady state 
part plus a transient. 
Plus steady state part is pretty easy to 
solve for and it's exactly the same way 
we solve for the steady state in a first 
order differential equation. 
We just take this value K I redivided by 
this last coefficient, a sub 2. 
And that's what happens to y of t as it 
becomes very large. 
It approaches this value. 
The transient, well the transient's a 
little bit more difficult than it was in 
the first order differential equation. 
Because we have 3 possible forms, 
depending on the roots of this 
polynomial. 
Okay. 
Where do I get that from? 
I can look up at the differential 
equation, and wherever I see a second 
derivative, I'm going to replace it with 
an s squared. 
That's a sub 1. 
My first derivative, I'm going to replace 
with an s. 
Plus a 2, equals 0. 
Those of you who have studied 
differential equations before might 
recognize this as a, as a characteristic 
equation. 
Those of you that have seen Laplace 
transform before may recognize it as a 
Laplace transform. 
In our case, we're just going to use. 
This define the solution. 
I'm not going to assume that you had any 
other background in differential 
equations than what I'm covering here. 
So the transient response again is 
governed by that polynomial and we've got 
three possible cases. 
If I've got real and distinct roots. 
Say, R1 and R2. 
Then I've got this case right here, and 
my solution, kind of my generic form of 
my transient response looks like this. 
So remember, I've got a steady state plus 
this part, here's my solution to the 
differential equation. 
Now, if R1 And r two are both less than 
zero. 
Then my transient suppose it starts up 
here with initial condition up here. 
It's going to go like this. 
It's going to look exponential versus 
time. 
And that's because one of these roots. 
It usually is, is much it decays much 
faster than the other one. 
So if this decays much faster compared to 
this one, then my response will look 
exponential. 
In the critically damped case, I have two 
roots that are equal, r and r. 
So my solution looks like this. 
So I get this, this sort of strange term 
in here. 
T times e to the r t. 
Well it turns out r is less than zero 
then the exponential goes to zero faster 
than t gets large. 
So my response for that case. 
Also looks exponential, and that's if the 
case where r is less than 0. 
We oftentimes call these stable systems, 
where the transient decays to 0. 
Now the under-damped case, that's when 
I've got complex roots, and I'll say the 
complex roots are equal to A plus or 
minus j v. 
Where j is the square root of minus one. 
By imaginary term. 
So in that case I get this sinusoidal. 
Where the real part, the real part goes 
as the exponential. 
And the imaginary part shows up as a 
frequency. 
In this case, if a is less than 0, then 
I've got e to that a t with a negative 
exponent. 
That's going to decay. 
And let me show you what this looks like. 
Given some initial condition, the 
transient Will look sort of like that. 
Where if I were to look at this 
amplitude, the amplitude would be 
exponential because it follows this. 
This looks like an amplitude of this sine 
way. 
We often call this a damped sign wave. 
So this is a damp sign wave and it's 
damped by this real part of my root. 
So my overdamped case looks exponential, 
my underdamped case looks oscillatory. 
In the cases we're going to be examining, 
it's, we're not going to be really 
worried about the critically damped case, 
because critically damped, we have to 
have your, your system perfectly match. 
So that you end up with real and equal 
roots. 
And the chances of that happening in a 
physical system aren't very good. 
Unless you really are tuning your system 
so with resistors that have tolerances to 
them. 
It's unlikely you actually come up with a 
critically damped system. 
So we're just going to worry about 
overdamped and underdamped. 
Lets look at some sample examples. 
Sample problems. 
Given this problem, the steady state 
value is one over four. 
And if I see, what happens is y gets very 
large as it approaches 0.25. 
So that a steady-state value. 
The transient is found by solving this 
differential equation, s squared plus 5 s 
plus 4 equals 0. 
If I find the roots of that, I get roots 
equal to minus 1 and minus 4. 
That means my transient is some constant, 
k 1, times e to the minus 1 times t, plus 
some constant k 2. 
E to the minus 4 times t. 
Well in this case you see that e to the 
minus 4 to k is much faster than e to 
minus t. 
So this system kind of dominates by this 
exponential here. 
And that's what we see right here. 
We see something that looks almost 
exponential. 
The, the transient has an exponential 
decay until it reaches a steady-state 
value. 
And, in fact, if I look at a value of 1, 
this, this looks to be, having pretty 
close to a time constant of 1, because 
that's when the transient decays to about 
gets to about two-thirds of the way to 
the steady-state value. 
So it's pretty close to a time constant 
of 1. 
So this case, real and distinct roots, 
it's overdamped. 
And it looks sort of exponential. 
Now if I look at this, the next case, the 
steady state. 
Just like it was in the previous case, 
it's equal to one over four. 
And we see that it does approach one over 
four here. 
Transient I have to solve s square plus 
.8 s. 
Plus four equals zero. 
Well the roots to this one are complex. 
It's equal to minus .4s .4 plus or minus 
1.96 j. 
It's complex so that means it's 
underdamped. 
And I serve the general form of the 
transient, is k, you know some constant 
case of one. 
E to the minus .4 t. 
Sin of 1.96. 
T Plus some angle. 
This is my sine wave, and you see that if 
I were to look at the amplitude here, the 
amplitude looks like this, it's a 
decaying exponential. 
With that decay rate. 
One thing I want to point out about this 
is, if I look at a case of a low value 
here, compared to the previous example, 
the lower the value here, it turns out 
the more oscillatory the system is. 
And there's a parameter here in this 
coefficient that we're going to call 
Damping factor or damping ratio. 
And so as the damping ratio goes down, as 
this coefficient goes down, gets 
smaller,then it becomes more oscillatory 
and it peaks up a little bit more So this 
is, this coefficient is proportional for 
the damping ratio and as damping ratio 
goes down then it becomes more 
oscillatory. 
[SOUND] And again it peaks out more. 
So in this case we've got complex roots. 
It's underdamped and the lower that 
value, the more it peaks up. 
In summary, we've examined in generic 
second order differential equations. 
And, we've talked about how it these 
apply to vibratory systems and RLC 
circuits. 
We've showed how to find the steady state 
solution. 
We've showed generic transient solutions 
to underdamped and overdamped responses. 
And then, we looked at characteristics of 
plots of underdamped and overdamped 
responses. 
Constant input applied at time equal to 
zero. 
In our next lesson, we will demonstrate 
RLC circuit equations, And their 
responses and show how we use these 
solutions to second order differential 
equations to analyze those sorts of 
circuits. 
Now please visit the forms And I'll see 
you online. 
Thank you.