36.5 Nonhomogeneous Differential Equations Problem Set

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

PROBLEM 1: What is the general solution of the differential equation below? 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 )* )

SOLUTION 1: 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 𝑦3 .



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We will determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 3𝑦 " − 4𝑦 = 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 ) + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the coefficients and order of the derivatives from the given equation we find: 𝑟 ) + 3𝑟 − 4 = 0 Where: • 𝑎=3 • 𝑏 = −4



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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 𝑎) term and 4𝑏 term are always used to determine the form of the solution. 𝑎) = 3

)

=9

4 𝑏 = 4 −4 = −16 We know that 9 > −16, and can classify the solution as over damped, and that the roots are characterized as distinct and real roots. Distinct and real roots characterized by the relationship 𝑎) > 4𝑏, where the solution is of the over damped form. Therefore, the two roots are real and different. 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 find the roots are: 𝑟- = −4

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𝑟) = 1 We know that for an over damped solution, each distinct real root is represented as: 𝑟= = 𝐶= 𝑒 >? * If the roots are real and the same, the solution is: 𝑦3 𝑥 = 𝐶- 𝑒 >@ * + 𝐶) 𝑥𝑒 >A * + ⋯ + 𝐶C 𝑥 =.- 𝑒 >? * Plugging in the calculated values for the roots, we find the general solution is expressed as: 𝑦3 𝑥 = 𝐶- 𝑒 ./* + 𝐶) 𝑒 * 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. 𝑦D"" + 3𝑦D" − 4𝑦D = 3𝑒 )* 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 𝑦D (𝑥) form. The procedure in this method is to assume for 𝑦D (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)

𝑦D (𝑥)

𝐴

𝐵

𝐴𝑒 J*

𝐵𝑒 J* ; 𝛼 ≠ 𝑟=

𝐴- sin 𝜔𝑥 + 𝐴) cos 𝜔𝑥

𝐵- sin 𝜔𝑥 + 𝐵) 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 an exponential function. Since the derivative of an exponential returns the same exponential, we could try to guess an exponential for the particular solution. It should cancel the entire equation, leaving us with a condition on its coefficient: 𝑦D 𝑥 = 𝐵𝑒 )* We can then calculate the first and second derivatives of the particular solution as: 𝑦D" 𝑥 = 2𝐵𝑒 )* 𝑦D"" = 4𝐵𝑒 )*



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In order to solve for the undetermined constants, we will substitute 𝑦D 𝑥 and the derivatives of the particular solution into our equation for a particular solution of the nonhomogeneous portion of the equation: 𝑦D"" + 3𝑦D" − 4𝑦D = 3𝑒 )* 4𝐵𝑒 )* + 3 2𝐵𝑒 )* − 4 𝐵𝑒 )* = 3𝑒 )* Simplifying and combining like terms, the equation is simplified to: 4𝐵 + 6𝐵 − 4𝐵 = 3 Solving the system of linear equations, we find:

𝐵=

1 2

Therefore, our particular solution is:

𝑦D (𝑥) =

1 )* 𝑒 2

We can plug our particular solution and homogeneous solution into the formula for the general solution and find: 1 𝑦 𝑥 = 𝑦3 𝑥 + 𝑦D 𝑥 = 𝐶- 𝑒 ./* + 𝐶) 𝑒 * + 𝑒 )* 2

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Therefore, the correct answer choice is 𝟏

D. 𝐲 𝐱 = 𝐂𝟏 𝐞.𝟒𝐱 + 𝐂𝟐 𝐞𝐱 + 𝐞𝟐𝐱 𝟐

PROBLEM 2: Which of the following is TRUE when it comes to rewriting the general solution for a nonhomogeneous differential equation? A. 𝑦 𝑥 = 𝑦3 (𝑥) − 𝑦D (𝑥) B. 𝑦 𝑥 = 𝑦3 (𝑥) + 𝑦D (𝑥) C. 𝑦 𝑥 = 𝑦3) (𝑥) + 𝑦D) (𝑥) D. 𝑦 𝑥 = 𝑦3 (𝑥) + 𝑦D) (𝑥)

SOLUTION 2: 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 = 𝑦3 𝑥 + 𝑦D 𝑥



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Where: • 𝑦3 (𝑥) is the complementary solution, which solves the complementary or homogeneous solution (the general solution of the corresponding homogeneous equation) • 𝑦D (𝑥) is any particular solution with 𝑓(𝑥) present

Therefore, the correct answer choice is B. 𝐲 𝐱 = 𝐲𝐡 𝐱 + 𝐲𝐩 𝐱 PROBLEM 3: What is the general solution of the differential equation below? 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 .)* +

` -) ` -) ` -) ` -)

𝑒 )* 𝑒 )* 𝑒 )* 𝑒 )*

SOLUTION 3: In order to solve this second order differential equation, we will use the method of undetermined coefficients. 𝑦 "" + 3𝑦 " + 2𝑦 = 5𝑒 )*



