35.5 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: What is the particular solution of the differential equation and initial conditions below? 𝑦 "" + 4𝑦 = 0; 𝑦 0 = 4, 𝑦 " 0 = 6 A. 𝑦) (𝑥) = 𝐶. sin 𝑥 + C3 cos 𝑥 B. 𝑦) (𝑥) = −𝐶. sin 2𝑥 − C3 cos 2𝑥 C. 𝑦) (𝑥) = −𝐶. sin 2𝑥 + C3 cos 𝑥 D. 𝑦) (𝑥) = 𝐶. sin 2𝑥 + C3 cos 2𝑥
SOLUTION 1: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 4𝑦 = 0 The FORMULA FOR THE CHARACTERISTIC EQUATION OF A SECOND-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION can be referenced under the
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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 3 + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the coefficients and order of the derivatives from the given equation we find: 𝑟3 + 4 = 0 Where: • 𝑎=0 • 𝑏=4 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = 0
3
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4 𝑏 = 4 4 = 16 We know that 0 < 16, and can class the solution as under damped, and that the roots are characterized as complex roots. An complex is characterized by the relationship 𝑎3 < 4𝑏, where the solution is of the under damped form. The two complex roots are imaginary and represented as: 𝑟. = 𝛼 + 𝑖𝛽 and 𝑟3 = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions. 𝑦) 𝑥 = 𝑒 BC (𝐶. sin 𝛽𝑥 + 𝐶3 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=
B 3
DEFB G 3
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Solving for “𝛼”, we plug in the values for 𝑎 = 0 to find:
𝛼=−
𝑎 0 =− =0 2 2
Solving for 𝛽, we plug in the values for 𝑎 = 0 𝑎𝑛𝑑 𝑏 = 4, to find:
4𝑏 − 𝑎3 𝛽= = 2
4(4) − (0)3 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: 𝑦) 𝑥 = 𝑒 BC (𝐶. sin 𝛽𝑥 + C3 cos 𝛽𝑥) 𝑦) 𝑥 = 𝑒
J C
(C. sin 2𝑥 + C3 cos 2𝑥)
Simplifying we find the general solution of the equation is: 𝑦) 𝑥 = (1)(𝐶. sin 2𝑥 + C3 cos 2𝑥) 𝑦) (𝑥) = 𝐶. sin 2𝑥 + C3 cos 2𝑥
Therefore, the correct answer choice is D.𝐲𝐡 (𝐱) = 𝐂𝟏 𝐬𝐢𝐧 𝟐𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝟐𝐱
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PROBLEM 2: What is the particular solution of the differential equation and initial conditions below? 𝑦 "" − 4𝑦 " + 4𝑦 = 0; 𝑦 0 = 4, 𝑦 " 0 = 5 A. 𝑦) 𝑥 = 4𝑒 3C − 3𝑥𝑒 3C B. 𝑦) 𝑥 = 4𝑒 3C + 3𝑥𝑒 3C C. 𝑦) 𝑥 = −4𝑒 3C − 3𝑥𝑒 3C D. 𝑦) 𝑥 = −4𝑒 3C + 3𝑥𝑒 3C
SOLUTION 2: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" − 4𝑦 " + 4𝑦 = 0 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 3 + 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 3 − 4𝑟 + 4 = 0 Where: • 𝑎 = −4 • 𝑏=4 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = −4
3
= 16
4 𝑏 = 4 4 = 16 We know that 16 = 16, and can class the solution as critically damped, and that the roots are characterized as distinct and real roots.
