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Welcome back, this is Dr. 
Ferri. 

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Today's lesson will be on second order 
differential equations. 

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In previous lessons we've looked at 
solutions to first order differential 

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equations and we've applied that to the 
analysis of RC and RL circuits. 

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So looking at our module on reactive 
circuits here. 

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that covers, that first order analysis 
covers up through RC and RL circuits. 

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So now we want to introduce second-order 
differential equations, so that we can 

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examine more complicated circuits, 
particularly ones that involve RLC 

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circuits. 
They include both inductors and 

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capacitors, as well as resistors. 
Our lesson objectives are to examine 

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second-order differential equations with 
a constant input. 

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We want to be able to determine the 
steady-state solution and determine the 

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type of transient response that we can 
get out of it. 

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And then we want to see what the actual 
response looks like if we plot it and 

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recognize the characteristics of that. 
So as we talked about before with first 

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order differential equations, And 
ordinary Differential equation is just an 

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equation with functions and variables of 
their derivatives. 

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Looking at this equation right here, 
we've got a second derivative, so that 

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makes it a second order differential 
equation and were just assuming that 

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we've got constant coefficients here. 
And that our forcing term here is just a 

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constant. 
Many physical systems are governed by 

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differential equations like this second 
order differential equations. 

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Where the forcing term is the input to 
our system. 

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And then the output is the solution to 
our differential equation. 

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Vibratory system and RLC circuit are 
examples of this. 

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In a vibratory system our forcing term, 
our actual physical force that we apply 

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at the system is the input. 
And then the solution to the differential 

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equation is a position. 
In our RLC circuit, our input is a 

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voltage, or current source, and our 
output is a voltage or current across a 

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particular element. 
So let's look at a generic second-order 

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differential equation. 
And I say generic so that we can apply it 

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to Rlc circuits, vibratory systems or 
anything that's governed by second order 

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differential equations. 
And in, in the system that were looking 

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at, were assuming that these are constant 
but unknown parameters, a1 and a2 and k 

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is also constant. 
We assume that we have known values for 

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initial conditions, so y of 0 is known, 
That's time equals zero and the first 

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derivative at time equals zero is also 
known. 

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Our solution is made up of a steady state 
part plus a transient. 

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Plus steady state part is pretty easy to 
solve for and it's exactly the same way 

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we solve for the steady state in a first 
order differential equation. 

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We just take this value K I redivided by 
this last coefficient, a sub 2. 

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And that's what happens to y of t as it 
becomes very large. 

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It approaches this value. 
The transient, well the transient's a 

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little bit more difficult than it was in 
the first order differential equation. 

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Because we have 3 possible forms, 
depending on the roots of this 

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polynomial. 
Okay. 

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Where do I get that from? 
I can look up at the differential 

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equation, and wherever I see a second 
derivative, I'm going to replace it with 

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an s squared. 
That's a sub 1. 

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My first derivative, I'm going to replace 
with an s. 

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Plus a 2, equals 0. 
Those of you who have studied 

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differential equations before might 
recognize this as a, as a characteristic 

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equation. 
Those of you that have seen Laplace 

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transform before may recognize it as a 
Laplace transform. 

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In our case, we're just going to use. 
This define the solution. 

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I'm not going to assume that you had any 
other background in differential 

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equations than what I'm covering here. 
So the transient response again is 

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governed by that polynomial and we've got 
three possible cases. 

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If I've got real and distinct roots. 
Say, R1 and R2. 

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Then I've got this case right here, and 
my solution, kind of my generic form of 

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my transient response looks like this. 
So remember, I've got a steady state plus 

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this part, here's my solution to the 
differential equation. 

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Now, if R1 And r two are both less than 
zero. 

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Then my transient suppose it starts up 
here with initial condition up here. 

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It's going to go like this. 
It's going to look exponential versus 

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time. 
And that's because one of these roots. 

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It usually is, is much it decays much 
faster than the other one. 

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So if this decays much faster compared to 
this one, then my response will look 

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exponential. 
In the critically damped case, I have two 

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roots that are equal, r and r. 
So my solution looks like this. 

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So I get this, this sort of strange term 
in here. 

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T times e to the r t. 
Well it turns out r is less than zero 

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then the exponential goes to zero faster 
than t gets large. 

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So my response for that case. 
Also looks exponential, and that's if the 

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case where r is less than 0. 
We oftentimes call these stable systems, 

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where the transient decays to 0. 
Now the under-damped case, that's when 

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I've got complex roots, and I'll say the 
complex roots are equal to A plus or 

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minus j v. 
Where j is the square root of minus one. 

