73.1 Nonhomogeneous Differential Equations Problem Set

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

PROBLEM 1: The general solution of the differential equation is best represented as: 𝑦 !! + 3𝑦 ! − 4𝑦 = 3𝑒 !! !

A. 𝑦 𝑥 = 𝐶! 𝑒 !!! − 𝐶! 𝑒 ! − ! 𝑒 !! !

B. 𝑦 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 ! − ! 𝑒 !! !

C. 𝑦 𝑥 = 𝐶! 𝑒 !!! − 𝐶! 𝑒 ! + 𝑒 !! ! !

D. 𝑦 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 ! + ! 𝑒 !!



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PROBLEM 2: The statement that is TRUE when it comes to writing the complete solution for a nonhomogeneous differential equation is: A. 𝑦 𝑥 = 𝑦! (𝑥) − 𝑦! (𝑥) B. 𝑦 𝑥 = 𝑦! (𝑥) + 𝑦! (𝑥) C. 𝑦 𝑥 = 𝑦!! (𝑥) + 𝑦!! (𝑥) D. 𝑦 𝑥 = 𝑦! (𝑥) + 𝑦!! (𝑥)



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PROBLEM 3: The complete solution of the differential equation below is best represented as: 𝑦 !! + 3𝑦 ! + 2𝑦 = 5𝑒 !! A. 𝑦 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 !! + B. 𝑦 𝑥 = 𝐶! 𝑒 !! + 𝐶! 𝑒 !!! + C. 𝑦 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 ! −

!" !

! !"

D. 𝑦 𝑥 = 𝐶! 𝑒 !! − 𝐶! 𝑒 !!! +



! !"

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𝑒 !! 𝑒 !!

𝑒 !!

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PROBLEM 4: The particular solution of the stated differential equation is best represented as: 𝑦 !! + 2𝑦 ! + 5𝑦 = cos(3𝑥) !

!

!

!

A. 𝑦! (𝑥) = − sin 3𝑥 − cos(3𝑥) B. 𝑦! (𝑥) =

! !"

sin 3𝑥 −

!

!

!

!

!

!

! !"

cos(3𝑥)

C. 𝑦! (𝑥) = ! sin 3𝑥 + ! cos(3𝑥) D. 𝑦! (𝑥) = sin 3𝑥 − cos(3𝑥)



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NONHOMOGENEOUS DIFFERENTIAL EQUATIONS | SOLUTIONS SOLUTION 1: The TOPIC of SECOND ORDER NONHOMOGENEOUS 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 HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. This same DIFFERENTIAL EQUATION can become of the NONHOMOGENEOUS form if the sum DOES NOT equal zero, but rather a nonzero forcing function of the INDEPENDENT VARIABLE. In formulaic terms, a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION can generally be written as: 𝑑! 𝑦 𝑑𝑦 + 𝑎 + 𝑏𝑦 = 𝑔(𝑥) 𝑑𝑥 ! 𝑑𝑥



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Or: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 𝑔(𝑥) The GENERAL FORM of the COMPLETE SOLUTION 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. The GENERAL SOLUTION, otherwise referred to as the COMPLETE SOLUTION, of a DIFFERENTIAL EQUATION is expressed as: 𝑦 𝑥 = 𝑦! 𝑥 + 𝑦! (𝑥) Where: • 𝑦! (𝑥) is the PARTICULAR SOLUTION with 𝑓 𝑥 present • 𝑦! (𝑥) is the HOMOGENEOUS SOLUTION corresponding to 𝑓 𝑥 = 0 It is important to always remember that the GENERAL SOLUTION for a NONHOMOGENEOUS DIFFERENTIAL EQUATION will be made up these two unique components, the PARTICULAR SOLUTION, which may also be referred to as the NONHOMOGENEOUS SOLUTION, and the HOMEGENEOUS SOLUTION. Now back to the problem statement, we are given the function: 𝑦 !! + 3𝑦 ! − 4𝑦 = 3𝑒 !!

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This DIFFERENTIAL EQUATION is both SECOND ORDER and NONHOMOGENEOUS. The process we will follow to define the solution for the function is as follows: 1. Define the HOMOGENEOUS SOLUTION 2. Define the PARTICULAR SOLUTION 3. Deploy the INITIAL CONDITIONS to determine any UNKNOWN COEFFICIENTS. Let’s determine the HOMOGENEOUS SOLUTION, 𝑦! 𝑥 : 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 determining the HOMOGENEOUS SOLUTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, we will ignore the DRIVING FUNCTION, g(x), on the right side of the equation and solve it as if it was HOMOGENEOUS. We will follow the same process outlined for HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS.