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The first step of this method is to find the homogeneous solution to the differential equation. We will call the homogeneous solution 𝑦3 . We will determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 3𝑦 " + 2𝑦 = 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 ) + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the coefficients and order of the derivatives from the given equation we find: 𝑟 ) + 3𝑟 + 2 = 0 Where: • 𝑎=3

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• 𝑏=2 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 𝑎) term and 4𝑏 term are always used to determine the form of the solution. 𝑎) = 3

)

=9

4 𝑏 =4 2 =8 We know that 9 > 8, and can classify the solution as over damped, and that the roots are characterized as distinct and real roots. Distinct and real roots characterized by the relationship 𝑎) > 4𝑏, where the solution is of the over damped form. Therefore, the two roots are real and different. Factoring the quadratic expression, we find the characteristic equation can be rewritten in factored terms as: 𝑟+2 𝑟+1 =0



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Solving for the roots of the polynomials, we find the roots are: 𝑟- = −2 𝑟) = −1 We know that for an over damped solution, each distinct real root is represented as: 𝑟= = 𝐶= 𝑒 >? * If the roots are real and the same, the solution is: 𝑦3 𝑥 = 𝐶- 𝑒 >@ * + 𝐶) 𝑥𝑒 >A * + ⋯ + 𝐶C 𝑥 =.- 𝑒 >? * Plugging in the calculated values for the roots, we find the general solution is expressed as: 𝑦3 𝑥 = 𝐶- 𝑒 .* + 𝐶) 𝑒 .)* 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. 𝑦D"" + 3𝑦D" + 2𝑦D = 5𝑒 )* Our guess for the particular solution depends on the functions on the right hand side of the above equation.



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When specific 𝑓(𝑥) forms result in a specific 𝑦D (𝑥) form. The procedure in this method is to assume for 𝑦D (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)

𝑦D (𝑥)

𝐴

𝐵

𝐴𝑒 J*

𝐵𝑒 J* ; 𝛼 ≠ 𝑟=

𝐴- sin 𝜔𝑥 + 𝐴) cos 𝜔𝑥

𝐵- sin 𝜔𝑥 + 𝐵) 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 an exponential function. Since the derivative of an exponential returns the same exponential, we could try to guess an exponential for the particular solution. It should cancel the entire equation, leaving us with a condition on its coefficient: 𝑦D 𝑥 = 𝐵𝑒 )* We can then calculate the first and second derivatives of the particular solution as: 𝑦D" 𝑥 = 2𝐵𝑒 )* 𝑦D"" = 4𝐵𝑒 )*



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In order to solve for the undetermined constants, we will substitute 𝑦D 𝑥 and the derivatives of the particular solution into our equation for a particular solution of the nonhomogeneous portion of the equation: 𝑦D"" + 3𝑦D" + 2𝑦D = 5𝑒 )* 4𝐵𝑒 )* + 3 2𝐵𝑒 )* + 2 𝐵𝑒 )* = 5𝑒 )* Simplifying and combining like terms, the equation is simplified to: 4𝐵 + 6𝐵 + 2𝐵 = 5 Solving the system of linear equations, we find:

𝐵=

1 2

Therefore, our particular solution is:

𝑦D (𝑥) =

5 )* 𝑒 12

We can plug our particular solution and homogeneous solution into the formula for the general solution and find:

𝑦 𝑥 = 𝑦3 𝑥 + 𝑦D 𝑥 = 𝐶- 𝑒 .* + 𝐶) 𝑒 .)* +



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Therefore, the correct answer choice is A. 𝐲 𝐱 = 𝐂𝟏 𝐞.𝐱 + 𝐂𝟐 𝐞.𝟐𝐱 + 𝟏 𝟐

𝐞𝟐𝐱

PROBLEM 4: What is the particular solution of the differential equation below? 𝑦 "" + 2𝑦 " + 5𝑦 = cos(3𝑥) c

-

/

)

c

-

/

)

A. 𝑦D (𝑥) = − sin 3𝑥 − cos(3𝑥) B. 𝑦D (𝑥) = − sin 3𝑥 + cos(3𝑥) c

-

/

)

c

-

/

)

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

SOLUTION 4: 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 𝑦3 . We will determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 2𝑦 " + 5𝑦 = 0

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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 ) + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the coefficients and order of the derivatives from the given equation we find: 𝑟 ) + 2𝑟 + 5 = 0 Where: • 𝑎=2 • 𝑏=5 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.



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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 𝑎) term and 4𝑏 term are always used to determine the form of the solution. 𝑎) = 2

)

=4

4 𝑏 = 4 5 = 20 We know that 4 < 20, and can classify the solution as under damped, and that the roots are characterized as complex roots. An complex is characterized by the relationship 𝑎) < 4𝑏, where the solution is of the under damped form. The two complex roots are imaginary and represented as: 𝑟- = 𝛼 + 𝑖𝛽 and 𝑟) = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions. 𝑦3 𝑥 = 𝑒 g* (𝐶- sin 𝛽𝑥 + 𝐶) 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=−



g )

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o

𝛽=

/h.g A )