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By imaginary term. 
So in that case I get this sinusoidal. 

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Where the real part, the real part goes 
as the exponential. 

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And the imaginary part shows up as a 
frequency. 

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In this case, if a is less than 0, then 
I've got e to that a t with a negative 

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exponent. 
That's going to decay. 

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And let me show you what this looks like. 
Given some initial condition, the 

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transient Will look sort of like that. 
Where if I were to look at this 

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amplitude, the amplitude would be 
exponential because it follows this. 

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This looks like an amplitude of this sine 
way. 

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We often call this a damped sign wave. 
So this is a damp sign wave and it's 

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damped by this real part of my root. 
So my overdamped case looks exponential, 

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my underdamped case looks oscillatory. 
In the cases we're going to be examining, 

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it's, we're not going to be really 
worried about the critically damped case, 

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because critically damped, we have to 
have your, your system perfectly match. 

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So that you end up with real and equal 
roots. 

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And the chances of that happening in a 
physical system aren't very good. 

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Unless you really are tuning your system 
so with resistors that have tolerances to 

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them. 
It's unlikely you actually come up with a 

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critically damped system. 
So we're just going to worry about 

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overdamped and underdamped. 
Lets look at some sample examples. 

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Sample problems. 
Given this problem, the steady state 

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value is one over four. 
And if I see, what happens is y gets very 

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large as it approaches 0.25. 
So that a steady-state value. 

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The transient is found by solving this 
differential equation, s squared plus 5 s 

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plus 4 equals 0. 
If I find the roots of that, I get roots 

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equal to minus 1 and minus 4. 
That means my transient is some constant, 

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k 1, times e to the minus 1 times t, plus 
some constant k 2. 

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E to the minus 4 times t. 
Well in this case you see that e to the 

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minus 4 to k is much faster than e to 
minus t. 

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So this system kind of dominates by this 
exponential here. 

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And that's what we see right here. 
We see something that looks almost 

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exponential. 
The, the transient has an exponential 

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decay until it reaches a steady-state 
value. 

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And, in fact, if I look at a value of 1, 
this, this looks to be, having pretty 

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close to a time constant of 1, because 
that's when the transient decays to about 

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gets to about two-thirds of the way to 
the steady-state value. 

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So it's pretty close to a time constant 
of 1. 

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So this case, real and distinct roots, 
it's overdamped. 

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And it looks sort of exponential. 
Now if I look at this, the next case, the 

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steady state. 
Just like it was in the previous case, 

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it's equal to one over four. 
And we see that it does approach one over 

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four here. 
Transient I have to solve s square plus 

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.8 s. 
Plus four equals zero. 

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Well the roots to this one are complex. 
It's equal to minus .4s .4 plus or minus 

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1.96 j. 
It's complex so that means it's 

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underdamped. 
And I serve the general form of the 

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transient, is k, you know some constant 
case of one. 

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E to the minus .4 t. 
Sin of 1.96. 

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T Plus some angle. 
This is my sine wave, and you see that if 

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I were to look at the amplitude here, the 
amplitude looks like this, it's a 

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decaying exponential. 
With that decay rate. 

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One thing I want to point out about this 
is, if I look at a case of a low value 

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here, compared to the previous example, 
the lower the value here, it turns out 

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the more oscillatory the system is. 
And there's a parameter here in this 

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coefficient that we're going to call 
Damping factor or damping ratio. 

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And so as the damping ratio goes down, as 
this coefficient goes down, gets 

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smaller,then it becomes more oscillatory 
and it peaks up a little bit more So this 

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is, this coefficient is proportional for 
the damping ratio and as damping ratio 

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goes down then it becomes more 
oscillatory. 

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[SOUND] And again it peaks out more. 
So in this case we've got complex roots. 

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It's underdamped and the lower that 
value, the more it peaks up. 

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In summary, we've examined in generic 
second order differential equations. 

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And, we've talked about how it these 
apply to vibratory systems and RLC 

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circuits. 
We've showed how to find the steady state 

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solution. 
We've showed generic transient solutions 

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to underdamped and overdamped responses. 
And then, we looked at characteristics of 

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plots of underdamped and overdamped 
responses. 

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Constant input applied at time equal to 
zero. 

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In our next lesson, we will demonstrate 
RLC circuit equations, And their 

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responses and show how we use these 
solutions to second order differential 

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equations to analyze those sorts of 
circuits. 

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Now please visit the forms And I'll see 
you online. 

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Thank you.