71.1 Homogenous Differential Equations Problem Set

HOMOGENEOUS DIFFERENTIAL EQUATIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: The particular solution of the differential equation, given the initial conditions stated, is best represented as: 𝑦 !! + 4𝑦 = 0 Given: 𝑦 0 =4 𝑦! 0 = 6 A. 𝑦! 𝑥 = −3 sin 𝑥 + 4cos 𝑥 B. 𝑦! (𝑥) = 3 sin 2𝑥 − 4cos 2𝑥 C. 𝑦! 𝑥 = −3 sin 2𝑥 − 4cos 𝑥 D. 𝑦! (𝑥) = 3 sin 2𝑥 + 4 cos 2𝑥



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PROBLEM 2: The particular solution of the differential equation, given the initial conditions stated, is most close to: 𝑦 !! − 4𝑦 ! + 4𝑦 = 0 Given: 𝑦 0 =4 𝑦! 0 = 5 A. 𝑦! 𝑥 = 4𝑒 !! − 3𝑥𝑒 !! B. 𝑦! 𝑥 = 4𝑒 !! + 3𝑥𝑒 !! C. 𝑦! 𝑥 = −4𝑒 !! − 3𝑥𝑒 !! D. 𝑦! 𝑥 = −4𝑒 !! + 3𝑥𝑒 !!



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PROBLEM 3: The general solution to the following second order differential equation, given that 𝐴 and 𝐵 are constants, is best written as: 𝑦 !! + 9𝑦 = 0 A. 𝑦 𝑥 = 𝐴𝑒 !! + 𝐵𝑥𝑒 !!! B. 𝑦 𝑥 = 𝐴𝑒 !! + 𝐵𝑒 !!! C. 𝑦 𝑥 = 𝐴𝑒 ! − 𝐵𝑥𝑒 !! D. 𝑦 𝑥 = 𝐴 cos 3𝑥 + 𝐵 sin 3𝑥



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PROBLEM 4: The particular solution of the differential equation, given the initial conditions, can be written best as: 𝑦 !! + 4𝑦′ + 4𝑦 = 0 Given: 𝑦 0 =1 𝑦! 0 = 0 A. 𝑦! 𝑥 = 1 − 2𝑥 𝑒 !! B. 𝑦! 𝑥 = 2 − 𝑥 𝑒 !!! C. 𝑦! 𝑥 = 2 + 𝑥 𝑒 !!! D. 𝑦! 𝑥 = 1 + 2𝑥 𝑒 !!!



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PROBLEM 5: The general solution of the differential equation is best represented by: 𝑦 !! − 8𝑦 ! + 16𝑦 = 0 A. 𝑦! 𝑥 = 𝐶! 𝑒 !! B. 𝑦! 𝑥 = 𝐶! + 𝐶! 𝑥 𝑒 !! C. 𝑦! 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 !! D. 𝑦! 𝑥 = 𝐶! 𝑒 !! + 𝐶! 𝑒 !!



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HOMOGENEOUS DIFFERENTIAL EQUATIONS | SOLUTIONS SOLUTION 1: 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. 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: 𝑦 !! + 4𝑦 = 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



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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. This CHARACTERISTIC EQUATION is simply the polynomial formed by replacing all the derivatives with an “r” variable raised to the power of their respective derivatives, such that: 𝑃 𝑟 = 𝑏! 𝑟 ! + 𝑏!!! 𝑟 !!! + ⋯ + 𝑏! 𝑟 + 𝑏! 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: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0 In route to defining our PARTICULAR SOLUTION, we first need to determine the general HOMOGENEOUS SOLUTION.



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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. The characteristic equation is given in the standard form as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: 𝑟 ! + 0𝑟 + 4 = 0 Where: • 𝑎=0 • 𝑏=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.



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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 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 “a” and “b”, we find that: 𝑎! = 0

!

=0

And: 4 𝑏 = 4 4 = 16 Concluding that: 0 < 16 Or: 𝑎! < 4 𝑏



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This tells us that we can classify the solution as UNDERDAMPED, and that the ROOTS are characterized as COMPLEX. The two complex roots are imaginary and represented as: 𝑟! = 𝛼 + 𝑖𝛽 𝑟! = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions, such that: 𝑦! 𝑥 = 𝑒 !" (C! sin 𝛽𝑥 + 𝐶! cos 𝛽𝑥) Where: • The real part of the complex root, “𝑎”, appears in the exponent, such that:

𝛼=−

𝑎 2

• The coefficient of the imaginary part, “𝑏”, appears in the sine and cosine terms of the general formula, such that: 4𝑏 − 𝑎! 𝛽= 2



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So let’s determine both 𝛼 and 𝛽:

𝛼=−

0 =0 2

𝛽=

4(4) − (0)! =2 2

And:

Plugging in these values in to the general form of an UNDERDAMPED SOLUTION, we get: 𝑦! 𝑥 = 𝑒

! !

