70.2 Homogenous Differential Equations Concept Overview
HOMOGENEOUS DIFFERENTIAL EQUATIONS | CONCEPT OVERVIEW The topic of HOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: A DIFFERENTIAL EQUATION is considered HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. If this is the case, and the sum of all these terms is equal to zero, then we employ the standard process in solving a HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION. A HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION can generally be written as: π! π¦ ππ¦ + π + ππ¦ = 0 ππ₯ ! ππ₯ Or: π¦ !! + ππ¦β² + ππ¦ = 0
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Both of these expressions are equivalent, the only difference being is that in the latter, we replaced the various derivatives strewn throughout the former formula with more simple single y-term variables. Illustrating this further, taking the original general formula above, we turned: π! π¦ ππ¦ + π + ππ¦ = 0 ππ₯ ! ππ₯ In to: π¦ !! + ππ¦β² + ππ¦ = 0 By replacing: π! π¦ = π¦β²β² ππ₯ ! ππ¦ = π¦β² ππ₯ Thatβs all, nothing complex to it, however, you may see these types of equations presented in both ways, so be comfortable with each. When we attempt to pursue a solution for any HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION, everything will start with a CHARACTERISTIC POLYNOMIAL, or what we are probably more comfortable calling it, a CHARACTERISIC EQUATION.
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The FORMULA for the CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, of a DIFFERENTIAL EQUATION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This CHARACTERISTIC EQUATION is simply the polynomial formed by replacing all the derivatives with an βrβ variable raised to the power of that respective derivative, such that: π π = π! π ! + π!!! π !!! + β― + π! π + π! This is presented in a very complicated manner in the NCEES Reference Handbook, but at the end of the day, this is what it comes down to, given: π¦ !! + ππ¦β² + ππ¦ = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: π ! + ππ + π = 0 HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS are most easily solved by finding the βπβ roots of the characteristic polynomial π(π). Each root of the CHARACTERISTIC EQUATION, π! , appears in the HOMOGENEOUS SOLUTION, π¦! π₯ , in the form π !! ! .
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When working with HOMOGENEOUS DIFFERENTIAL EQUATIONS to the FIRST ORDER, we will use the standard process illustrated in previous reviews. However, when presented a HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION, the solution process is different, yet, not any more complex. The DRIVING FORMULAS highlighting the GENERAL SOLUTIONS for SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. There are THREE CASES that we will be concerned with when EVALUATING the ROOTS of the CHARACTERISTIC EQUATION, they are: 1. REAL AND DISTINCT ROOTS 2. REAL AND EQUAL ROOTS 3. COMPLEX ROOTS We classify each of the roots, and further the basic solution, based on the DISCRIMINANT of the CHARACTERISTIC EQUATION.
1.
REAL AND DISTINCT ROOTS
A REAL and DISTINCT ROOT is characterized by the relationship: π! > 4π
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Each distinct real root is represented as π! = πΆ! π !! ! . The STANDARD FORM for the HOMOGENEOUS SOLUTION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 30 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When we are dealing with a HOMOGENEOUS DIFFERENTIAL EQUATION and the ROOTS are both REAL AND DISTINCT, which will be the most common case, the HOMOGENEOUS SOLUTION can be generally written as: π¦! π₯ + πΆ! π !! ! + πΆ! π !! ! + β― + πΆ! π !! ! + β― + πΆ! π !! ! Where: β’ π! is the ππ‘β distinct root of the CHARACTERISTIC POLYNOMIAL π(π₯) β’ πΆ! , πΆ! , πΆ! , β¦ , πΆ! are all real constants β’ πΆ! β 0 This solution is of the OVERDAMPED form.
2.