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As a quick review, we will proceed as follows: 1. Define the CHARACTERISTIC POLYNOMIAL (also known as the CHARACTERISTIC EQUATION) 2. Determine the ROOTS 3. Classify the ROOTS as: a. REAL and DISTINCT b. REAL and EQUAL c. COMPLEX 4. Write the GENERAL SOLUTION based on the CLASSIFICATION of the ROOTS Taking the initial function, let’s rewrite it, ignoring the driving function, giving us: 𝑦 !! + 3𝑦 ! − 4𝑦 = 0 This is now 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.



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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 This 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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Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: 𝑟 ! + 3𝑟 − 4 = 0 Where: • 𝑎=3 • 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 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.



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Developing a relationship between our coefficients “a” and “b”, we find that: 𝑎! = 3

!

=9

And: 4 𝑏 = 4 −4 = −16 Concluding that: 9 > −16 Or: 𝑎! > 4 𝑏 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: 𝑟+4 𝑟−1 =0 Solving for the roots of the polynomials, we get: 𝑟! = −4 𝑟! = 1

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We know that for an OVERDAMPED solution, each distinct real root is represented as: 𝑟! = 𝐶! 𝑒 !! ! If the roots are real and the same, the solution is: 𝑦! 𝑥 = 𝐶! 𝑒 !! ! + 𝐶! 𝑥𝑒 !! ! + ⋯ + 𝐶! 𝑥 !!! 𝑒 !! ! Plugging in the calculated values for the roots, we find the general solution is expressed as: 𝑦! 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 ! This is the HOMOGENEOUS SOLUTION to our COMPLETE SOLUTION. Let’s move on to defining the PARTICULAR SOLUTION. The PARTICULAR SOLUTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is defined by using the METHOD OF UNDETERMINED COEFFICIENTS. This is a method, which in practice, generally requires that we “guess” the PARTICULAR SOLUTION based on the DRIVING FUNCTION on the right side of the equation. We then will take this “guess”, DIFFERENTIATE it a few times, and plug those results back in to the left side of the equation in place of the original DERIVATIVE forms.

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At this point, we will take this new formula, combine the like terms to simplify it, and then set the UNDETERMINED COEFFICIENTS (which can also be thought simply as UNKNOWN COEFFICIENTS) on the left side of the equation to their corresponding values on the right side of the equation. This will provide us a SET OF EQUATIONS that we can then proceed in solving for each UNKNOWN, defining our PARTICULAR SOLUTION. Fortunately, on this exam, there will be no “guessing” needed as we are provided a TABLE highlighting the various FORMS of a PARTICULAR SOLUTION based on the corresponding DRIVING FUNCTION. This TABLE 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 few specific forms of the function 𝑓(𝑥), as it is being referred to as in the NCEES Reference Handbook, but we are using 𝑔(𝑥) for clarity purposes, and it’s resulting PARTICULAR SOLUTION, 𝑦! (𝑥), are defined as: 𝑓(𝑥)

𝑦! (𝑥)

𝐴

𝐵

𝐴𝑒 !"

𝐵𝑒 !" ; 𝛼 ≠ 𝑟!

𝐴! sin 𝜔𝑥 + 𝐴! cos 𝜔𝑥

𝐵! sin 𝜔𝑥 + 𝐵! cos 𝜔𝑥

While we are provided this table in the reference handbook, it is often not clear what the particular form of the solution should be.

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Here are a few guidelines that are not in the NCEES REFERENCE HANDBOOK, but may guide you when working problems involving the METHOD OF UNDETERMINED COEFFICIENTS: • If f x is constant, then y! = Cx • If f x is linear, then y! = B! x + B! • If f x is quadratic, then y! = Ax ! + Bx + C • If f x features e!" , then y! = Be!" • If f x features sin cx or cos cx , then y! = B! sin cx + B! cos(cx) The take away from the table is to match the DRIVING FUNCTION in the left column with the PARTICULAR SOLUTION function in the right column. To find the values of the UNDETERMINED COEFFICIENTS, substitute the assumed form of 𝑦! (𝑥) and it’s associated DERIVATIVES in to the differential equation and equate the UNKNOWN COEFFICIENTS based on the values that correspond to those terms on the opposite side of the equation.