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

𝛼=−

𝑎 2 = − = −1 2 2

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

4𝑏 − 𝑎) 𝛽= = 2

4(5) − (2)) 4 = =2 2 2

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: 𝑦3 𝑥 = 𝑒 g* (𝐶- sin 𝛽𝑥 + C) cos 𝛽𝑥) The general homogeneous solution is then: 𝑦3 𝑥 = 𝑒

.- *

(C- sin 2𝑥 + C) cos 2𝑥)

The two complex roots are imaginary and represented as: 𝑟- = −1 + 2𝑖 and 𝑟) = −1 − 2𝑖



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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. 𝑦D"" + 2𝑦D" + 5𝑦D = cos(3𝑥) Our guess for the particular solution depends on the functions on the right hand side of the above equation. When a specific 𝑓(𝑥) forms result in a specific 𝑦D (𝑥) form. The procedure in this method is to assume for 𝑦D (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)

𝑦D (𝑥)

𝐴

𝐵

𝐴𝑒 J*

𝐵𝑒 J* ; 𝛼 ≠ 𝑟=

𝐴- sin 𝜔𝑥 + 𝐴) cos 𝜔𝑥

𝐵- sin 𝜔𝑥 + 𝐵) 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 a trigonometric function. Based on this scenario, a good first guess would be a solution that contains a sine term and cosine term:

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𝑦D 𝑥 = 𝐵- sin 𝜔𝑥 + 𝐵) cos 𝜔𝑥 We replace 𝜔 with the coefficient term as given in the differential equation problem statement: 𝑦D 𝑥 = 𝐵- sin 3𝑥 + 𝐵) cos 3𝑥 Where: • 𝐵- 𝑎𝑛𝑑 𝐵) are undetermined constants. 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. 𝑦D 𝑥 = 𝐵- sin 3𝑥 + 𝐵) cos 3𝑥 𝑦D ′ 𝑥 = 3𝐵- cos 3𝑥 − 3𝐵) sin 3𝑥 𝑦D"" 𝑥 = −9𝐵- sin 3𝑥 − 12𝐵) cos 3𝑥 In order to solve for the undetermined constants, we will substitute 𝑦D 𝑥 and the derivatives of the particular solution into our equation for a particular solution of the nonhomogeneous portion of the equation: 𝑦D"" + 2𝑦D" + 5𝑦D = cos(3𝑥)



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Substituting 𝑦D 𝑥 and its derivatives we find: −9𝐵- sin 3𝑥 − 12𝐵) cos(3𝑥) + 2 3𝐵- cos(3𝑥) − 3𝐵) sin(3𝑥) + 5 𝐵- sin(3𝑥) + 𝐵) cos(3𝑥) = cos(3𝑥) Simplifying and combining like terms, the equation is simplified to: −9𝐵- sin 3𝑥 − 12𝐵) cos 3𝑥 + 6𝐵- cos 3𝑥 − 6𝐵) sin 3𝑥 + 5𝐵- sin 3𝑥 + 5𝐵) cos(3𝑥) = cos (3𝑥) Further simplifying the like terms, we find: −9𝐵- sin 3𝑥 + 5𝐵- sin(3𝑥) + −12𝐵) cos 3𝑥 + 5𝐵) cos(3𝑥) + 6𝐵- cos(3𝑥) + + −6𝐵) sin(3𝑥) = cos(3𝑥) We can finally re-write the equation as: −4𝐵- sin 3𝑥 − 7𝐵) cos 3𝑥 + 6𝐵- cos 3𝑥 − 6𝐵) sin 3𝑥 = cos(3𝑥) From this equation we can create two equations to represent each variable as it is related on each side of the equation: 𝑆𝑖𝑛𝑒 𝑇𝑒𝑟𝑚𝑠: − 4𝐵- sin 3𝑥 − 6𝐵) sin 3𝑥 = 0 −4𝐵- − 6𝐵) = 0 𝐶𝑜𝑠𝑖𝑛𝑒 𝑇𝑒𝑟𝑚𝑠: − 7𝐵) cos 3𝑥 + 6𝐵- cos 3𝑥 = cos(3𝑥)

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6𝐵- − 7𝐵) = 1 We can solve the linear system of equations by using algebraic substitution. To do this we first solve for 𝐵- in terms of 𝐵) : −4𝐵- = 6𝐵) 6 𝐵- = − 𝐵) = 1.5𝐵) 4 Plugging in the equation for 𝐵- , we are able to solve for 𝐵) : 6𝐵- − 7𝐵) = 1 6 1.5𝐵) − 7𝐵) = 1 Solving for 𝐵) , we find:

𝐵) =

1 2

Plugging this in our equation relating both coefficients, we find: 6 𝐵- = − 𝐵) = 1.5𝐵) 4



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𝐵- = −

3 1 3 =− 2 2 4

Solving the system of linear equations, we find: 𝐵- = −3 𝐵) = 5 Therefore, our particular solution is: 3 1 𝑦D (𝑥) = − sin 3𝑥 + cos(3𝑥) 4 2 𝟑

Therefore, the correct answer choice is B. 𝐲𝐩 (𝐱) = − 𝐬𝐢𝐧 𝟑𝐱 + 𝟒

𝟏 𝟐

𝐜𝐨𝐬(𝟑𝐱)



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