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An equal and real root is characterized by the relationship 𝑎3 = 4𝑏, where the solution is of the critically damped form. The two roots are real and the same, such that they can be referred to as the same variable. Repeated real roots are represented as 𝑟. = 𝑟3 , where 𝑟. and 𝑟3 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.,3 = 2 If the root 𝑟. = 𝑟3 , then 𝐶3 𝑒 XG C is replaced with 𝐶3 𝑥𝑒 XY C . If the roots are real and the same, the solution is: 𝑦) 𝑥 = (𝐶. + 𝐶3 𝑥 + ⋯ + 𝐶[ 𝑥 \F. )𝑒 XC Plugging in the calculated value for the roots, we find the general solution is expressed as: 𝑦) 𝑥 = 𝐶. 𝑒 3C + 𝐶3 𝑥𝑒 3C
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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 of 𝑦 0 = 4, 𝑦 " 0 = 5 First we take the general solution and use the initial condition 𝑦 0 = 4, which has values of 𝑥 = 0 and 𝑦 = 4: 𝑦) 𝑥 = 𝐶. 𝑒 3C + 𝐶3 𝑥𝑒 3C Plugging in our values to the general solution we find: 4 = 𝐶. 𝑒 3(J) + 𝐶3 (0)𝑒 3(J) 4 = 𝐶. 𝑒 J + 0 We can then solve for 𝐶. as the second constant was eliminated from the equation: 𝐶. = 4 In order to use the second initial condition, we need to calculate the derivative of the general solution: 𝑦)" 𝑥 = 2𝐶. 𝑒 3C + 𝐶3 (2𝑥𝑒 3C + 𝑒 3C )
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We then plug in the initial condition 𝑦 " 0 = 5, which has values of 𝑥 = 0 and 𝑦 = 5, as well as the calculated constant value of 𝐶. = 4: 5 = 2 4 𝑒 3(J) + 𝐶3 2 0 𝑒 3
J
+ 𝑒3
J
Simplifying the equation, we find the equation is rewritten as: 5 = 8 1 + 𝐶3 0 + 1 Solving for the miss constant 𝐶3 , we find: 𝐶3 = −3 Plugging the values for 𝐶. and 𝐶3 into the general solution, we find the particular solution is: 𝑦) 𝑥 = 4𝑒 3C − 3𝑥𝑒 3C
Therefore, the correct answer choice is A. 𝐲𝐡 𝐱 = 𝟒𝐞𝟐𝐱 − 𝟑𝐱𝐞𝟐𝐱
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PROBLEM 3: What is the general solution to the following second-order differential equation if 𝐴 and 𝐵 are constants? 𝑦 "" + 9𝑦 = 0 A. 𝑦 𝑥 = 𝐴𝑒 eC + 𝐵𝑥𝑒 FeC B. 𝑦 𝑥 = 𝐴𝑒 eC + 𝐵𝑒 FeC C. 𝑦 𝑥 = 𝐴𝑒 C − 𝐵𝑥𝑒 FC D. 𝑦 𝑥 = 𝐴 cos 3𝑥 + 𝐵 sin 3𝑥
SOLUTION 3: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 9𝑦 = 0 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 3 + 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: 𝑟3 + 9 = 0 Where: • 𝑎=0 • 𝑏 = 9 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = 0
3
=0
4 𝑏 = 4 9 = 36 We know that 0 < 36, and can class the solution as under damped, and that the roots are characterized as complex roots. A complex root is characterized by the relationship 𝑎3 < 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 𝑟3 = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions. 𝑦) 𝑥 = 𝑒 BC (𝐶. sin 𝛽𝑥 + 𝐶3 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=
B 3
DEFB G 3
Solving for “𝛼”, we plug in the vales for 𝑎 = 0 to find:
𝛼=−
𝑎 0 =− =0 2 2
Solving for 𝛽, we plug in the vales for 𝑎 = 0 and 𝑏 = 9, to find:
4𝑏 − 𝑎3 𝛽= = 2
4(9) − (0)3 6 = =3 2 2
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Plugging in the calculated 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 BC (C. sin 𝛽𝑥 + C3 cos 𝛽𝑥) 𝑦) 𝑥 = 𝑒
J C
(C. sin 3𝑥 + C3 cos 3𝑥)
Simplifying we find the general solution of the equation is: 𝑦) 𝑥 = (1)(𝐶. sin 3𝑥 + C3 cos 3𝑥) Plugging in the constants of A and B in lieu of 𝐶. and 𝐶3 , we find the general solution is: 𝑦) (𝑥) = 𝐴 sin 3𝑥 + B cos 3𝑥