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Laying out a rough process, we have: 1. Define the general PARTICULAR SOLUTION by observing the DRIVING FUNCTION and using the TABLE presented to us in the NCEES REFERENCE HANDBOOK 2. DIFFERENTIATE this generalized PARTICULAR SOLUTION up the maximum order of the equation we are solving. In this case, we are working a SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, so we will take up through the SECOND DERIVATIVE. 3. INSERT the DERIVATIVES in to the DIFFERENTIAL EQUATION replacing the original DERIVATIVE terms on the left side. 4. Combine LIKE terms 5. Set the UNDETERMINED COEFFICIENTS as they relate to the terms on the left side of the equation to the corresponding COEFFICIENTS on the right side of the equation and solve. So let’s start with the first step here. The driving function of our SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is: 3𝑒 !!



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Flipping over to the TABLE on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we find that: 𝑓(𝑥)

𝑦! (𝑥)

𝐴

𝐵

𝐴𝑒 !"

𝐵𝑒 !" ; 𝛼 ≠ 𝑟!

𝐴! sin 𝜔𝑥 + 𝐴! cos 𝜔𝑥

𝐵! sin 𝜔𝑥 + 𝐵! cos 𝜔𝑥

The second row represents our DRIVING FUNCTION, so our initial “guess” for our PARTICULAR SOLUTION can be written as: 𝑦! 𝑥 = 𝐵𝑒 !" We can further define this PARTICULAR SOLUTION by grabbing the constants from the original DRIVING FUNCTION, such that: 𝑦! 𝑥 = 𝐵𝑒 !! As it stands right now, 𝐵 is the only UNDETERMINED COEFFICIENTS. We now need to DIFFERENTIATE this generalized PARTICULAR SOLUTION up the maximum order of the equation we are solving. Again, in this case, we are working a SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, so we will take up through the SECOND DERIVATIVE, giving us: 𝑦! ′(𝑥) = 2𝐵𝑒 !!

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𝑦! ′′(𝑥) = 4𝐵𝑒 !! Our next move is to INSERT the DERIVATIVES in to the DIFFERENTIAL EQUATION replacing the original DERIVATIVE terms on the left side. Our original function reads: 𝑦 !! + 3𝑦 ! − 4𝑦 = 3𝑒 !! We will replace the terms on the left with our DIFFERENTIATED terms of our general PARTICULAR SOLUTION, such that: 𝑦!!! + 3𝑦!! − 4𝑦! = 3𝑒 !! Carrying out this substitution, we get: 4𝐵𝑒 !! + 3 2𝐵𝑒 !! − 4 𝐵𝑒 !! = 3𝑒 !! Simplifying, we get: 4𝐵 + 6𝐵 − 4𝐵 = 3 Combining the LIKE terms: 6𝐵 = 3



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Typically, this is the point where we would set the UNDETERMINED COEFFICIENTS as they relate to the terms on the left side of the equation to the corresponding COEFFICIENTS on the right side of the equation and solve. There isn’t much we need to do, we just need to solve for B, which gives us:

𝐵=

1 2

Therefore, our PARTICULAR SOLUTION is: !

𝑦! (𝑥) = ! 𝑒 !! Revisiting our HOMOGENEOUS SOLUTION, we derived: 𝑦! 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 ! The COMPLETE SOLUTION will be the combination of these two components, such that: 𝑦(𝑥) = 𝑦! (𝑥) + 𝑦! (𝑥) Which gives us: !

𝑦 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 ! + ! 𝑒 !! 𝟏

The correct answer choice is D. 𝐲 𝐱 = 𝐂𝟏 𝐞!𝟒𝐱 + 𝐂𝟐 𝐞𝐱 + 𝟐 𝐞𝟐𝐱

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SOLUTION 2: The TOPIC of SECOND ORDER NONHOMOGENEOUS 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 HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. This same DIFFERENTIAL EQUATION can become of the NONHOMOGENEOUS form if the sum DOES NOT equal zero, but rather a nonzero forcing function of the INDEPENDENT VARIABLE. In formulaic terms, a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION can generally be written as: 𝑑! 𝑦 𝑑𝑦 + 𝑎 + 𝑏𝑦 = 𝑔(𝑥) 𝑑𝑥 ! 𝑑𝑥 Or: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 𝑔(𝑥) The GENERAL FORM of the COMPLETE SOLUTION of a DIFFERENTIAL EQUATION can be referenced under the topic of DIFFERENTIAL EQUATIONS on