(C! sin 2𝑥 + C! cos 2𝑥)

Simplifying a bit: 𝑦! 𝑥 = (1)(𝐶! sin 2𝑥 + C! cos 2𝑥) And further, we find the general solution of the equation is: 𝑦! (𝑥) = 𝐶! sin 2𝑥 + C! cos 2𝑥 Now that we have the general solution of the equation, we can use the initial conditions to solve for the particular solution.



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We are given the INITIAL CONDITIONS: 𝑦 0 =4 𝑦! 0 = 6 First we take the general solution and use the initial condition 𝑦 0 = 4, giving us: 4 = 𝐶! sin 2(0) + C! cos 2(0) Which simplifies to: 4 = 𝐶! sin 0 + 𝐶! cos 0 We know that the sin 0 = 0, so: 4 = 𝐶! cos 0 Rearranging and solving for 𝐶! , we get: 4 = 𝐶! In order to deploy the second initial condition, we need to first calculate the derivative of the general solution, which is: 𝑦′! 𝑥 = 𝐶! 2cos 2𝑥 − 2C! sin 2𝑥



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Plugging in the initial condition 𝑦 ! 0 = 6, we get: 6 = 𝐶! 2cos 2(0) − 2(4)sin 2(0) Again, since sin 0 = 0, we can simplify the formula to read: 6 = 𝐶! 2 (1) Rearranging and solving for the missing constant 𝐶! , we find: 𝐶! = 3 With our unknown constants 𝐶! and 𝐶! now defined, we can plug them in to the general solution to define the PARTICULAR SOLUTION as: 𝑦! 𝑥 = 3 sin 2𝑥 + 4cos 2𝑥 The correct answer choice is D. 𝐲𝐡 (𝐱) = 𝟑 𝐬𝐢𝐧 𝟐𝐱 + 𝟒𝐜𝐨𝐬 𝟐𝐱

SOLUTION 2: 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.



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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: 𝑦 !! − 4𝑦 ! + 4𝑦 = 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 “r” variable raised to the power of their respective derivatives, such that: 𝑃 𝑟 = 𝑏! 𝑟 ! + 𝑏!!! 𝑟 !!! + ⋯ + 𝑏! 𝑟 + 𝑏! 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: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 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: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: 𝑟 ! − 4𝑟 + 4 = 0 Where: • 𝑎 = −4 • 𝑏=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 “a” and “b”, we find that: 𝑎! = −4

!

= 16

And: 4 𝑏 = 4 4 = 16 Concluding that 16 = 16 Or: 𝑎! = 4 𝑏 This tells us that we can classify the solution as CRITICALLY DAMPED, and that the ROOTS are characterized as REAL and EQUAL. Repeated real roots are represented as 𝑟! = 𝑟! , where 𝑟! and 𝑟! are real numbers. Factoring the quadratic expression, we find the characteristic equation can be rewritten in factored terms as: 𝑟−2 𝑟−2 =0

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Solving for the roots of the polynomials, we find the roots are both: 𝑟!,! = 2 If the roots 𝑟! = 𝑟! , then the second term, 𝐶! 𝑒 !! ! , is replaced with 𝐶! 𝑥𝑒 !! ! …take note of the additional x variable and the 𝑟! exponent. When roots are REAL AND UNIQUE, the general solution is written as: 𝑦! 𝑥 = (𝐶! + 𝐶! 𝑥 + ⋯ + 𝐶! 𝑥 !!! )𝑒 !" Plugging in the values that we have defined for the roots, we find that the general solution is expressed as: 𝑦! 𝑥 = 𝐶! 𝑒 !! + 𝐶! 𝑥𝑒 !! Now that we have the general solution of the equation, we can use the initial conditions to solve for the particular solution. We are given the INITIAL CONDITIONS: 𝑦 0 =4 𝑦! 0 = 5 First we take the general solution and use the initial condition 𝑦 0 = 4, giving us: 4 = 𝐶! 𝑒 !(!) + 𝐶! (0)𝑒 !(!)

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The second term disappears, giving us: 4 = 𝐶! 𝑒 ! + 0 Rearranging and solving for 𝐶! , we get: 𝐶! = 4 Let’s now determine our second unknown constant, 𝐶! . We will use our second initial condition, however, to do so, we must first determine the first derivative of the general solution, which is: 𝑦!! 𝑥 = 2𝐶! 𝑒 !! + 𝐶! (2𝑥𝑒 !! + 𝑒 !! ) Plugging in the initial condition 𝑦 ! 0 = 5, we get: 5 = 2 4 𝑒 !(!) + 𝐶! 2 0 𝑒 !

!

+ 𝑒!

!