Therefore, the correct answer choice is D.𝐲𝐡 (𝐱) = 𝐀 𝐬𝐢𝐧 𝟑𝐱 + 𝐁 𝐜𝐨𝐬 𝟑𝐱
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PROBLEM 4: What is the particular solution of the differential equation and initial conditions below? 𝑦 "" + 4𝑦′ + 4𝑦 = 0; 𝑦 0 = 1, 𝑦 " 0 = 0 A. 𝑦) 𝑥 = 1 − 2𝑥 𝑒 3C B. 𝑦) 𝑥 = 2 − 𝑥 𝑒 F3C C. 𝑦) 𝑥 = 2 + 𝑥 𝑒 F3C D. 𝑦) 𝑥 = 1 + 2𝑥 𝑒 F3C
SOLUTION 4: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 4𝑦′ + 4𝑦 = 0 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 3 + 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 3 + 4𝑟 + 4 = 0 Where: • 𝑎=4 • 𝑏=4 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = 4
3
= 16
4 𝑏 = 4 4 = 16 We know that 16 = 16, and can class the solution as critically damped, and that the roots are characterized as distinct and real roots. An equal and real root is characterized by the relationship 𝑎3 = 4𝑏, where the solution is of the critically damped form.
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The two roots are real and the same, such that they can be referred to as the same variable. Repeated real roots are represented as 𝑟. = 𝑟3 , where 𝑟. and 𝑟3 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.,3 = −2 If the root 𝑟. = 𝑟3 , then 𝐶3 𝑒 XG C is replaced with 𝐶3 𝑥𝑒 XY C . If the roots are real and the same, the solution is: 𝑦) 𝑥 = (𝐶. + 𝐶3 𝑥 + ⋯ + 𝐶[ 𝑥 \F. )𝑒 XC Plugging in the calculated value for the roots, we find the general solution is expressed as: 𝑦) 𝑥 = 𝐶. 𝑒 F3C + 𝐶3 𝑥𝑒 F3C 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 of 𝑦 0 = 1, 𝑦 " 0 = 0 First we take the general solution and use the initial condition 𝑦 0 = 1, which has values of 𝑥 = 0 and 𝑦 = 1: 𝑦) 𝑥 = 𝐶. 𝑒 F3C + 𝐶3 𝑥𝑒 F3C Plugging in our values to the general solution we find: 1 = 𝐶. 𝑒 3(J) + 𝐶3 (0)𝑒 3(J) 1 = 𝐶. 𝑒 J + 0 We can then solve for 𝐶. as the second constant was eliminated from the equation: 𝐶. = 1 In order to use the second initial condition, we need to calculate the derivative of the general solution: 𝑦)" 𝑥 = 𝐶. + 𝐶3 𝑥 𝑒 F3C −2 + 𝐶3 (𝑒 F3C ) We then plug in the initial condition 𝑦 " 0 = 0, which has values of 𝑥 = 0 and 𝑦 = 0, as well as the calculated constant value of 𝐶. = 1: 0 = 1 + 𝐶3 0
𝑒 F3
J
−2 + 𝐶3 (𝑒 F3C )
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Simplifying the equation, we find the equation is rewritten as: 0 = −2 1 + 𝐶3 Solving for the miss constant 𝐶3 , we find: 𝐶3 = 2 Plugging the values for 𝐶. and 𝐶3 into the general solution, we find the particular solution is: 𝑦) 𝑥 = 1 + 2𝑥 𝑒 F3C
Therefore, the correct answer choice is D. 𝐲𝐡 𝐱 = 𝟏 + 𝟐𝐱 𝐞F𝟐𝐱
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PROBLEM 5: What is the general solution of the differential equation below? 𝑦 "" − 8𝑦 " + 16𝑦 = 0 A. 𝑦) 𝑥 = 𝐶. 𝑒 DC B. 𝑦) 𝑥 = 𝐶. + 𝐶3 𝑥 𝑒 DC C. 𝑦) 𝑥 = 𝐶. 𝑒 FDC + 𝐶3 𝑒 DC D. 𝑦) 𝑥 = 𝐶. 𝑒 3C + 𝐶3 𝑒 DC
SOLUTION 5: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" − 8𝑦 " + 16𝑦 = 0 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 3 + 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 3 − 8𝑟 + 16 = 0 Where: • 𝑎 = −8 • 𝑏 = 16 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = −8
3
= 64
4 𝑏 = 4 16 = 64 We know that 64 = 64, and can class the solution as critically damped, and that the roots are characterized as distinct and real roots. An equal and real root is characterized by the relationship 𝑎3 = 4𝑏, where the solution is of the critically damped form.