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page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The GENERAL SOLUTION, otherwise referred to as the COMPLETE SOLUTION, of a DIFFERENTIAL EQUATION is expressed as: 𝑦 𝑥 = 𝑦! 𝑥 + 𝑦! (𝑥) Where: • 𝑦! (𝑥) is the PARTICULAR SOLUTION with 𝑓 𝑥 present • 𝑦! (𝑥) is the HOMOGENEOUS SOLUTION corresponding to 𝑓 𝑥 = 0 It is important to always remember that the GENERAL SOLUTION for a NONHOMOGENEOUS DIFFERENTIAL EQUATION will be made up these two unique components, the PARTICULAR SOLUTION, which may also be referred to as the NONHOMOGENEOUS SOLUTION, and the HOMEGENEOUS SOLUTION. The correct answer choice is B. 𝐲 𝐱 = 𝐲𝐡 𝐱 + 𝐲𝐩 𝐱

SOLUTION 3: The TOPIC of SECOND ORDER NONHOMOGENEOUS 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 HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. This same DIFFERENTIAL EQUATION can become of the NONHOMOGENEOUS form if the sum DOES NOT equal zero, but rather a nonzero forcing function of the INDEPENDENT VARIABLE. In formulaic terms, a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION can generally be written as: 𝑑! 𝑦 𝑑𝑦 + 𝑎 + 𝑏𝑦 = 𝑔(𝑥) 𝑑𝑥 ! 𝑑𝑥 Or: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 𝑔(𝑥) The GENERAL FORM of the COMPLETE SOLUTION 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. The GENERAL SOLUTION, otherwise referred to as the COMPLETE SOLUTION, of a DIFFERENTIAL EQUATION is expressed as: 𝑦 𝑥 = 𝑦! 𝑥 + 𝑦! (𝑥)

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Where: • 𝑦! (𝑥) is the PARTICULAR SOLUTION with 𝑓 𝑥 present • 𝑦! (𝑥) is the HOMOGENEOUS SOLUTION corresponding to 𝑓 𝑥 = 0 It is important to always remember that the GENERAL SOLUTION for a NONHOMOGENEOUS DIFFERENTIAL EQUATION will be made up these two unique components, the PARTICULAR SOLUTION, which may also be referred to as the NONHOMOGENEOUS SOLUTION, and the HOMEGENEOUS SOLUTION. Now back to the problem statement, we are given the function: 𝑦 !! + 3𝑦 ! + 2𝑦 = 5𝑒 !! This DIFFERENTIAL EQUATION is both SECOND ORDER and NONHOMOGENEOUS. The process we will follow to define the solution for the function is as follows: 1. Define the HOMOGENEOUS SOLUTION 1. Define the PARTICULAR SOLUTION 2. Deploy the INITIAL CONDITIONS to determine any UNKNOWN COEFFICIENTS.



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Let’s determine the HOMOGENEOUS SOLUTION, 𝑦! 𝑥 : 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 determining the HOMOGENEOUS SOLUTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, we will ignore the DRIVING FUNCTION, g(x), on the right side of the equation and solve it as if it was HOMOGENEOUS. We will follow the same process outlined for HIGHER ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS. As a quick review, we will proceed as follows: 1. Define the CHARACTERISTIC POLYNOMIAL (also known as the CHARACTERISTIC EQUATION) 2. Determine the ROOTS 3. Classify the ROOTS as: a. REAL and DISTINCT b. REAL and EQUAL c. COMPLEX



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4. Write the GENERAL SOLUTION based on the CLASSIFICATION of the ROOTS Taking the initial function, let’s rewrite it, ignoring the driving function, giving us: 𝑦 !! + 3𝑦 ! + 2𝑦 = 0 This is now 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. 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!



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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 This 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. Plugging in the values for the COEFFICIENTS and ORDER of the derivatives from the equation that we are working, we get: 𝑟 ! + 3𝑟 + 2 = 0 Where: • 𝑎=3 • 𝑏=2 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! = 3

!