Simplifying this equation, we rewrite it as: 5 = 8 1 + 𝐶! 0 + 1 Rearranging and solving for the unknown constant 𝐶! , we get: 𝐶! = −3

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With our unknown constants 𝐶! and 𝐶! now defined, we can plug them in to the general solution to define the PARTICULAR SOLUTION as: 𝑦! 𝑥 = 4𝑒 !! − 3𝑥𝑒 !! The correct answer choice is A. 𝐲𝐡 𝐱 = 𝟒𝐞𝟐𝐱 − 𝟑𝐱𝐞𝟐𝐱

SOLUTION 3: 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. 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: 𝑦 !! + 9𝑦 = 0 This is a SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATION.

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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. This CHARACTERISTIC EQUATION is simply the polynomial formed by replacing all the derivatives with an “r” variable raised to the power of their respective derivatives, such that: 𝑃 𝑟 = 𝑏! 𝑟 ! + 𝑏!!! 𝑟 !!! + ⋯ + 𝑏! 𝑟 + 𝑏! 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: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0



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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. The characteristic equation is given in the standard form as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: 𝑟 ! + 0𝑟 + 9 = 0 Where: • 𝑎=0 • 𝑏=9 The FORMULAS highlighting the various FORMS of SOLUTIONS for SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced

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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 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 “a” and “b”, we find that: 𝑎! = 0

!

=0

And: 4 𝑏 = 4 9 = 36 Concluding that: 0 < 36



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Or: 𝑎! < 4 𝑏 This tells us that we can classify the solution as UNDERDAMPED, and that the ROOTS are characterized as COMPLEX. The two complex roots are imaginary and represented as: 𝑟! = 𝛼 + 𝑖𝛽 𝑟! = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions, such that: 𝑦! 𝑥 = 𝑒 !" (C! sin 𝛽𝑥 + 𝐶! cos 𝛽𝑥) Where: • The real part of the complex root, “𝑎”, appears in the exponent, such that:

𝛼=−

𝑎 2

• The coefficient of the imaginary part, “𝑏”, appears in the sine and cosine terms of the general formula, such that:



4𝑏 − 𝑎! 𝛽= 2 Made with

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So let’s determine both 𝛼 and 𝛽:

𝛼=−

0 =0 2

𝛽=

4(9) − (0)! =3 2

And:

Plugging in these values in to the general form of an underdamped solution, we get: 𝑦! 𝑥 = 𝑒

! !

(C! sin 3𝑥 + C! cos 3𝑥)

Simplifying a bit: 𝑦! 𝑥 = (1)(𝐶! sin 3𝑥 + C! cos 3𝑥) And further, we find the general solution of the equation is: 𝑦! (𝑥) = 𝐶! sin 3𝑥 + C! cos 3𝑥 The fact that we are told that our constants are A and B, rather than 𝐶! and C! makes no difference in our solution, therefore, we can rewrite this general solution to read: 𝑦! (𝑥) = 𝐴 sin 3𝑥 + B cos 3𝑥



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The correct answer choice is D. 𝐲𝐡 (𝐱) = 𝐀 𝐬𝐢𝐧 𝟑𝐱 + 𝐁 𝐜𝐨𝐬 𝟑𝐱

SOLUTION 4: 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. 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: 𝑦 !! + 4𝑦 ! + 4𝑦 = 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. Made with

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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 “r” variable raised to the power of their respective derivatives, such that: 𝑃 𝑟 = 𝑏! 𝑟 ! + 𝑏!!! 𝑟 !!! + ⋯ + 𝑏! 𝑟 + 𝑏! 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: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 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.



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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. The characteristic equation is given in the standard form as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: 𝑟 ! + 4𝑟 + 4 = 0 Where: • 𝑎=4 • 𝑏=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.



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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 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 “a” and “b”, we find that: 𝑎! = 4

!

= 16

And: 4 𝑏 = 4 4 = 16 Concluding that: 16 = 16 Or: 𝑎! = 4 𝑏



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This tells us that we can classify the solution as CRITICALLY DAMPED, and that the ROOTS are characterized as COMPLEX. We know that 16 = 16, and can class the solution as critically damped, and that the roots are characterized as REAL AND UNIQUE. Repeated real roots are represented as 𝑟! = 𝑟! , where 𝑟! and 𝑟! are real numbers. Factoring the quadratic expression, we find the characteristic equation can be rewritten in factored terms as: 𝑟+2 𝑟+2 =0 Solving for the roots of the polynomials, we find the roots are both: 𝑟!,! = −2 If the roots 𝑟! = 𝑟! , then the second term, 𝐶! 𝑒 !! ! , is replaced with 𝐶! 𝑥𝑒 !! ! …take note of the additional x variable and the 𝑟! exponent. When roots are REAL AND UNIQUE, the general solution is written as: 𝑦! 𝑥 = (𝐶! + 𝐶! 𝑥 + ⋯ + 𝐶! 𝑥 !!! )𝑒 !"