REAL AND EQUAL ROOTS
A REAL AND EQUAL ROOT is characterized by the relationship: π! = 4π
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The two roots are real and the same, such that they can be considered the same variable. Repeated real roots are represented as π! = π! , where π! and π! are real numbers. If the root π! = π! , then the second term πΆ! π !! ! is replaced with πΆ! π₯π !! ! β¦note the added x variable and the exponent of π! . If the roots are real and the same, the solution is: π¦! π₯ = (πΆ! + πΆ! π₯ + β― + πΆ! π₯ !!! )π !" This solution is of the CRITICALLY DAMPED form.
3.
COMPLEX ROOTS
A COMPLEX ROOT is characterized by the relationship: π! < 4π The two complex roots are imaginary and represented as: π! = πΌ + ππ½ π! = πΌ β ππ½
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If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions, such that: π¦! π₯ = π !" (C! sin π½π₯ + πΆ! cos π½π₯) Where: β’ The real part of the complex root, βπβ, appears in the exponent, such that:
πΌ=β
π 2
β’ The coefficient of the imaginary part, βπβ, appears in the sine and cosine terms of the general formula, such that: 4π β π! π½= 2 This solution is of the UNDERDAMPED form. When it comes down to developing the structure for the HOMOGENEOUS SOLUTION, it all starts with determining the ROOTS after defining the CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQAUTION.
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FIRST ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS: The TOPIC of FIRST ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with CONSTANT COEFFICEINTS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A FIRST ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION is a differential equation in which each term contains either the function or one of its derivatives. Generally this type of DIFFERENTIAL EQUATION can be written in the form: ππ¦ + ππ¦ = 0 ππ₯ Or in CHARACTERISTIC FORM as: π+π =0 Where: a = real constant
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Thereβs not much to be done to determine the roots of this CHARACTERISTIC EQUATION, in fact, it simply comes down to this: π = βπ From this standard form of the differential equation, the solution can be generally represented as: π¦ = πΆπ !!" Where: β’ πΆ is a constant that satisfies the initial conditions
SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS: The TOPIC of SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with CONSTANT COEFFICIENTS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with constant coefficients are commonly represented in the standard form: π! π¦ ππ¦ + π + ππ¦ = 0 ππ₯ ! ππ₯
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Or equivalently as: π¦ !! + ππ¦β² + ππ¦ = 0 The same process of establishing the CHARACTERISTIC EQUATION as our first step remains the same, such that: π ! + ππ + π = 0 The characteristic equation of a SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION is represented as a quadratic function, therefore, we can use the quadratic formula to solve for the roots of characteristic equation, such that:
π!,!
βπ Β± π! β 4π = 2
Once these roots are defined, we will deploy our knowledge of the three cases of solutions that we mentioned earlier.
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HOMOGENEOUS DIFFERENTIAL EQUATIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The particular solution of the differential equation, given the stated initial conditions, is best represented as: π¦ !! + 5π¦ ! + 4π¦ = 0 Given: π¦ 0 =1 π¦ ! 0 = β7 A. π¦! π₯ = βπ !! β 2π !!! B. π¦! π₯ = βπ !! + 2π !!! C. π¦! π₯ = π !! β 2π !!! D. π¦! π₯ = π !! + 2π !!!