=9

And: 4 𝑏 =4 2 =8 Concluding that: 9>8



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Or: 𝑎! > 4 𝑏 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: 𝑟+2 𝑟+1 =0 Solving for the roots of the polynomials, we get: 𝑟! = −2 𝑟! = −1 We know that for an OVERDAMPED solution, each distinct real root is represented as: 𝑟! = 𝐶! 𝑒 !! ! If the roots are real and the same, the solution is: 𝑦! 𝑥 = 𝐶! 𝑒 !! ! + 𝐶! 𝑥𝑒 !! ! + ⋯ + 𝐶! 𝑥 !!! 𝑒 !! !



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Plugging in the calculated values for the roots, we find the general solution is expressed as: 𝑦! 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 !! This is the HOMOGENEOUS SOLUTION to our COMPLETE SOLUTION. Let’s move on to defining the PARTICULAR SOLUTION. The PARTICULAR SOLUTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is defined by using the METHOD OF UNDETERMINED COEFFICIENTS. This is a method, which in practice, generally requires that we “guess” the PARTICULAR SOLUTION based on the DRIVING FUNCTION on the right side of the equation. We then will take this “guess”, DIFFERENTIATE it a few times, and plug those results back in to the left side of the equation in place of the original DERIVATIVE forms. At this point, we will take this new formula, combine the like terms to simplify it, and then set the UNDETERMINED COEFFICIENTS (which can also be thought simply as UNKNOWN COEFFICIENTS) on the left side of the equation to their corresponding values on the right side of the equation. This will provide us a SET OF EQUATIONS that we can then proceed in solving for each UNKNOWN, defining our PARTICULAR SOLUTION.



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Fortunately, on this exam, there will be no “guessing” needed as we are provided a TABLE highlighting the various FORMS of a PARTICULAR SOLUTION based on the corresponding DRIVING FUNCTION. This TABLE 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 few specific forms of the function 𝑓(𝑥), as it is being referred to as in the NCEES Reference Handbook, but we are using 𝑔(𝑥) for clarity purposes, and it’s resulting PARTICULAR SOLUTION, 𝑦! (𝑥), are defined as:

𝑓(𝑥)

𝑦! (𝑥)

𝐴

𝐵

𝐴𝑒 !"

𝐵𝑒 !" ; 𝛼 ≠ 𝑟!

𝐴! sin 𝜔𝑥 + 𝐴! cos 𝜔𝑥

𝐵! sin 𝜔𝑥 + 𝐵! cos 𝜔𝑥

While we are provided this table in the reference handbook, it is often not clear what the particular form of the solution should be. Here are a few guidelines that are not in the NCEES REFERENCE HANDBOOK, but may guide you when working problems involving the METHOD OF UNDETERMINED COEFFICIENTS: • If f x is constant, then y! = Cx

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• If f x is linear, then y! = B! x + B! • If f x is quadratic, then y! = Ax ! + Bx + C • If f x features e!" , then y! = Be!" • If f x features sin cx or cos cx , then y! = B! sin cx + B! cos(cx) The take away from the table is to match the DRIVING FUNCTION in the left column with the PARTICULAR SOLUTION function in the right column. To find the values of the UNDETERMINED COEFFICIENTS, substitute the assumed form of 𝑦! (𝑥) and it’s associated DERIVATIVES in to the differential equation and equate the UNKNOWN COEFFICIENTS based on the values that correspond to those terms on the opposite side of the equation. Laying out a rough process, we have: 1. Define the general PARTICULAR SOLUTION by observing the DRIVING FUNCTION and using the TABLE presented to us in the NCEES REFERENCE HANDBOOK 2. DIFFERENTIATE this generalized PARTICULAR SOLUTION up the maximum order of the equation we are solving. In this case, we are working a SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, so we will take up through the SECOND DERIVATIVE.

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3. INSERT the DERIVATIVES in to the DIFFERENTIAL EQUATION replacing the original DERIVATIVE terms on the left side. 4. Combine LIKE terms 5. Set the UNDETERMINED COEFFICIENTS as they relate to the terms on the left side of the equation to the corresponding COEFFICIENTS on the right side of the equation and solve. So let’s start with the first step here. The driving function of our SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is: 5𝑒 !! Flipping over to the TABLE on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we find that:



𝑓(𝑥)

𝑦! (𝑥)

𝐴

𝐵

𝐴𝑒 !"

