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**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

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**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

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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

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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

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**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

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**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

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**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:

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**Complementary Solution**

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

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**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

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**Particular and Complementary Solutions**

Particular Integral Solution Complementary Solution

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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

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**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:

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**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 - +

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**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:

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**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

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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)

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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

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**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

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**General Solution The solution will now take the standard form:**

a can be directly determined K1 and K2 depend on vc(0) and vc(¥)

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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:

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**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

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**Solve Differential Equation**

Now solve for K1 and K2 Replace to solve differential equation for vc(t)

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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

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**Plot Capacitor Voltage**

For First-order RC circuits the Time Constant t = 1/RC

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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

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