SOLUTION: The TOPIC of SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with CONSTANT COEFFICIENTS can be referenced under the topic of
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DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A differential equation is considered a HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. If this is the case, and the sum of all these terms is equal to zero, then we employ the standard process in solving the HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION. In this problem, we are given the function: π¦ !! + 5π¦ ! + 4π¦ = 0 This is a SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATION. When we attempt to pursue a solution for any HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION, everything will start with a CHARACTERISTIC POLYNOMIAL, or what we are probably more comfortable calling it, a CHARACTERISIC EQUATION. The FORMULA for the CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, of a DIFFERENTIAL EQUATION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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This CHARACTERISTIC EQUATION is simply the polynomial formed by replacing all the derivatives with an βrβ variable raised to the power of their respective derivatives, such that: π π = π! π ! + π!!! π !!! + β― + π! π + π! This is presented in a very complicated manner in the NCEES Reference Handbook, but at the end of the day, it all comes down to this, given: π¦ !! + ππ¦β² + ππ¦ = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: π ! + ππ + π = 0 In route to defining our PARTICULAR SOLUTION, we first need to determine the general HOMOGENEOUS SOLUTION. We will then apply our INITIAL CONDITIONS to further establish the details we need to conclude the final form of our PARTICULAR SOLUTION. The FORMULA for the CHARACTERISTIC EQUATION of a SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The characteristic equation is given in the standard form as: π ! + ππ + π = 0 Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: π ! + 5π + 4 = 0 Where: β’ π=5 β’ π=4 The FORMULAS highlighting the various FORMS of SOLUTIONS for SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When EVALUATING the ROOTS of the CHARACTERISTIC EQUATION, there will be generally 3 cases in which we may encounter, they are: 1. REAL AND DISTINCT ROOTS 2. REAL AND EQUAL ROOTS 3. COMPLEX ROOTS
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We classify each of the roots, and further the form of the solution, based on the elements of the DISCRIMINANT based on the CHARACTERISTIC EQUATION. Developing a relationship between our coefficients βaβ and βbβ, we find that: π! = 5
!
= 25
And: 4 π = 4 4 = 16 Concluding that: 25 > 16 Or: π! > 4 π This tells us that we can classify the solution as OVERDAMPED, and that the ROOTS are characterized as REAL AND DISTINCT. Factoring the quadratic expression, we find the characteristic equation can be rewritten in factored terms as: π+1 π+4 =0
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Solving for the roots of the polynomials, we get: π! = β1 π! = β4 We know that for an OVERDAMPED solution, each distinct real root is represented as: π! = πΆ! π !! ! If the roots are real and the same, the solution is: π¦! π₯ = πΆ! π !! ! + πΆ! π₯π !! ! + β― + πΆ! π₯ !!! π !! ! Plugging in the calculated values for the roots, we find the general solution is expressed as: π¦! π₯ = πΆ! π !! + πΆ! π !!! Now that we have the general HOMOGENEOUS SOLUTION of the equation, we can use the initial conditions to solve of the particular solution. We are given the INITIAL CONDITIONS: π¦ 0 =1 π¦ ! 0 = β7
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In order to solve for the two coefficients, we will set up two equations, where we plug in each initial condition, and use a linear system of equations to solve for the value of each coefficient. First we take the general solution and use the initial condition π¦(0) = 1, giving us: π¦! π₯ = πΆ! π !! + πΆ! π !!! Plugging in our values to the general solution we find that: 1 = πΆ! π !(!) + πΆ! π !!(!) Which simplifies to: 1 = πΆ! + πΆ! In order to use the second initial condition, we need to define the derivative of the general solution. Doing so gives us: π¦!! π₯ = βπ₯πΆ! π !! β 4πΆ! π !!! We can now plug in the initial condition π¦β 0 = β7, giving us: β7 = βπΆ! π !
!
β 4πΆ! π !!
!