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The two roots are real and the same, such that they can be referred to as the same variable. Repeated real roots are represented as 𝑟. = 𝑟3 , where 𝑟. and 𝑟3 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.,3 = 4 If the root 𝑟. = 𝑟3 , then 𝐶3 𝑒 XG C is replaced with 𝐶3 𝑥𝑒 XY C . If the roots are real and the same, the solution is: 𝑦) 𝑥 = (𝐶. + 𝐶3 𝑥 + ⋯ + 𝐶[ 𝑥 \F. )𝑒 XC Plugging in the calculated value for the roots, we find the general solution is expressed as: 𝑦) 𝑥 = 𝐶. 𝑒 DC + 𝐶3 𝑥𝑒 DC 𝑦) 𝑥 = 𝐶. + 𝐶3 𝑥 𝑒 DC
The correct answer is B. 𝐲𝐡 𝐱 = 𝐂𝟏 + 𝐂𝟐 𝐱 𝐞𝟒𝐱
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PROBLEM 1: What is the particular solution of the differential equation and initial conditions below? 𝑦 "" + 4𝑦 = 0; 𝑦 0 = 4, 𝑦 " 0 = 6 A. 𝑦) (𝑥) = 𝐶. sin 𝑥 + C3 cos 𝑥 B. 𝑦) (𝑥) = −𝐶. sin 2𝑥 − C3 cos 2𝑥 C. 𝑦) (𝑥) = −𝐶. sin 2𝑥 + C3 cos 𝑥 D. 𝑦) (𝑥) = 𝐶. sin 2𝑥 + C3 cos 2𝑥
SOLUTION 1: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 4𝑦 = 0 The FORMULA FOR THE CHARACTERISTIC EQUATION OF A SECOND-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION can be referenced under the
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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 3 + 𝑎𝑟 + 𝑏 = 0 Plugging in the values for the coefficients and order of the derivatives from the given equation we find: 𝑟3 + 4 = 0 Where: • 𝑎=0 • 𝑏=4 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = 0
3
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4 𝑏 = 4 4 = 16 We know that 0 < 16, and can class the solution as under damped, and that the roots are characterized as complex roots. An complex is characterized by the relationship 𝑎3 < 4𝑏, where the solution is of the under damped form. The two complex roots are imaginary and represented as: 𝑟. = 𝛼 + 𝑖𝛽 and 𝑟3 = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions. 𝑦) 𝑥 = 𝑒 BC (𝐶. sin 𝛽𝑥 + 𝐶3 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=
B 3
DEFB G 3
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Solving for “𝛼”, we plug in the values for 𝑎 = 0 to find:
𝛼=−
𝑎 0 =− =0 2 2
Solving for 𝛽, we plug in the values for 𝑎 = 0 𝑎𝑛𝑑 𝑏 = 4, to find:
4𝑏 − 𝑎3 𝛽= = 2
4(4) − (0)3 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: 𝑦) 𝑥 = 𝑒 BC (𝐶. sin 𝛽𝑥 + C3 cos 𝛽𝑥) 𝑦) 𝑥 = 𝑒
J C
(C. sin 2𝑥 + C3 cos 2𝑥)
Simplifying we find the general solution of the equation is: 𝑦) 𝑥 = (1)(𝐶. sin 2𝑥 + C3 cos 2𝑥) 𝑦) (𝑥) = 𝐶. sin 2𝑥 + C3 cos 2𝑥
Therefore, the correct answer choice is D.𝐲𝐡 (𝐱) = 𝐂𝟏 𝐬𝐢𝐧 𝟐𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝟐𝐱
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PROBLEM 2: What is the particular solution of the differential equation and initial conditions below? 𝑦 "" − 4𝑦 " + 4𝑦 = 0; 𝑦 0 = 4, 𝑦 " 0 = 5 A. 𝑦) 𝑥 = 4𝑒 3C − 3𝑥𝑒 3C B. 𝑦) 𝑥 = 4𝑒 3C + 3𝑥𝑒 3C C. 𝑦) 𝑥 = −4𝑒 3C − 3𝑥𝑒 3C D. 𝑦) 𝑥 = −4𝑒 3C + 3𝑥𝑒 3C
SOLUTION 2: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" − 4𝑦 " + 4𝑦 = 0 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 3 + 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 3 − 4𝑟 + 4 = 0 Where: • 𝑎 = −4 • 𝑏=4 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = −4
3
= 16
4 𝑏 = 4 4 = 16 We know that 16 = 16, and can class the solution as critically damped, and that the roots are characterized as distinct and real roots.
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An equal and real root is characterized by the relationship 𝑎3 = 4𝑏, where the solution is of the critically damped form. The two roots are real and the same, such that they can be referred to as the same variable. Repeated real roots are represented as 𝑟. = 𝑟3 , where 𝑟. and 𝑟3 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.,3 = 2 If the root 𝑟. = 𝑟3 , then 𝐶3 𝑒 XG C is replaced with 𝐶3 𝑥𝑒 XY C . If the roots are real and the same, the solution is: 𝑦) 𝑥 = (𝐶. + 𝐶3 𝑥 + ⋯ + 𝐶[ 𝑥 \F. )𝑒 XC Plugging in the calculated value for the roots, we find the general solution is expressed as: 𝑦) 𝑥 = 𝐶. 𝑒 3C + 𝐶3 𝑥𝑒 3C
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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 of 𝑦 0 = 4, 𝑦 " 0 = 5 First we take the general solution and use the initial condition 𝑦 0 = 4, which has values of 𝑥 = 0 and 𝑦 = 4: 𝑦) 𝑥 = 𝐶. 𝑒 3C + 𝐶3 𝑥𝑒 3C Plugging in our values to the general solution we find: 4 = 𝐶. 𝑒 3(J) + 𝐶3 (0)𝑒 3(J) 4 = 𝐶. 𝑒 J + 0 We can then solve for 𝐶. as the second constant was eliminated from the equation: 𝐶. = 4 In order to use the second initial condition, we need to calculate the derivative of the general solution: 𝑦)" 𝑥 = 2𝐶. 𝑒 3C + 𝐶3 (2𝑥𝑒 3C + 𝑒 3C )
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We then plug in the initial condition 𝑦 " 0 = 5, which has values of 𝑥 = 0 and 𝑦 = 5, as well as the calculated constant value of 𝐶. = 4: 5 = 2 4 𝑒 3(J) + 𝐶3 2 0 𝑒 3
J
+ 𝑒3
J
Simplifying the equation, we find the equation is rewritten as: 5 = 8 1 + 𝐶3 0 + 1 Solving for the miss constant 𝐶3 , we find: 𝐶3 = −3 Plugging the values for 𝐶. and 𝐶3 into the general solution, we find the particular solution is: 𝑦) 𝑥 = 4𝑒 3C − 3𝑥𝑒 3C
Therefore, the correct answer choice is A. 𝐲𝐡 𝐱 = 𝟒𝐞𝟐𝐱 − 𝟑𝐱𝐞𝟐𝐱
