Skip to main content

Math Advanced Math Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

The particular solution to the differential equation y ″ − y = 2 t + 4 using undetermined coefficients and variation of parameters and explain which method is quicker. The particular solution to the differential equation y ″ − y = 2 t + 4 is y p ( t ) = − 2 t − 4 _ . Procedure used: 1. Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = C t m e r t , where m is a nonnegative integer, is of the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t with: (i) s = 0 , if r is not a root of the associated auxiliary equation; (ii) s = 1 , if r is a simple root of the associated auxiliary equation; (iii) s = 2 , if r is a double root of the associated auxiliary equation. 2. Method of variation of parameters: The following are the steps to obtain a particular solution to a y ″ + b y ′ + c y = f . (i) Find two linearly independent solutions { y 1 ( t ) , y 2 ( t ) } to the corresponding homogenous equation and take y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . (ii) Determine v 1 ( t ) and v 2 ( t ) by solving the below system for v ′ 1 ( t ) and v ′ 2 ( t ) and integrating. y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a This can be simplified using Cramer’s rule as, v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] and v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] (iii) Substitute v 1 ( t ) and v 2 ( t ) into the expression for y p ( t ) to obtain a particular solution. Calculation: The given differential equation is, y ″ − y = 2 t + 4 . The corresponding auxiliary equation is r 2 − 1 = 0 . Simplify r 2 − 1 = 0 as, r 2 = 1 r = ± 1 That is, the roots are r 1 = − 1 and r 2 = 1 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 e − t + c 2 e t . Obtain the particular solution using the method of undetermined coefficients as follows: The RHS of the differential equation y ″ − y = 2 t + 4 has two terms 2 t and 4 . Comparing the term 2 t with the form C t m e r t , it is observed that m = 1 and r = 0 . Since r = 0 is not a root of the associated auxiliary equation, s = 0 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t for s = 0 , m = 1 and r = 0 will be, y p ( t ) = t 0 ( A 1 t + A 0 ) e 0 t = A 1 t + A 0 Similarly, comparing the term 4 with the form C t m e r t , it is observed that m = 0 and r = 0 . Since r = 0 is not a root of the associated auxiliary equation, s = 0 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t for s = 0 , m = 0 and r = 0 will be, y p ( t ) = t 0 ( A 0 ) e 0 t = A 0 Combine the particular solution forms for the individual terms and rename the coefficients as, y p ( t ) = ( A t + B ) + C . Since there are two constants, replace with a single term D as, y p ( t ) = A t + D . Obtain the first and second order derivatives of y p ( t ) = A t + D as follows: Differentiate y p ( t ) = A t + D with respect to t as, y ′ p ( t ) = A . Differentiate y ′ p ( t ) = A with respect to t as, y ″ p ( t ) = 0 . Thus, y ″ − y for y p will be, y ″ p − y p = 0 − ( A t + D ) = − A t − D Setting − A t − D = 2 t + 4 and comparing the coefficients result in, − A = 2 and − D = 4 . On simplification, A = − 2 and D = − 4 . Thus, the particular solution y p ( t ) = A t + D becomes y p ( t ) = − 2 t − 4 . Obtain the particular solution using the method of variation of parameters as follows: Compare y ″ − y = 2 t + 4 with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = − 1 and f = 2 t + 4 . Since the general solution for the homogenous equation y ″ − y = 0 is y ( t ) = c 1 e − t + c 2 e t , the two independent solutions are e − t and e t . Let y 1 ( t ) = e − t and y 2 ( t ) = e t . Then, y ′ 1 ( t ) = − e − t and y ′ 2 ( t ) = e t . Set the particular solution y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) as y p ( t ) = v 1 ( t ) e − t + v 2 ( t ) e t . That is, a = 1 , y 1 ( t ) = e − t , y 2 ( t ) = e t , y ′ 1 ( t ) = − e − t , y ′ 2 ( t ) = e t and f = 2 t + 4 . The equation v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 1 ( t ) = − ( 2 t + 4 ) e t 1 [ e − t ( e t ) − ( − e − t ) e t ] = − ( 2 t + 4 ) e t [ 1 + 1 ] = − ( t + 2 ) e t The equation v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 2 ( t ) = ( 2 t + 4 ) e − t 1 [ e − t ( e t ) − ( − e − t ) e t ] = ( 2 t + 4 ) e − t [ 1 + 1 ] = ( t + 2 ) e − t Integrating v ′ 2 ( t ) = ( t + 2 ) e − t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ ( t + 2 ) e − t d t = ( t + 2 ) ∫ e − t d t − ∫ − e − t d t = − ( t + 2 ) e − t − e − t + C 1 = − ( t + 3 ) e − t + C 1 Integrating v ′ 1 ( t ) = − ( t + 2 ) e t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) d t = ∫ − ( t + 2 ) e t d t = − ( t + 2 ) ∫ e t d t − ∫ − e t d t = − ( t + 2 ) e t + e t + C 2 = − ( t + 1 ) e t + C 2 Since particular solution is to be obtained, take both C 1 and C 2 zero for simplicity. That is, v 1 ( t ) = − ( t + 1 ) e t and v 2 ( t ) = − ( t + 3 ) e − t . Thus, the particular solution y p ( t ) = v 1 ( t ) e − t + v 2 ( t ) e t becomes, y p ( t ) = − ( t + 1 ) e t e − t − ( t + 3 ) e − t e t = − ( t + 1 ) − ( t + 3 ) = − 2 t − 4 Therefore, the particular solution to the differential equation y ″ − y = 2 t + 4 is y p ( t ) = − 2 t − 4 _ . With the number of steps to obtain a particular solution, the method of undetermined coefficients gives quick result than the method of variation of parameters.