𝐵𝑒 !" ; 𝛼 ≠ 𝑟!

𝐴! sin 𝜔𝑥 + 𝐴! cos 𝜔𝑥

𝐵! sin 𝜔𝑥 + 𝐵! cos 𝜔𝑥

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The second row represents our DRIVING FUNCTION, so our initial “guess” for our PARTICULAR SOLUTION can be written as: 𝑦! 𝑥 = 𝐵𝑒 !" We can further define this PARTICULAR SOLUTION by grabbing the constants from the original DRIVING FUNCTION, such that: 𝑦! 𝑥 = 𝐵𝑒 !! As it stands right now, 𝐵 is the only UNDETERMINED COEFFICIENTS. We now need to DIFFERENTIATE this generalized PARTICULAR SOLUTION up the maximum order of the equation we are solving. Again, in this case, we are working a SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, so we will take up through the SECOND DERIVATIVE, giving us: 𝑦! ′(𝑥) = 2𝐵𝑒 !! 𝑦! ′′(𝑥) = 4𝐵𝑒 !! Our next move is to INSERT the DERIVATIVES in to the DIFFERENTIAL EQUATION replacing the original DERIVATIVE terms on the left side. Our original function reads: 𝑦 !! + 3𝑦 ! + 2𝑦 = 5𝑒 !!

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We will replace the terms on the left with our DIFFERENTIATED terms of our general PARTICULAR SOLUTION, such that: 𝑦!!! + 3𝑦!! + 2𝑦! = 5𝑒 !! Carrying out this substitution, we get: 4𝐵𝑒 !! + 3 2𝐵𝑒 !! + 2 𝐵𝑒 !! = 5𝑒 !! Simplifying, we get: 4𝐵 + 6𝐵 + 2𝐵 = 5 Combining the LIKE terms: 12𝐵 = 5 Typically, this is the point where we would set the UNDETERMINED COEFFICIENTS as they relate to the terms on the left side of the equation to the corresponding COEFFICIENTS on the right side of the equation and solve. There isn’t much we need to do, we just need to solve for B, which gives us:

𝐵=



5 12

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Therefore, our PARTICULAR SOLUTION is:

𝑦! (𝑥) =

5 !! 𝑒 12

Revisiting our HOMOGENEOUS SOLUTION, we derived: 𝑦! 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 !! The COMPLETE SOLUTION will be the combination of these two components, such that: 𝑦(𝑥) = 𝑦! (𝑥) + 𝑦! (𝑥) Which gives us:

𝑦 𝑥 = 𝐶! 𝑒 !!! + 𝐶! 𝑒 !! +

5 !! 𝑒 12 𝟓

The correct answer choice is A. 𝐲 𝐱 = 𝐂𝟏 𝐞!𝟐𝐱 + 𝐂𝟐 𝐞!𝐱 + 𝟏𝟐 𝐞𝟐𝐱

SOLUTION 4: The TOPIC of SECOND ORDER NONHOMOGENEOUS 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 HIGHER ORDER and HOMOGENEOUS if all the terms contain the DEPENDENT VARIABLE or an associated DERIVATIVE to an ORDER higher than one. This same DIFFERENTIAL EQUATION can become of the NONHOMOGENEOUS form if the sum DOES NOT equal zero, but rather a nonzero forcing function of the INDEPENDENT VARIABLE. In formulaic terms, a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION can generally be written as: 𝑑! 𝑦 𝑑𝑦 + 𝑎 + 𝑏𝑦 = 𝑔(𝑥) 𝑑𝑥 ! 𝑑𝑥 Or: 𝑦 !! + 𝑎𝑦′ + 𝑏𝑦 = 𝑔(𝑥) The GENERAL FORM of the COMPLETE SOLUTION 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. The GENERAL SOLUTION, otherwise referred to as the COMPLETE SOLUTION, of a DIFFERENTIAL EQUATION is expressed as: 𝑦 𝑥 = 𝑦! 𝑥 + 𝑦! (𝑥)