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PROBLEM 3: What is the general solution to the following second-order differential equation if 𝐴 and 𝐵 are constants? 𝑦 "" + 9𝑦 = 0 A. 𝑦 𝑥 = 𝐴𝑒 eC + 𝐵𝑥𝑒 FeC B. 𝑦 𝑥 = 𝐴𝑒 eC + 𝐵𝑒 FeC C. 𝑦 𝑥 = 𝐴𝑒 C − 𝐵𝑥𝑒 FC D. 𝑦 𝑥 = 𝐴 cos 3𝑥 + 𝐵 sin 3𝑥
SOLUTION 3: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 9𝑦 = 0 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 3 + 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: 𝑟3 + 9 = 0 Where: • 𝑎=0 • 𝑏 = 9 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = 0
3
=0
4 𝑏 = 4 9 = 36 We know that 0 < 36, and can class the solution as under damped, and that the roots are characterized as complex roots. A complex root is characterized by the relationship 𝑎3 < 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 𝑟3 = 𝛼 − 𝑖𝛽 If the roots are complex, it is convenient to write solution terms corresponding to complex roots with trigonometric functions. 𝑦) 𝑥 = 𝑒 BC (𝐶. sin 𝛽𝑥 + 𝐶3 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=
B 3
DEFB G 3
Solving for “𝛼”, we plug in the vales for 𝑎 = 0 to find:
𝛼=−
𝑎 0 =− =0 2 2
Solving for 𝛽, we plug in the vales for 𝑎 = 0 and 𝑏 = 9, to find:
4𝑏 − 𝑎3 𝛽= = 2
4(9) − (0)3 6 = =3 2 2
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Plugging in the calculated 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 BC (C. sin 𝛽𝑥 + C3 cos 𝛽𝑥) 𝑦) 𝑥 = 𝑒
J C
(C. sin 3𝑥 + C3 cos 3𝑥)
Simplifying we find the general solution of the equation is: 𝑦) 𝑥 = (1)(𝐶. sin 3𝑥 + C3 cos 3𝑥) Plugging in the constants of A and B in lieu of 𝐶. and 𝐶3 , we find the general solution is: 𝑦) (𝑥) = 𝐴 sin 3𝑥 + B cos 3𝑥
Therefore, the correct answer choice is D.𝐲𝐡 (𝐱) = 𝐀 𝐬𝐢𝐧 𝟑𝐱 + 𝐁 𝐜𝐨𝐬 𝟑𝐱
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PROBLEM 4: What is the particular solution of the differential equation and initial conditions below? 𝑦 "" + 4𝑦′ + 4𝑦 = 0; 𝑦 0 = 1, 𝑦 " 0 = 0 A. 𝑦) 𝑥 = 1 − 2𝑥 𝑒 3C B. 𝑦) 𝑥 = 2 − 𝑥 𝑒 F3C C. 𝑦) 𝑥 = 2 + 𝑥 𝑒 F3C D. 𝑦) 𝑥 = 1 + 2𝑥 𝑒 F3C
SOLUTION 4: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" + 4𝑦′ + 4𝑦 = 0 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 3 + 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 3 + 4𝑟 + 4 = 0 Where: • 𝑎=4 • 𝑏=4 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = 4
3
= 16
4 𝑏 = 4 4 = 16 We know that 16 = 16, and can class the solution as critically damped, and that the roots are characterized as distinct and real roots. An equal and real root is characterized by the relationship 𝑎3 = 4𝑏, where the solution is of the critically damped form.