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Plugging in the values that we have defined for the roots, we find that the general solution is expressed as: 𝑦! 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑥𝑒 !!! Now that we have the general solution of the equation, we can use the initial conditions to solve for the particular solution. We are given the INITIAL CONDITIONS: 𝑦 0 =1 𝑦! 0 = 0 First we take the general solution and use the initial condition 𝑦 0 = 1, giving us: 1 = 𝐶! 𝑒 !!(!) + 𝐶! (0)𝑒 !!(!) The second term disappears, giving us: 1 = 𝐶! 𝑒 ! + 0 Rearranging and solving for 𝐶! , we get: 𝐶! = 1 Let’s now determine our second unknown constant, 𝐶! .

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We will use our second initial condition, however, to do so, we must first determine the first derivative of the general solution, which is: 𝑦!! 𝑥 = 𝐶! + 𝐶! 𝑥 𝑒 !!! −2 + 𝐶! (𝑒 !!! ) Plugging in the initial condition 𝑦 ! 0 = 0, along with the value we have calculated for 𝐶! , we get: 0 = 1 + 𝐶! 0

𝑒 !!

!

−2 + 𝐶! (𝑒 !!! )

Simplifying the equation, we can rewrite it as: 0 = −2 1 + 𝐶! Rearranging and solving for the unknown constant 𝐶! , we get: 𝐶! = 2 Plugging the values for 𝐶! and 𝐶! into the general solution, we find the PARTICULAR SOLUTION is: 𝑦! 𝑥 = 𝑒 !!! + 2𝑥𝑒 !!! We don’t see this solution as an option given to us in the problem statement. Many would panic at this point with all the work that has been done…but don’t stop, don’t clam up, look for a potentially slightly different way to present this solution, the NCEES needs to prove that you are willing to take this additional step.

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This equation can be written as: 𝑦! 𝑥 = 1 + 2𝑥 𝑒 !!! The correct answer choice is D. 𝐲𝐡 𝐱 = 𝟏 + 𝟐𝐱 𝐞!𝟐𝐱

SOLUTION 5: 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. 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: 𝑦 !! − 8𝑦 ! + 16𝑦 = 0 This is a SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATION.

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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. This CHARACTERISTIC EQUATION is simply the polynomial formed by replacing all the derivatives with an “r” variable raised to the power of their respective derivatives, such that: 𝑃 𝑟 = 𝑏! 𝑟 ! + 𝑏!!! 𝑟 !!! + ⋯ + 𝑏! 𝑟 + 𝑏! 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: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 0 The CHARACTERISTIC POLYNOMIAL, or CHARACTERISTIC EQUATION, is written as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0



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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. The characteristic equation is given in the standard form as: 𝑟 ! + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: 𝑟 ! − 8𝑟 + 16 = 0 Where: • 𝑎 = −8 • 𝑏 = 16 The FORMULAS highlighting the various FORMS of SOLUTIONS for SECOND ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced

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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 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 “a” and “b”, we find that: 𝑎! = −8

!

= 64

And: 4 𝑏 = 4 16 = 64 Concluding that: 64 = 64



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Or: 𝑎! = 4 𝑏 This tells us that we can classify the solution as CRITICALLY DAMPED, and that the ROOTS are characterized as REAL AND UNIQUE. Repeated real roots are represented as 𝑟! = 𝑟! , where 𝑟! and 𝑟! are real numbers. Factoring the quadratic expression, we find the characteristic equation can be rewritten in factored terms as: 𝑟−4 𝑟−4 =0 Solving for the roots of the polynomials, we find the roots are both: 𝑟!,! = 4 If the roots 𝑟! = 𝑟! , then the second term, 𝐶! 𝑒 !! ! , is replaced with 𝐶! 𝑥𝑒 !! ! …take note of the additional x variable and the 𝑟! exponent. When roots are REAL AND UNIQUE, the general solution is written as: 𝑦! 𝑥 = (𝐶! + 𝐶! 𝑥 + ⋯ + 𝐶! 𝑥 !!! )𝑒 !"



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Plugging in the values that we have defined for the roots, we find that the general solution is expressed as: 𝑦! 𝑥 = 𝐶! 𝑒 !! + 𝐶! 𝑥𝑒 !! We don’t see this solution as an option given to us in the problem statement. Many would panic at this point with all the work that has been done…but don’t stop, don’t clam up, look for a potentially slightly different way to present this solution. This equation can be written as: 𝑦! 𝑥 = (𝐶! + 𝐶! 𝑥)𝑒 !! The correct answer is B. 𝐲𝐡 𝐱 = 𝐂𝟏 + 𝐂𝟐 𝐱 𝐞𝟒𝐱



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