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Where: • 𝑦! (𝑥) is the PARTICULAR SOLUTION with 𝑓 𝑥 present • 𝑦! (𝑥) is the HOMOGENEOUS SOLUTION corresponding to 𝑓 𝑥 = 0 It is important to always remember that the GENERAL SOLUTION for a NONHOMOGENEOUS DIFFERENTIAL EQUATION will be made up these two unique components, the PARTICULAR SOLUTION, which may also be referred to as the NONHOMOGENEOUS SOLUTION, and the HOMEGENEOUS SOLUTION. Now back to the problem statement, we are given the function: 𝑦 !! + 2𝑦 ! + 5𝑦 = cos(3𝑥) This DIFFERENTIAL EQUATION is both SECOND ORDER and NONHOMOGENEOUS. The process we will follow to define the solution for the function is as follows: 1. Define the HOMOGENEOUS SOLUTION 2. Define the PARTICULAR SOLUTION 3. Deploy the INITIAL CONDITIONS to determine any UNKNOWN COEFFICIENTS.



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In this problem, we don’t need the first step or the third step, we are asked specifically for the PARTICULAR SOLUTION. The PARTICULAR SOLUTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is defined by using the METHOD OF UNDETERMINED COEFFICIENTS. This is a method, which in practice, generally requires that we “guess” the PARTICULAR SOLUTION based on the DRIVING FUNCTION on the right side of the equation. We then will take this “guess”, DIFFERENTIATE it a few times, and plug those results back in to the left side of the equation in place of the original DERIVATIVE forms. At this point, we will take this new formula, combine the like terms to simplify it, and then set the UNDETERMINED COEFFICIENTS (which can also be thought simply as UNKNOWN COEFFICIENTS) on the left side of the equation to their corresponding values on the right side of the equation. This will provide us a SET OF EQUATIONS that we can then proceed in solving for each UNKNOWN, defining our PARTICULAR SOLUTION. Fortunately, on this exam, there will be no “guessing” needed as we are provided a TABLE highlighting the various FORMS of a PARTICULAR SOLUTION based on the corresponding DRIVING FUNCTION.



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This TABLE 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 few specific forms of the function 𝑓(𝑥), as it is being referred to as in the NCEES Reference Handbook, but we are using 𝑔(𝑥) for clarity purposes, and it’s resulting PARTICULAR SOLUTION, 𝑦! (𝑥), are defined as: 𝑓(𝑥)

𝑦! (𝑥)

𝐴

𝐵

𝐴𝑒 !"

𝐵𝑒 !" ; 𝛼 ≠ 𝑟!

𝐴! sin 𝜔𝑥 + 𝐴! cos 𝜔𝑥

𝐵! sin 𝜔𝑥 + 𝐵! cos 𝜔𝑥

While we are provided this table in the reference handbook, it is often not clear what the particular form of the solution should be. Here are a few guidelines that are not in the NCEES REFERENCE HANDBOOK, but may guide you when working problems involving the METHOD OF UNDETERMINED COEFFICIENTS: • If f x is constant, then y! = Cx • If f x is linear, then y! = B! x + B! • If f x is quadratic, then y! = Ax ! + Bx + C



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• If f x features e!" , then y! = Be!" • If f x features sin cx or cos cx , then y! = B! sin cx + B! cos(cx) The take away from the table is to match the DRIVING FUNCTION in the left column with the PARTICULAR SOLUTION function in the right column. To find the values of the UNDETERMINED COEFFICIENTS, substitute the assumed form of 𝑦! (𝑥) and it’s associated DERIVATIVES in to the differential equation and equate the UNKNOWN COEFFICIENTS based on the values that correspond to those terms on the opposite side of the equation. Laying out a rough process, we have: 1. Define the general PARTICULAR SOLUTION by observing the DRIVING FUNCTION and using the TABLE presented to us in the NCEES REFERENCE HANDBOOK 2. DIFFERENTIATE this generalized PARTICULAR SOLUTION up the maximum order of the equation we are solving. In this case, we are working a SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, so we will take up through the SECOND DERIVATIVE. 3. INSERT the DERIVATIVES in to the DIFFERENTIAL EQUATION replacing the original DERIVATIVE terms on the left side. 4. Combine LIKE terms

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5. Set the UNDETERMINED COEFFICIENTS as they relate to the terms on the left side of the equation to the corresponding COEFFICIENTS on the right side of the equation and solve. So let’s start with the first step here. The driving function of our SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is: cos(3𝑥) Flipping over to the TABLE on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we find that: 𝑓(𝑥)

𝑦! (𝑥)

𝐴

𝐵

𝐴𝑒 !"