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Which simplifies to: β7 = βπΆ! β 4πΆ! With two equations and two unknowns now established, we can set up our system of equations to solve for the two undefined constants. There are many ways to solve this system, in which, we will go down the route of first summing both of the equations together to isolate πΆ! , such that: 1 = πΆ! + πΆ! β7 = βπΆ! β 4πΆ! SUMMING these two equations we get: 1 + β7 = πΆ! + βπΆ! + πΆ! + (β4πΆ! ) Simplifying: β6 = β3πΆ! Solving for πΆ! : πΆ! = 2
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Plugging the value for πΆ! into the first equation, we can solve for πΆ! , such that: 1 = πΆ! + 2 Rearranging and solving for πΆ! , we get: πΆ! = β1 Plugging the values for πΆ! and πΆ! into the general solution, we find that the PARTICULAR SOLUTION is: π¦! π₯ = βπ !! + 2π !!! The correct answer choice is B. π²π‘ π± = βπ!π± + ππ!ππ±
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CONCEPT INTRO: A DIFFERENTIAL EQUATION is considered HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. If this is the case, and the sum of all these terms is equal to zero, then we employ the standard process in solving a HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION. A HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION can generally be written as: π! π¦ ππ¦ + π + ππ¦ = 0 ππ₯ ! ππ₯ Or: π¦ !! + ππ¦β² + ππ¦ = 0
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Both of these expressions are equivalent, the only difference being is that in the latter, we replaced the various derivatives strewn throughout the former formula with more simple single y-term variables. Illustrating this further, taking the original general formula above, we turned: π! π¦ ππ¦ + π + ππ¦ = 0 ππ₯ ! ππ₯ In to: π¦ !! + ππ¦β² + ππ¦ = 0 By replacing: π! π¦ = π¦β²β² ππ₯ ! ππ¦ = π¦β² ππ₯ Thatβs all, nothing complex to it, however, you may see these types of equations presented in both ways, so be comfortable with each. When we attempt to pursue a solution for any HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION, everything will start with a CHARACTERISTIC POLYNOMIAL, or what we are probably more comfortable calling it, a CHARACTERISIC EQUATION.
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The FORMULA for the CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, of a DIFFERENTIAL EQUATION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This CHARACTERISTIC EQUATION is simply the polynomial formed by replacing all the derivatives with an βrβ variable raised to the power of that respective derivative, such that: π π = π! π ! + π!!! π !!! + β― + π! π + π! This is presented in a very complicated manner in the NCEES Reference Handbook, but at the end of the day, this is what it comes down to, given: π¦ !! + ππ¦β² + ππ¦ = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: π ! + ππ + π = 0 HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS are most easily solved by finding the βπβ roots of the characteristic polynomial π(π). Each root of the CHARACTERISTIC EQUATION, π! , appears in the HOMOGENEOUS SOLUTION, π¦! π₯ , in the form π !! ! .
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When working with HOMOGENEOUS DIFFERENTIAL EQUATIONS to the FIRST ORDER, we will use the standard process illustrated in previous reviews. However, when presented a HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION, the solution process is different, yet, not any more complex. The DRIVING FORMULAS highlighting the GENERAL SOLUTIONS for SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. There are THREE CASES that we will be concerned with when EVALUATING the ROOTS of the CHARACTERISTIC EQUATION, they are: 1. REAL AND DISTINCT ROOTS 2. REAL AND EQUAL ROOTS 3. COMPLEX ROOTS We classify each of the roots, and further the basic solution, based on the DISCRIMINANT of the CHARACTERISTIC EQUATION.
1.
REAL AND DISTINCT ROOTS
A REAL and DISTINCT ROOT is characterized by the relationship: π! > 4π
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Each distinct real root is represented as π! = πΆ! π !! ! . The STANDARD FORM for the HOMOGENEOUS SOLUTION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 30 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When we are dealing with a HOMOGENEOUS DIFFERENTIAL EQUATION and the ROOTS are both REAL AND DISTINCT, which will be the most common case, the HOMOGENEOUS SOLUTION can be generally written as: π¦! π₯ + πΆ! π !! ! + πΆ! π !! ! + β― + πΆ! π !! ! + β― + πΆ! π !! ! Where: β’ π! is the ππ‘β distinct root of the CHARACTERISTIC POLYNOMIAL π(π₯) β’ πΆ! , πΆ! , πΆ! , β¦ , πΆ! are all real constants β’ πΆ! β 0 This solution is of the OVERDAMPED form.
2.
REAL AND EQUAL ROOTS
A REAL AND EQUAL ROOT is characterized by the relationship: π! = 4π
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The two roots are real and the same, such that they can be considered the same variable. Repeated real roots are represented as π! = π! , where π! and π! are real numbers. If the root π! = π! , then the second term πΆ! π !! ! is replaced with πΆ! π₯π !! ! β¦note the added x variable and the exponent of π! . If the roots are real and the same, the solution is: π¦! π₯ = (πΆ! + πΆ! π₯ + β― + πΆ! π₯ !!! )π !" This solution is of the CRITICALLY DAMPED form.
