36. Nonhomogeneous Differential Equations Concept Overview

NONHOMOGENEOUS DIFFERENTIAL EQUATIONS | CONCEPT OVERVIEW The topic of NONHOMOGENEOUS DIFFERENTIAL EQUATIONS can be referenced on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

CONCEPT INTRO: In a NONHOMOGENEOUS DIFFERENTIAL EQUATION, the sum of the derivative terms is equal to a nonzero forcing function of the independent variable. When the equation is a NONHOMOGENEOUS differential equation, 𝑓 𝑥 ≠ 0 𝑜𝑟 𝑥 ≠ 0, the roots are real and different. The first step in determining the general solution is to determine the corresponding homogeneous solution. We covered this method in the previous review on homogeneous differential equations. The next step is to then determine the particular solution, by using the METHOD OF UNDETERMINED COEFFICIENTS, which is a systematic way to determine the general form/type of the particular solution 𝑦(𝑡) based on the nonhomogeneous term in the given differential equation. The values are then substituted back into the original equation and the like terms are combined.

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The like coefficients from each side are set against each other and each unknown coefficient is then solved for. If values are able to be determined, then the original guess was correct, if not, the original values are wrong. We then combine the homogeneous and nonhomogeneous (particular) solutions to obtain the overall solution of the differential equation. The homogeneous equation corresponding to a nonhomogeneous equation is known as a reduced equation or complementary equation. The FORMULA FOR THE GENERAL 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 complete solution to the nonhomogeneous differential equation is the sum of a particular solution of the nonhomogeneous equation and the general solution of the homogeneous equation: 𝑦 𝑥 = 𝑦. 𝑥 + 𝑦0 𝑥 Where: • 𝑦. (𝑥) is the complementary solution, which solves the complementary or homogeneous solution (the general solution of the corresponding homogeneous equation) • 𝑦0 (𝑥) is any particular solution with 𝑓(𝑥) present



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If 𝑓 𝑥 has 𝑒 23 4 terms in the solution, then RESONANCE is manifested or is present to some extent.. Initial values are used to evaluate any unknown coefficients in the complementary solution after 𝑦. (𝑥) and 𝑦0 (𝑥) have been combined. The particular solution will not have any unknown coefficients. PARTICULAR SOLUTION 𝒚𝒑 𝒙 : The PARTICULATION SOLUTION is formed from the sum of 𝑓(𝑥) and its derivatives, each term of which has an unknown constant coefficient to be determined by substituting the particular solution into the differential equation. Higher order of multiplicity improves higher powers of “𝑥”. The TABLE FOR THE FORMS OF A PARTICULAR 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. If 𝑓(𝑥) has 𝑒 23 4 terms, then resonance is manifested. Furthermore, specific 𝑓(𝑥) forms result in specific 𝑦0 (𝑥) forms, some of which are: The METHOD OF UNDETERMINED COEFFICIENTS is a way to obtain a particular solution of the nonhomogeneous equation. The idea is to detect repeating patterns in the derivatives of the nonhomogeneous terms and to set up the particular solution as a linear combination of the patterns with undetermined coefficients.

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When specific 𝑓(𝑥) forms result in a specific 𝑦0 (𝑥) form. The procedure in this method is to assume for 𝑦0 (𝑥) an expression based on 𝑓(𝑥) and its derivatives, and containing certain undetermined coefficients. The TABLE FOR THE FORMS OF PARTICULAR SOLUTIONS FOR FUNCTIONS 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. 𝑓(𝑥)

𝑦0 (𝑥)

