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S. Awad, Ph.D.
M. Corless, M.S.E.E.
E.C.E. Department
University of Michigan-Dearborn
Differential Equations
Math Review with Matlab:
First Order Constant Coefficient Linear
Differential Equations
U of M-Dearborn ECE DepartmentMath Review with Matlab
2
Differential Equations: First Order Systemsdt
tdx )(
First Order Constant Coefficient Linear
Differential Equations
First Order Differential Equations
General Solution of a First Order Constant Coefficient Differential Equation
Electrical Applications
RC Application Example
U of M-Dearborn ECE DepartmentMath Review with Matlab
3
Differential Equations: First Order Systemsdt
tdx )(
First Order D.E. A General First Order Linear Constant Coefficient
Differential Equation of x(t) has the form:
)()()(
tftxdt
tdx
Where is a constant and the function f(t) is given
U of M-Dearborn ECE DepartmentMath Review with Matlab
4
Differential Equations: First Order Systemsdt
tdx )(
In general the coefficient of dx/dt is normalized to 1
Properties
The DE is a linear combination of x(t) and its derivative x(t) and its derivative are multiplied by constants
)()()(
tftxdt
tdx
A General First Order Linear Constant Coefficient DE of x(t) has the properties:
2)(
dt
tdx There are no cross products
U of M-Dearborn ECE DepartmentMath Review with Matlab
5
Differential Equations: First Order Systemsdt
tdx )(
SOLUTION
SOLUTION
SOLUTION
Fundamental Theorem
)()()(
tftxdt
tdx
A fundamental theorem of differential equations states that given a differential equation of the form below where x(t)=xp(t) is any solution to:
0)()(
txdt
tdx
and x(t)=xc(t) is any solution to the homogenous equation
Then x(t) = xp(t)+xc(t) is also a solution to the original DE
)()()(
tftxdt
tdx
)()( txtx p
)()( txtx c
)()()( txtxtx cp
U of M-Dearborn ECE DepartmentMath Review with Matlab
6
Differential Equations: First Order Systemsdt
tdx )(
f(t) = Constant Solution If f(t) = (some constant) the general solution to the
differential equation consists of two parts that are obtained by solving the two equations:
)()(
txdt
tdxp
p
0)()(
txdt
tdxc
c
xp(t) = Particular Integral
Solution
xc(t) = Complementary
Solution
U of M-Dearborn ECE DepartmentMath Review with Matlab
7
Differential Equations: First Order Systemsdt
tdx )(
Particular Integral Solution
Since the right-hand side is a constant, it is reasonable to assume that xp(t) must also be a constant
)()(
txdt
tdxp
p
1)( Ktxp
Substituting yields:
1K
U of M-Dearborn ECE DepartmentMath Review with Matlab
8
Differential Equations: First Order Systemsdt
tdx )(
Complementary Solution To solve for xc(t) rearrange terms
0 ) () (
t xdt
t dxc
c
dt
t dx
t xc
c
) (
) (
1 Which is equivalent to:
) ( lnt xdt
dc c t t xc ) ( ln
Integrating both sides:
c t c tce e e t x
) (
tce K t x
2 ) (
Taking the exponential of both sides:
Resulting in:
U of M-Dearborn ECE DepartmentMath Review with Matlab
9
Differential Equations: First Order Systemsdt
tdx )(
First Order Solution Summary A General First-Order Constant Coefficient
Differential Equation of the form:
)()(
txdt
tdx and are constants
teKKtx 21)(
Has a General Solution of the form
and are constants
U of M-Dearborn ECE DepartmentMath Review with Matlab
10
Differential Equations: First Order Systemsdt
tdx )(
Particular and Complementary Solutions
1)( Ktxp tc eKtx 2)(
Particular Integral Solution Complementary Solution
)()()(
)( 21
txtxtx
eKKtx
cp
t
U of M-Dearborn ECE DepartmentMath Review with Matlab
11
Differential Equations: First Order Systemsdt
tdx )(
Determining K1 and K2 In certain applications it may be possible to directly
determine the constants K1 and K2
teKKtx 21)(
210
21)0( KKeKKx
The second by taking the limit as t approaches infinity
12121 0)()( KKKeKKtxLimxt
The first relationship can be seen by evaluating for t=0
U of M-Dearborn ECE DepartmentMath Review with Matlab
12
Differential Equations: First Order Systemsdt
tdx )(
Solution Summary By rearranging terms, we see that given particular conditions, the
solution to:
)()(
txdt
tdx and are constants
Takes the form:
teKKtx 21)(
)()0(
)(
2
1
xxK
xK
U of M-Dearborn ECE DepartmentMath Review with Matlab
13
Differential Equations: First Order Systemsdt
tdx )(
A Resistor has a linear relationship between voltage and current governed by Ohm’s Law
Electrical Applications Basic electrical elements such as resistors (R), capacitors (C), and
inductors (L) are defined by their voltage and current relationships
Rtitv RR )()(
)(tiR R
)(tvR
U of M-Dearborn ECE DepartmentMath Review with Matlab