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The two roots are real and the same, such that they can be referred to as the same variable. Repeated real roots are represented as 𝑟. = 𝑟3 , where 𝑟. and 𝑟3 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.,3 = −2 If the root 𝑟. = 𝑟3 , then 𝐶3 𝑒 XG C is replaced with 𝐶3 𝑥𝑒 XY C . If the roots are real and the same, the solution is: 𝑦) 𝑥 = (𝐶. + 𝐶3 𝑥 + ⋯ + 𝐶[ 𝑥 \F. )𝑒 XC Plugging in the calculated value for the roots, we find the general solution is expressed as: 𝑦) 𝑥 = 𝐶. 𝑒 F3C + 𝐶3 𝑥𝑒 F3C 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 of 𝑦 0 = 1, 𝑦 " 0 = 0 First we take the general solution and use the initial condition 𝑦 0 = 1, which has values of 𝑥 = 0 and 𝑦 = 1: 𝑦) 𝑥 = 𝐶. 𝑒 F3C + 𝐶3 𝑥𝑒 F3C Plugging in our values to the general solution we find: 1 = 𝐶. 𝑒 3(J) + 𝐶3 (0)𝑒 3(J) 1 = 𝐶. 𝑒 J + 0 We can then solve for 𝐶. as the second constant was eliminated from the equation: 𝐶. = 1 In order to use the second initial condition, we need to calculate the derivative of the general solution: 𝑦)" 𝑥 = 𝐶. + 𝐶3 𝑥 𝑒 F3C −2 + 𝐶3 (𝑒 F3C ) We then plug in the initial condition 𝑦 " 0 = 0, which has values of 𝑥 = 0 and 𝑦 = 0, as well as the calculated constant value of 𝐶. = 1: 0 = 1 + 𝐶3 0
𝑒 F3
J
−2 + 𝐶3 (𝑒 F3C )
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Simplifying the equation, we find the equation is rewritten as: 0 = −2 1 + 𝐶3 Solving for the miss constant 𝐶3 , we find: 𝐶3 = 2 Plugging the values for 𝐶. and 𝐶3 into the general solution, we find the particular solution is: 𝑦) 𝑥 = 1 + 2𝑥 𝑒 F3C
Therefore, the correct answer choice is D. 𝐲𝐡 𝐱 = 𝟏 + 𝟐𝐱 𝐞F𝟐𝐱
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PROBLEM 5: What is the general solution of the differential equation below? 𝑦 "" − 8𝑦 " + 16𝑦 = 0 A. 𝑦) 𝑥 = 𝐶. 𝑒 DC B. 𝑦) 𝑥 = 𝐶. + 𝐶3 𝑥 𝑒 DC C. 𝑦) 𝑥 = 𝐶. 𝑒 FDC + 𝐶3 𝑒 DC D. 𝑦) 𝑥 = 𝐶. 𝑒 3C + 𝐶3 𝑒 DC
SOLUTION 5: The first step is to determine the corresponding homogeneous solution to the given differential equation: 𝑦 "" − 8𝑦 " + 16𝑦 = 0 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 3 + 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 3 − 8𝑟 + 16 = 0 Where: • 𝑎 = −8 • 𝑏 = 16 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 plug in the values for the variables “𝑎” and “𝑏”, and then look to determine the form of the solution. We know that the 𝑎3 term and 4𝑏 term are always used to determine the form of the solution. 𝑎3 = −8
3
= 64
4 𝑏 = 4 16 = 64 We know that 64 = 64, and can class the solution as critically damped, and that the roots are characterized as distinct and real roots. An equal and real root is characterized by the relationship 𝑎3 = 4𝑏, where the solution is of the critically damped form.
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The two roots are real and the same, such that they can be referred to as the same variable. Repeated real roots are represented as 𝑟. = 𝑟3 , where 𝑟. and 𝑟3 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.,3 = 4 If the root 𝑟. = 𝑟3 , then 𝐶3 𝑒 XG C is replaced with 𝐶3 𝑥𝑒 XY C . If the roots are real and the same, the solution is: 𝑦) 𝑥 = (𝐶. + 𝐶3 𝑥 + ⋯ + 𝐶[ 𝑥 \F. )𝑒 XC Plugging in the calculated value for the roots, we find the general solution is expressed as: 𝑦) 𝑥 = 𝐶. 𝑒 DC + 𝐶3 𝑥𝑒 DC 𝑦) 𝑥 = 𝐶. + 𝐶3 𝑥 𝑒 DC
The correct answer is B. 𝐲𝐡 𝐱 = 𝐂𝟏 + 𝐂𝟐 𝐱 𝐞𝟒𝐱
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