𝐴

𝐵

𝐴𝑒 ;4

𝐵𝑒 ;4 ; 𝛼 ≠ 𝑟>

𝐴? sin 𝜔𝑥 + 𝐴D cos 𝜔𝑥

𝐵? sin 𝜔𝑥 + 𝐵D cos 𝜔𝑥

While we are provided this table in the reference handbook, it is not clear what the particular form of the function should be. Here are a few guidelines that are not in the reference handbook, but guidelines when working problems involving the method of undetermined coefficients: 𝐼𝑓 𝑓 𝑥 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑡ℎ𝑒𝑛 𝑦0 = 𝐶x 𝐼𝑓 𝑓 𝑥 𝑖𝑠 𝑙𝑖𝑛𝑒𝑎𝑟, 𝑡ℎ𝑒𝑛 𝑦0 = 𝐵? 𝑥 + 𝐵D 𝐼𝑓 𝑓 𝑥 𝑖𝑠 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐, 𝑡ℎ𝑒𝑛 𝑦0 = 𝐴𝑥 D + 𝐵𝑥 + 𝐶 𝐼𝑓 𝑓 𝑥 𝑓𝑒𝑎𝑡𝑢𝑟𝑒𝑠 𝑒 S4 , 𝑡ℎ𝑒𝑛 𝑦0 = 𝐵𝑒 S4



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𝐼𝑓 𝑓 𝑥 𝑓𝑒𝑎𝑡𝑢𝑟𝑒𝑠 sin 𝑐𝑥 𝑜𝑟 cos 𝑐𝑥 , 𝑡ℎ𝑒𝑛 𝑦0 = 𝐵? sin 𝑐𝑥 + 𝐵D cos(𝑐𝑥) If the independent variable is time “𝑡”, then transient dynamic solutions are implied. Note that if 𝑓 𝑥 = 𝑒 ;4 , and the term 𝑒 ;4 also appears in the homogeneous solution, the particular solution has the form 𝐴𝑥𝑒 ;4 . The particular solution can be read from the table above if the forcing function is one of the forms given. Of course, the coefficients, 𝐴T and 𝐵T are not known. These coefficients are the UNDETERMINED COEFFICIENTS. The take away from the table is to match the 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 𝑦0 (𝑥) in the differential equation and equate coefficients of corresponding types. FIRST-ORDER INSTRUMENTS: The FORMULA FOR THE DIFFERENTIAL EQUATION OF A FIRST-ORDER INSTRUMENTS 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 INSTRUMENT has an output which is given by a nonhomogeneous first order linear differential equation:

𝜏

𝑑𝑦 + 𝑦 = 𝐾𝑥 𝑡 𝑑𝑡 Made with

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Where: • 𝜏 is the time constant of the instrument, which is a measure of the time delay in the response to changes of input. The time constant is given in units of time • 𝑦(𝑡) is the response of the system output to the forcing function • 𝑥(𝑡) is the forcing function • 𝐾 is the gain of the system The initial condition for the first-order system is given as: 𝑦 0 = 𝐾𝐴 Thermometers for measuring temperature are first-order instruments. The time constant of a measurement of temperature is determined by the thermal capacity of the thermometer and the thermal contact between the thermometer and the body whose temperature is being measured. A cup anemometer for measuring wind speed is also a first order instrument. The time constant depends on the anemometer's moment of inertia. A first-order system subjected to a constant force applied instantaneously at the initial time 𝑡 = 0 is represented by the step function:

𝑥 𝑡 =



𝐴 𝑡 < 0 𝐵 𝑡 < 0

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The solution to the step function is:

