EGR 1101: Unit 12 Lecture #1
Differential Equations
(Sections 10.1 to 10.4 of Rattan/Klingbeil text)
Linear ODE with Constant
Coefficients

Given independent variable t and
dependent variable y(t), a linear ordinary
differential equation with constant
coefficients is an equation of the form
n
An
d y
dt
n
 ...  A1
dy
dt
 A0 y ( t )  f ( t )
where A0, A1, …, An, are constants.
Some Examples

Examples of linear ordinary differential
equation with constant coefficients:
dy
2
 y 8
dt
2
d y
dt
2
5
d y
dt
 6 y  3t
dt
4
3
dy
4
3

d y
dt
3
8
dy
dt
 sin t
Forcing Function

In the equation
n
An

d y
dt
n
 ...  A1
dy
dt
 A0 y ( t )  f ( t )
the function f(t) is called the forcing
function.
It can be a constant (including 0) or a
function of t, but it cannot be a function of
y.
Solving Linear ODEs with
Constant Coefficients

•
Solving one of these equations means
finding a function y(t) that satisfies the
equation.
You already know how to solve some of
these equations, such as
dy
2
dt
•
But many equations are more complicated
and cannot be solved just by integrating.
A Procedure for Solving Linear
ODEs with Constant Coefficients

We’ll use a four-step procedure for solving
this type of equation:
1.
2.
3.
4.

Find the transient solution.
Find the steady-state solution.
Find the total solution by adding the results of
Steps 1 and 2.
Apply initial conditions (if given) to evaluate
unknown constants that arose in the previous
steps.
See pages 371-372 in Rattan/Klingbeil
textbook.
Forcing Function = 0?


If the forcing function (the right-hand side
of your differential equation) is equal to 0,
then the steady-state solution is also 0.
In such cases, you get to skip straight from
Step 1 to Step 3!
Some Equations that Our
Procedure Can’t Handle


Nonlinear differential equations
2
 dy 
3
2
  ( 7 y )  sin t
 dt 
Partial differential equations
2

y
t
7
y
x
t
2
Diff eqs whose coefficients depend on y or t
2t
dy
dt
7y  0
MATLAB Commands

Without initial conditions:
>>dsolve('2*Dy + y = 8')

With initial conditions:
>>dsolve('2*Dy + y = 8', 'y(0)=5')
MATLAB Commands

Without initial conditions:
>>dsolve('D2y+5*Dy+6*y=3*t')

With initial conditions:
>>dsolve('D2y+5*Dy+6*y=3*t', 'y(0)=0',
'Dy(0)=0')
Today’s Examples
1.
2.
Leaking bucket with constant inflow rate
and bucket initially empty
Leaking bucket with zero inflow and
bucket initially filled to a given level
EGR 1101: Unit 12 Lecture #2
First-Order Differential Equations
in Electrical Systems
(Section 10.4 of Rattan/Klingbeil text)
Review: Procedure

Steps in solving a linear ordinary
differential equation with constant
coefficients:
1.
2.
3.
4.
Find the transient solution.
Find the steady-state solution.
Find the total solution by adding the results of
Steps 1 and 2.
Apply initial conditions (if given) to evaluate
unknown constants that arose in the previous
steps.
Forcing Function = 0?


Recall that if the forcing function (the righthand side of your differential equation) is
equal to 0, then the steady-state solution is
also 0.
In such cases, you get to skip straight from
Step 1 to Step 3.
Today’s Examples
1.
2.
Series RC circuit with constant source
voltage
First-order low-pass filter
Exponentially Saturating Function
A function of the form
𝑓 𝑡 = 𝐾(1 − 𝑒 −𝑡/𝜏 )
where K and  are constants, is called an
exponentially saturating function.
 At t = 0, f(t) = 0.
 As t  , f(t)  K.

Exponentially Saturating Function:
Time Constant



In 𝑓 𝑡 = 𝐾(1 − 𝑒 −𝑡/𝜏 ), the quantity  is
called the time constant.
The time constant is a measure of how
quickly or slowly the function rises.
The greater  is, the more slowly the
function approaches its limiting value K.
Time Constant Rules of Thumb

For 𝑓 𝑡 = 𝐾(1 − 𝑒 −𝑡/𝜏 ),



When t = , f(t)  0.632  K.
(After one time constant, the function has risen
to about 63.2% of its limiting value.)
When t = 5  , f(t)  0.993  K.
(After five time constants, the function has risen
to about 99.3% of its limiting value.)
See next slide for graph.
Exponentially Saturating Function:
Graph
Exponentially Decaying Function
A function of the form
𝑓 𝑡 = 𝐾𝑒 −𝑡/𝜏
where K and  are constants, is called an
exponentially decaying function.
 At t = 0, f(t) = K.
 As t  , f(t)  0.

Exponentially Decaying Function:
Time Constant



In 𝑓 𝑡 = 𝐾𝑒 −𝑡/𝜏 , the quantity  is called the
time constant.
The time constant is a measure of how
quickly or slowly the function falls.
The greater  is, the more slowly the
function approaches 0.
Time Constant Rules of Thumb

For 𝑓 𝑡 = 𝐾𝑒 −𝑡/𝜏 ,



When t = , f(t)  0.368  K.
(After one time constant, the function has fallen
to about 36.8% of its initial value.)
When t = 5  , f(t)  0.007  K.
(After five time constants, the function has fallen
to about 0.7% of its initial value.)
See next slide for graph.
Exponentially Decaying Function:
Graph
Low-Pass and High-Pass Filters


A low-pass filter is a circuit that passes
low-frequency signals and blocks highfrequency signals.
A high-pass filter is a circuit that does
just the opposite: it blocks low-frequency
signals and passes high-frequency
signals.