14
Differential Equations: First Order Systemsdt
tdx )(
Capacitors and Inductors
The current and voltage relationship for a capacitor C is given by:
dt
tvdCti c
c
)()(
The current and voltage relationship for an inductor L is given by:
dt
tidLtv L
L
)()(
A first-order differential equation is used to describe electrical circuits containing a single memory storage elements like a capacitors or inductor
)(tiC
)(tvCC
)(tiL
)(tvL
L
U of M-Dearborn ECE DepartmentMath Review with Matlab
15
Differential Equations: First Order Systemsdt
tdx )(
RC Application Example Example: For the circuit below, determine an
equation for the voltage across the capacitor for t>0. Assume that the capacitor is initially discharged and the switch closes at time t=0
R
C
CvDCVCi
Rv
0t
U of M-Dearborn ECE DepartmentMath Review with Matlab
16
Differential Equations: First Order Systemsdt
tdx )(
Plan of Attack
Write a first-order differential equation for the circuit for time t>0
The solution will be of the form K1+K2e-t
These constants can be found by: Determining Determining vc(0)
Determining vc()
Finally graph the resulting vc(t)
U of M-Dearborn ECE DepartmentMath Review with Matlab
17
Differential Equations: First Order Systemsdt
tdx )(
Equation for t > 0
Use KVL and Ohm’s Law to write an equation describing the circuit after the switch closes
Kirchhoff’s Voltage Law (KVL) states that the sum of the voltages around a closed loop must equal zero
R
C
CvDCVCi
Rv
0tDCCC
DCCC
DCCR
VtvtRi
VtvtRi
Vtvtv
)()(
0)()(
0)()()(
Ohm’s Law states that the voltage across a resistor is directly proportional to the current through it, V=IR
U of M-Dearborn ECE DepartmentMath Review with Matlab
18
Differential Equations: First Order Systemsdt
tdx )(
Differential Equation Since we want to solve for vc(t), write the
differential equation for the circuit in terms of vc(t)
DCCc
DCCC
Vtvdt
tdvCR
VtvtRi
)()(
)()( Replace i = Cdv/dt for capacitor current voltage relationship
Rearrange terms to put DE in Standard Form
RC
V
RC
tv
dt
tdv DCCc )()(
U of M-Dearborn ECE DepartmentMath Review with Matlab
19
Differential Equations: First Order Systemsdt
tdx )(
General Solution
The solution will now take the standard form:
RC
V
RC
tv
dt
tdv DCCc )()(
)()(
txdt
tdxteKKtx 21)(
RC
1
can be directly determined
K1 and K2 depend on vc(0) and vc()
U of M-Dearborn ECE DepartmentMath Review with Matlab
20
Differential Equations: First Order Systemsdt
tdx )(
Initial Condition A physical property of a capacitor is that voltage cannot
change instantaneously across it
Before the switch closes, the capacitor was initially discharged, therefore:
Vvc 0)0(
Therefore voltage is a continuous function of time and the limit as t approaches 0 from the right vc(0-) is the same as t approaching from the left vc(0+)
)0()0( cc vv
Substituting gives: Vvc 0)0(
U of M-Dearborn ECE DepartmentMath Review with Matlab
21
Differential Equations: First Order Systemsdt
tdx )(
Steady State Condition As t approaches infinity, the capacitor will fully charge to
the source VDC voltage
No current will flow in the circuit because there will be no potential difference across the resistor, vR() = 0 V
DCc Vv )( R
C
DCC Vv )(DCV
0)( Ci
Rv
t
U of M-Dearborn ECE DepartmentMath Review with Matlab
22
Differential Equations: First Order Systemsdt
tdx )(
Solve Differential Equation
Now solve for K1 and K2
Vvc 0)0( DCc Vv )(
RC
1
)()0(
)(
2
1
cc
c
vvK
vK
DC
DC
VK
VK
2
1
Replace to solve differential equation for vc(t)
tc eKKtv 21)( RC
t
DCDCc eVVtv
)(
U of M-Dearborn ECE DepartmentMath Review with Matlab
23
Differential Equations: First Order Systemsdt
tdx )(
Time Constant When analyzing electrical circuits the constant 1/ is
called the Time Constant
The time constant determines the rate at which the decaying exponential goes to zero
t
c eKKtv
21)( 1
K1 = Steady State Solution = Time Constant
Hence the time constant determines how long it takes to reach the steady state constant value of K1
U of M-Dearborn ECE DepartmentMath Review with Matlab
24
Differential Equations: First Order Systemsdt
tdx )(
Plot Capacitor Voltage For First-order RC circuits the Time Constant = 1/RC
RCt
DCDCc eVVtv
)(
U of M-Dearborn ECE DepartmentMath Review with Matlab
25
Differential Equations: First Order Systemsdt
tdx )(
Summary Discussed general form of a first order
constant coefficient differential equation
Proved general solution to a first order constant coefficient differential equation
Applied general solution to analyze a resistor and capacitor electrical circuit