𝑦 𝑡 = 𝐾𝐴 + (𝐾𝐵 − 𝐾𝐴) 1 − exp

−𝑡 𝜏

or

𝑡 𝐾𝐵 − 𝐾𝐴 = ln 𝜏 𝐾𝐵 − 𝑦

CONCEPT EXAMPLE: What is the general solution of the differential equation below given the two initial conditions? 𝑦 ^^ + 6𝑦 ^ + 13𝑦 = 9 cos 2𝑥 − 87 sin 2𝑥 ; 𝑦 0 = 9, 𝑦 ^ 0 = −4 A. 𝑦 𝑥 = 𝑒 fg4 4 cos 2𝑥 + 7 sin(2𝑥) − 5 cos 2𝑥 − 3 sin(2𝑥) B. 𝑦 𝑥 = 𝑒 fg4 4 cos 2𝑥 + 7 sin(2𝑥) + 5 cos 2𝑥 − 3 sin(2𝑥) C. 𝑦 𝑥 = 𝑒 fg4 4 cos 2𝑥 + 7 sin(2𝑥) + 5 cos 2𝑥 + 3 sin(2𝑥) D. 𝑦 𝑥 = 𝑒 g4 4 cos 2𝑥 − 7 sin(2𝑥) + 5 cos 2𝑥 − 3 sin(2𝑥)



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SOLUTION: In order to solve this second order differential equation, we will use the method of undetermined coefficients. The first step of this method is to find the homogeneous solution to the differential equation. We will call the homogeneous solution 𝑦. . We will determine the corresponding homogeneous solution to the given differential equation: 𝑦 ^^ + 6𝑦 ^ + 13𝑦 = 0 Notice that the above equation is simply the left hand side of our initial problem set equal to zero. 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 of: 𝑟 D + 𝑎𝑟 + 𝑏 = 0



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Plugging in the values for the coefficients and order of the derivatives from the given equation we find: 𝑟 D + 6𝑟 + 13 = 0 Where: • 𝑎=6 • 𝑏 = 13 The FORMULAS FOR THE 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. Calculating the discriminant of the quadratic expression, we find plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎D term and 4𝑏 term are always used to determine the form of the solution. 𝑎D = 6

D

= 36

4 𝑏 = 4 13 = 52 We know that 36 < 52, and can class the solution as under damped, and that the roots are characterized as complex roots. An complex is characterized by the relationship 𝑎D < 4𝑏, where the solution is of the under damped form.

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The two complex roots are imaginary and represented as: 𝑟? = 𝛼 + 𝑖𝛽 and 𝑟D = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions. 𝑦. 𝑥 = 𝑒 k4 (𝐶? sin 𝛽𝑥 + 𝐶D cos 𝛽𝑥) Where: o

The real part of the complex root, “𝑎”, appears in the exponent.

o

The coefficients of the imaginary part, “𝑏”, appears in the sine and cosine terms.

o

𝛼=−

o

𝛽=

k D

lmfk n D

Solving for 𝛼, we plug in the vales for 𝑎 = 36 to find:

𝛼=−

𝑎 6 = − = −3 2 2

Solving for 𝛽, we plug in the vales for 𝑎 − 0 𝑎𝑛𝑑 𝑏 = 4, to find:

4𝑏 − 𝑎D 𝛽= = 2



4(13) − (6)D 4 = =2 2 2

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Plugging in the calculate values of 𝛼 𝑎𝑛𝑑 𝛽 into the under damped form of the general solution, we find the general solution of the equation is expressed as: 𝑦. 𝑥 = 𝑒 k4 (𝐶? sin 𝛽𝑥 + CD cos 𝛽𝑥) The general homogeneous solution is then: 𝑦. 𝑥 = 𝑒

fg 4

(C? sin 2𝑥 + CD cos 2𝑥)

The two complex roots are imaginary and represented as: 𝑟? = −3 + 2𝑖 and 𝑟D = −3 − 2𝑖 The second step is to use the method of undetermined coefficients to solve the nonhomogeneous solution of the differential equation. The first step in this process is to guess a particular solution of our given problem. 𝑦0^^ + 6𝑦0^ + 13𝑦0 = 9 cos 2𝑥 − 87 sin(2𝑥) Our guess for the particular solution depends on the functions on the right hand side of the above equation. When specific 𝑓(𝑥) forms result in a specific 𝑦0 (𝑥) form. The procedure in this method is to assume for 𝑦0 (𝑥) an expression based on 𝑓(𝑥) and its derivatives, and containing certain undetermined coefficients.