𝐵𝑒 !" ; 𝛼 ≠ 𝑟!

𝐴! sin 𝜔𝑥 + 𝐴! cos 𝜔𝑥

𝐵! sin 𝜔𝑥 + 𝐵! cos 𝜔𝑥

The third row represents our DRIVING FUNCTION, so our initial “guess” for our PARTICULAR SOLUTION can be written as: 𝐵! sin 𝜔𝑥 + 𝐵! cos 𝜔𝑥



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We can further define this PARTICULAR SOLUTION by grabbing the constants from the original DRIVING FUNCTION, such that: 𝐵! sin 3𝑥 + 𝐵! cos 3𝑥 As it stands right now, 𝐵! and 𝐵! are our UNDETERMINED COEFFICIENTS. We now need to DIFFERENTIATE this generalized PARTICULAR SOLUTION up the maximum order of the equation we are solving. Again, in this case, we are working a SECOND ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, so we will take up through the SECOND DERIVATIVE, giving us: 𝑦! ′ 𝑥 = 3𝐵! cos 3𝑥 − 3𝐵! sin 3𝑥 𝑦!!! 𝑥 = −9𝐵! sin 3𝑥 − 9𝐵! cos 3𝑥 Our next move is to INSERT the DERIVATIVES in to the DIFFERENTIAL EQUATION replacing the original DERIVATIVE terms on the left side. Our original function reads: 𝑦 !! + 2𝑦 ! + 5𝑦 = cos 3𝑥 We will replace the terms on the left with our DIFFERENTIATED terms of our general PARTICULAR SOLUTION, such that: 𝑦!!! + 2𝑦!! + 5𝑦! = cos(3𝑥)

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Carrying out this substitution, we get: −9𝐵! sin 3𝑥 − 9𝐵! cos(3𝑥) + 2 3𝐵! cos(3𝑥) − 3𝐵! sin(3𝑥) + 5 𝐵! sin(3𝑥) + 𝐵! cos(3𝑥) = cos(3𝑥) Expanding our terms further: −9𝐵! sin 3𝑥 − 9𝐵! cos 3𝑥 + 6𝐵! cos 3𝑥 − 6𝐵! sin 3𝑥 + 5𝐵! sin 3𝑥 +5𝐵! cos(3𝑥) = cos (3𝑥) Grouping together and combining the LIKE terms: −9𝐵! sin 3𝑥 + 5𝐵! sin(3𝑥) + −9𝐵! cos 3𝑥 + 5𝐵! cos(3𝑥) + 6𝐵! cos(3𝑥) + −6𝐵! sin(3𝑥) = cos(3𝑥) And simplifying: −4𝐵! sin 3𝑥 − 4𝐵! cos 3𝑥 + 6𝐵! cos 3𝑥 − 6𝐵! sin 3𝑥 = cos(3𝑥) We will now set the UNDETERMINED COEFFICIENTS as they relate to the terms on the left side of the equation to the corresponding COEFFICIENTS on the right side of the equation and solve. Doing this we get: COSINE TERMS: −4𝐵! + 6𝐵! = 1

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SINE TERMS: −4𝐵! − 6𝐵! = 0 We can now solve this system of linear equations, starting with the SINE TERMS, we can rearrange the formula such that: −4𝐵! = 6𝐵! Establishing the relationship: 6 𝐵! = − 𝐵! = −1.5𝐵! 4 Now jumping to the COSINE TERMS, we have: 6𝐵! − 4𝐵! = 1 Plugging in the relationship we developed between the two UNKNOWN COEFFICIENTS 𝐵! and 𝐵! , we get: 6 −1.5𝐵! − 4𝐵! = 1 Which gives us: 5𝐵! = 1 And:

𝐵! = −

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Taking the value, we plug it in to quantify our second UNKNOWN VARIABLE, giving us:

𝐵! =

3 26

Therefore, our PARTICULAR SOLUTION is:

𝑦! (𝑥) =

3 1 sin 3𝑥 − cos(3𝑥) 26 13

The correct answer choice is B. 𝐲𝐩 (𝐱) =



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𝟑 𝟐𝟔

𝐬𝐢𝐧 𝟑𝐱 −

𝟏 𝟏𝟑

𝐜𝐨𝐬(𝟑𝐱)

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