3.
COMPLEX ROOTS
A COMPLEX ROOT is characterized by the relationship: π! < 4π The two complex roots are imaginary and represented as: π! = πΌ + ππ½ π! = πΌ β ππ½
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If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions, such that: π¦! π₯ = π !" (C! sin π½π₯ + πΆ! cos π½π₯) Where: β’ The real part of the complex root, βπβ, appears in the exponent, such that:
πΌ=β
π 2
β’ The coefficient of the imaginary part, βπβ, appears in the sine and cosine terms of the general formula, such that: 4π β π! π½= 2 This solution is of the UNDERDAMPED form. When it comes down to developing the structure for the HOMOGENEOUS SOLUTION, it all starts with determining the ROOTS after defining the CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQAUTION.
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FIRST ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS: The TOPIC of FIRST ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with CONSTANT COEFFICEINTS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A FIRST ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION is a differential equation in which each term contains either the function or one of its derivatives. Generally this type of DIFFERENTIAL EQUATION can be written in the form: ππ¦ + ππ¦ = 0 ππ₯ Or in CHARACTERISTIC FORM as: π+π =0 Where: a = real constant
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Thereβs not much to be done to determine the roots of this CHARACTERISTIC EQUATION, in fact, it simply comes down to this: π = βπ From this standard form of the differential equation, the solution can be generally represented as: π¦ = πΆπ !!" Where: β’ πΆ is a constant that satisfies the initial conditions
SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS: The TOPIC of SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with CONSTANT COEFFICIENTS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with constant coefficients are commonly represented in the standard form: π! π¦ ππ¦ + π + ππ¦ = 0 ππ₯ ! ππ₯
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Or equivalently as: π¦ !! + ππ¦β² + ππ¦ = 0 The same process of establishing the CHARACTERISTIC EQUATION as our first step remains the same, such that: π ! + ππ + π = 0 The characteristic equation of a SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION is represented as a quadratic function, therefore, we can use the quadratic formula to solve for the roots of characteristic equation, such that:
π!,!
βπ Β± π! β 4π = 2
Once these roots are defined, we will deploy our knowledge of the three cases of solutions that we mentioned earlier.
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HOMOGENEOUS DIFFERENTIAL EQUATIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The particular solution of the differential equation, given the stated initial conditions, is best represented as: π¦ !! + 5π¦ ! + 4π¦ = 0 Given: π¦ 0 =1 π¦ ! 0 = β7 A. π¦! π₯ = βπ !! β 2π !!! B. π¦! π₯ = βπ !! + 2π !!! C. π¦! π₯ = π !! β 2π !!! D. π¦! π₯ = π !! + 2π !!!