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The TABLE FOR THE FORMS OF PARTICULAR SOLUTIONS FOR FUNCTIONS 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. 𝑓(𝑥)

𝑦0 (𝑥)

𝐴

𝐵

𝐴𝑒 ;4

𝐵𝑒 ;4 ; 𝛼 ≠ 𝑟>

𝐴? sin 𝜔𝑥 + 𝐴D cos 𝜔𝑥

𝐵? sin 𝜔𝑥 + 𝐵D cos 𝜔𝑥

Looking at the above table we look at a list of common initial guesses for particular solution. In our equation, the right hand side consists of trigonometric functions. Based on this scenario, a good first guess would be a solution that contains a sine term and cosine term: 𝑦0 𝑥 = 𝐵? sin 𝜔𝑥 + 𝐵D cos 𝜔𝑥 We replace “𝜔” with the coefficient term as given in the differential equation problem statement: 𝑦0 𝑥 = 𝐵? sin 2𝑥 + 𝐵D cos 2𝑥 Where: • 𝐵? 𝑎𝑛𝑑 𝐵D are undetermined constants.



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We will then calculate the first and second derivatives of the particular solution that we will later use with the given initial conditions of the differential equation to solve for the undetermined constants. 𝑦0 𝑥 = 𝐵? sin 2𝑥 + 𝐵D cos 2𝑥 𝑦0 ′ 𝑥 = 2𝐵? cos 2𝑥 − 2𝐵D sin 2𝑥 𝑦0^^ (𝑥) = −4𝐵? sin 2𝑥 − 4𝐵D cos 2𝑥 In order to solve for the undetermined constants, we will substitute 𝑦0 𝑥 and the derivatives of the particular solution into our equation for a particular solution of the nonhomogeneous portion of the equation: 𝑦0^^ + 6𝑦0^ + 13𝑦0 = 9 cos 2𝑥 − 87 sin(2𝑥) Substituting 𝑦0 𝑥 and its derivatives we find: −4𝐵? sin 2𝑥 − 4𝐵D cos 2𝑥 + 6 2𝐵? cos 2𝑥 − 2𝐵D sin 2𝑥 + 13 𝐵? sin 2𝑥 + 𝐵D cos 2𝑥 = 9 𝑐𝑜𝑠 (2𝑥) − 87 𝑠𝑖𝑛 (2𝑥) Simplifying and combining like terms, the equation is simplified to: 9𝐵D + 12𝐵? cos(2𝑥) + 9𝐵? − 12𝐵D sin 2𝑥 = 9 cos 2𝑥 − 87 sin(2𝑥)



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From this equation we can create two equations to represent each variable as it is related on each side of the equation: 𝐶𝑜𝑠𝑖𝑛𝑒 𝑇𝑒𝑟𝑚𝑠: 9𝐵D + 12𝐵? =9 𝑆𝑖𝑛𝑒 𝑇𝑒𝑟𝑚𝑠: 9𝐵? − 12𝐵? =-87 We solve the system of linear equations by setting the coefficient of the sine function on the left hand side equal to the coefficient of the sine on the right hand side. We then do to same process for the cosine terms. Solving the system of linear equations, we find: 𝐵? = −3 𝐵D = 5 Therefore, our particular solution is: 𝑦0 = −3 sin 2𝑥 + 5 cos(2𝑥) You can double check this result by substituting it into the particular solution’s differential equation. Our complete solution for the initial value problem is: 𝑦 = 𝑦. + 𝑦0 = 𝐶? 𝑒 fg4 cos 2𝑥 + 𝐶D 𝑒 fgt sin 2𝑥 − 3 sin 2𝑥 + 5 cos(2𝑥)



