Differential Equations

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

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

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

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

Fundamental Theorem • 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

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

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:

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

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

Particular and Complementary Solutions Particular Integral Solution Complementary Solution

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

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

+ - 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

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 inductorL is given by:

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

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)

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

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

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

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:

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

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

Time Constant • When analyzing electrical circuits the constant 1/a is called the Time Constantt 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

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

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

Differential Equations