1. Differential Equations
3/31/2017
Math Review with Matlab:
Differential Equations
First Order Constant Coefficient Linear Differential Equations
S. Awad, Ph.D.
M. Corless, M.S.E.E.
E.C.E. Department
University of Michigan-Dearborn

2. 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

3. First Order D.E.
A General First Order Linear Constant Coefficient Differential Equation of x(t) has the form:
Where a is a constant and the function f(t) is given

4. Properties
A General First Order Linear Constant Coefficient DE of x(t) has the properties:
The DE is a linear combination of x(t) and its derivative
x(t) and its derivative are multiplied by constants
There are no cross products
In general the coefficient of dx/dt is normalized to 1

5. Fundamental Theorem SOLUTION SOLUTION SOLUTION
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:
SOLUTION
and x(t)=xc(t) is any solution to the homogenous equation
SOLUTION
Then x(t) = xp(t)+xc(t) is also a solution to the original DE
SOLUTION

6. f(t) = Constant Solution
If f(t) = b (some constant) the general solution to the differential equation consists of two parts that are obtained by solving the two equations:
xp(t) = Particular Integral
Solution
xc(t) = Complementary
Solution

7. Particular Integral Solution
Since the right-hand side is a constant, it is reasonable to assume that xp(t) must also be a constant
Substituting yields:

8. Complementary Solution
To solve for xc(t) rearrange terms
Which is equivalent to:
Integrating both sides:
Taking the exponential of both sides:
Resulting in:

9. First Order Solution Summary
A General First-Order Constant Coefficient Differential Equation of the form:
a and b are constants
Has a General Solution of the form
K1 and K2 are constants

10. Particular and Complementary Solutions
Particular Integral Solution
Complementary Solution

11. Determining K1 and K2
In certain applications it may be possible to directly determine the constants K1 and K2
The first relationship can be seen by evaluating for t=0
The second by taking the limit as t approaches infinity

12. Solution Summary Takes the form:
By rearranging terms, we see that given particular conditions, the solution to:
a and b are constants
Takes the form:

13. Electrical Applications
Basic electrical elements such as resistors (R), capacitors (C), and inductors (L) are defined by their voltage and current relationships
A Resistor has a linear relationship between voltage and current governed by Ohm’s Law - +

14. Capacitors and Inductors
A first-order differential equation is used to describe electrical circuits containing a single memory storage elements like a capacitors or inductor
The current and voltage relationship for a capacitor C is given by:
The current and voltage relationship for an inductor L is given by:

15. 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

16. 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-at
These constants can be found by:
Determining a
Determining vc(0)
Determining vc(¥)
Finally graph the resulting vc(t)

17. Equation for t > 0
Kirchhoff’s Voltage Law (KVL) states that the sum of the voltages around a closed loop must equal zero
Ohm’s Law states that the voltage across a resistor is directly proportional to the current through it, V=IR
Use KVL and Ohm’s Law to write an equation describing the circuit after the switch closes

18. Differential Equation
Since we want to solve for vc(t), write the differential equation for the circuit in terms of vc(t)
Replace i = Cdv/dt for capacitor current voltage relationship
Rearrange terms to put DE in Standard Form

19. General Solution The solution will now take the standard form:
a can be directly determined
K1 and K2 depend on vc(0) and vc(¥)

20. Initial Condition
A physical property of a capacitor is that voltage cannot change instantaneously across it
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+)
Before the switch closes, the capacitor was initially discharged, therefore:
Substituting gives:

21. 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

22. Solve Differential Equation
Now solve for K1 and K2
Replace to solve differential equation for vc(t)

23. Time Constant
When analyzing electrical circuits the constant 1/a is called the Time Constant t
K1 = Steady State Solution
t = Time Constant
The time constant determines the rate at which the decaying exponential goes to zero
Hence the time constant determines how long it takes to reach the steady state constant value of K1

24. Plot Capacitor Voltage
For First-order RC circuits the Time Constant t = 1/RC

25. 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