SOLUTION: The TOPIC of SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS with CONSTANT COEFFICIENTS can be referenced under the topic of
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DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A differential equation is considered a HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. If this is the case, and the sum of all these terms is equal to zero, then we employ the standard process in solving the HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION. In this problem, we are given the function: π¦ !! + 5π¦ ! + 4π¦ = 0 This is a SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATION. When we attempt to pursue a solution for any HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATION, everything will start with a CHARACTERISTIC POLYNOMIAL, or what we are probably more comfortable calling it, a CHARACTERISIC EQUATION. The FORMULA for the CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, of a DIFFERENTIAL EQUATION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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This CHARACTERISTIC EQUATION is simply the polynomial formed by replacing all the derivatives with an βrβ variable raised to the power of their respective derivatives, such that: π π = π! π ! + π!!! π !!! + β― + π! π + π! This is presented in a very complicated manner in the NCEES Reference Handbook, but at the end of the day, it all comes down to this, given: π¦ !! + ππ¦β² + ππ¦ = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: π ! + ππ + π = 0 In route to defining our PARTICULAR SOLUTION, we first need to determine the general HOMOGENEOUS SOLUTION. We will then apply our INITIAL CONDITIONS to further establish the details we need to conclude the final form of our PARTICULAR SOLUTION. The FORMULA for the CHARACTERISTIC EQUATION of a SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The characteristic equation is given in the standard form as: π ! + ππ + π = 0 Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: π ! + 5π + 4 = 0 Where: β’ π=5 β’ π=4 The FORMULAS highlighting the various FORMS of SOLUTIONS for SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced under the topic of DIFFERENTIAL EQUATIONS on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When EVALUATING the ROOTS of the CHARACTERISTIC EQUATION, there will be generally 3 cases in which we may encounter, they are: 1. REAL AND DISTINCT ROOTS 2. REAL AND EQUAL ROOTS 3. COMPLEX ROOTS
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We classify each of the roots, and further the form of the solution, based on the elements of the DISCRIMINANT based on the CHARACTERISTIC EQUATION. Developing a relationship between our coefficients βaβ and βbβ, we find that: π! = 5
!
= 25
And: 4 π = 4 4 = 16 Concluding that: 25 > 16 Or: π! > 4 π This tells us that we can classify the solution as OVERDAMPED, and that the ROOTS are characterized as REAL AND DISTINCT. Factoring the quadratic expression, we find the characteristic equation can be rewritten in factored terms as: π+1 π+4 =0
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Solving for the roots of the polynomials, we get: π! = β1 π! = β4 We know that for an OVERDAMPED solution, each distinct real root is represented as: π! = πΆ! π !! ! If the roots are real and the same, the solution is: π¦! π₯ = πΆ! π !! ! + πΆ! π₯π !! ! + β― + πΆ! π₯ !!! π !! ! Plugging in the calculated values for the roots, we find the general solution is expressed as: π¦! π₯ = πΆ! π !! + πΆ! π !!! Now that we have the general HOMOGENEOUS SOLUTION of the equation, we can use the initial conditions to solve of the particular solution. We are given the INITIAL CONDITIONS: π¦ 0 =1 π¦ ! 0 = β7
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In order to solve for the two coefficients, we will set up two equations, where we plug in each initial condition, and use a linear system of equations to solve for the value of each coefficient. First we take the general solution and use the initial condition π¦(0) = 1, giving us: π¦! π₯ = πΆ! π !! + πΆ! π !!! Plugging in our values to the general solution we find that: 1 = πΆ! π !(!) + πΆ! π !!(!) Which simplifies to: 1 = πΆ! + πΆ! In order to use the second initial condition, we need to define the derivative of the general solution. Doing so gives us: π¦!! π₯ = βπ₯πΆ! π !! β 4πΆ! π !!! We can now plug in the initial condition π¦β 0 = β7, giving us: β7 = βπΆ! π !
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β 4πΆ! π !!
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Which simplifies to: β7 = βπΆ! β 4πΆ! With two equations and two unknowns now established, we can set up our system of equations to solve for the two undefined constants. There are many ways to solve this system, in which, we will go down the route of first summing both of the equations together to isolate πΆ! , such that: 1 = πΆ! + πΆ! β7 = βπΆ! β 4πΆ! SUMMING these two equations we get: 1 + β7 = πΆ! + βπΆ! + πΆ! + (β4πΆ! ) Simplifying: β6 = β3πΆ! Solving for πΆ! : πΆ! = 2
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Plugging the value for πΆ! into the first equation, we can solve for πΆ! , such that: 1 = πΆ! + 2 Rearranging and solving for πΆ! , we get: πΆ! = β1 Plugging the values for πΆ! and πΆ! into the general solution, we find that the PARTICULAR SOLUTION is: π¦! π₯ = βπ !! + 2π !!! The correct answer choice is B. π²π‘ π± = βπ!π± + ππ!ππ±
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