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In order to find 𝐶? 𝑎𝑛𝑑 𝐶D , we will use the given initial conditions of 𝑦 0 = 9, 𝑦 ^ 0 = −4 Note that it is extremely important that you do no try to solve for 𝐶? 𝑎𝑛𝑑 𝐶D without first finding the particular solution. You will get the wrong values for 𝐶? 𝑎𝑛𝑑 𝐶D . First we take the general solution and use the initial condition 𝑦 0 = 9, which has values of 𝑥 = 0 and 𝑦 = 9: 𝑦(𝑥) = 𝑦. + 𝑦0 = 𝐶? 𝑒 fg4 cos 2𝑥 + 𝐶D 𝑒 fgt sin 2𝑥 − 3 sin 2𝑥 + 5 cos(2𝑥) Plugging in the value of 𝑥 = 0, we find: 𝑦 0 = 𝐶? 𝑒 fg

v

cos 2 0

+ 𝐶D 𝑒 fg

v

sin 2 0

− 3 sin 2 0

+ 5 cos(2 0 )

Simplifying and evaluating the trig functions we find: 𝑦 0 = 𝐶? 𝑒 v cos 0 + 𝐶D 𝑒 v sin 0 − 3 sin 0 + 5 cos 0 𝑦 0 = 𝐶? 1 1 + 𝐶D 1 0 − 3 0 + 5 𝑦 0 = 𝐶? + 5 Plugging in the value of of 𝑦 = 9, we can solve for of 𝐶? : 9 = 𝐶? + 5

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𝐶? = 4 We then plug in the initial condition 𝑦 ^ 0 = −4, which has values of 𝑥 = 0 and 𝑦 = −4, as well as the calculated constant value of 𝐶? = 4. We calculate the first derivative as: 𝑦(𝑥) = 𝐶? 𝑒 fg4 cos 2𝑥 + 𝐶D 𝑒 fgt sin 2𝑥 − 3 sin 2𝑥 + 5 cos(2𝑥) 𝑦 ^ 𝑥 = −3𝑒 fg4 𝐶? cos 2𝑥 + 𝐶D sin 2𝑥

+ 𝑒 fg4 −2𝐶? sin 2𝑥 + 2𝐶D cos 2𝑥

−

10 sin 2𝑥 − 6 cos(2𝑥) Plugging in the value of 𝑥 = 0, we find: 𝑦 ^ 𝑥 = −3𝑒 fg(v) 𝐶? cos 2(0) + 𝐶D sin 2(0) + 𝑒 fg(v) −2𝐶? sin 2(0) + 2𝐶D cos 2(0) − 10 sin 2(0) − 6 cos(2(0)) Simplifying we rewrite the equation as: 𝑦 ^ 𝑥 = −3𝑒 v 𝐶? cos 0 + 𝐶D sin 0 + 𝑒 v −2𝐶? sin 0 + 2𝐶D cos 0 − 10 sin 0 − 6 cos 0 𝑦 ^ 𝑥 = −3 1 𝐶? 1 + 𝐶D 0 + 1 −2𝐶? 0 + 2𝐶D 1 − 10 0 − 6(1) 𝑦 ^ 𝑥 = −3 𝐶? + 2𝐶D − 6



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Plugging in our known value of 𝐶? = 4 and value of of 𝑦 = −4: −4 = −3 4 + 2𝐶D − 6 Solving for 𝐶D we find: 𝐶D = 7 Therefore, the solution to our initial value problem is: 𝑦 𝑥 = 𝑒 fg4 4 cos 2𝑥 + 7 sin(2𝑥) + 5 cos 2𝑥 − 3 sin(2𝑥)

Therefore, the correct answer choice is B. 𝒚 𝒙 = 𝒆f𝟑𝒙 𝟒 𝒄𝒐𝒔 𝟐𝒙 + 𝟕 𝒔𝒊𝒏(𝟐𝒙) + 𝟓 𝒄𝒐𝒔 𝟐𝒙 − 𝟑 𝒔𝒊𝒏(𝟐𝒙)



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