The particular solution to the differential equation y ″ − y = 2 t + 4 using undetermined coefficients and variation of parameters and explain which method is quicker. The particular solution to the differential equation y ″ − y = 2 t + 4 is y p ( t ) = − 2 t − 4 _ . Procedure used: 1. Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = C t m e r t , where m is a nonnegative integer, is of the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t with: (i) s = 0 , if r is not a root of the associated auxiliary equation; (ii) s = 1 , if r is a simple root of the associated auxiliary equation; (iii) s = 2 , if r is a double root of the associated auxiliary equation. 2. Method of variation of parameters: The following are the steps to obtain a particular solution to a y ″ + b y ′ + c y = f . (i) Find two linearly independent solutions { y 1 ( t ) , y 2 ( t ) } to the corresponding homogenous equation and take y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . (ii) Determine v 1 ( t ) and v 2 ( t ) by solving the below system for v ′ 1 ( t ) and v ′ 2 ( t ) and integrating. y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a This can be simplified using Cramer’s rule as, v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] and v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] (iii) Substitute v 1 ( t ) and v 2 ( t ) into the expression for y p ( t ) to obtain a particular solution. Calculation: The given differential equation is, y ″ − y = 2 t + 4 . The corresponding auxiliary equation is r 2 − 1 = 0 . Simplify r 2 − 1 = 0 as, r 2 = 1 r = ± 1 That is, the roots are r 1 = − 1 and r 2 = 1 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 e − t + c 2 e t . Obtain the particular solution using the method of undetermined coefficients as follows: The RHS of the differential equation y ″ − y = 2 t + 4 has two terms 2 t and 4 . Comparing the term 2 t with the form C t m e r t , it is observed that m = 1 and r = 0 . Since r = 0 is not a root of the associated auxiliary equation, s = 0 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t for s = 0 , m = 1 and r = 0 will be, y p ( t ) = t 0 ( A 1 t + A 0 ) e 0 t = A 1 t + A 0 Similarly, comparing the term 4 with the form C t m e r t , it is observed that m = 0 and r = 0 . Since r = 0 is not a root of the associated auxiliary equation, s = 0 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t for s = 0 , m = 0 and r = 0 will be, y p ( t ) = t 0 ( A 0 ) e 0 t = A 0 Combine the particular solution forms for the individual terms and rename the coefficients as, y p ( t ) = ( A t + B ) + C . Since there are two constants, replace with a single term D as, y p ( t ) = A t + D . Obtain the first and second order derivatives of y p ( t ) = A t + D as follows: Differentiate y p ( t ) = A t + D with respect to t as, y ′ p ( t ) = A . Differentiate y ′ p ( t ) = A with respect to t as, y ″ p ( t ) = 0 . Thus, y ″ − y for y p will be, y ″ p − y p = 0 − ( A t + D ) = − A t − D Setting − A t − D = 2 t + 4 and comparing the coefficients result in, − A = 2 and − D = 4 . On simplification, A = − 2 and D = − 4 . Thus, the particular solution y p ( t ) = A t + D becomes y p ( t ) = − 2 t − 4 . Obtain the particular solution using the method of variation of parameters as follows: Compare y ″ − y = 2 t + 4 with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = − 1 and f = 2 t + 4 . Since the general solution for the homogenous equation y ″ − y = 0 is y ( t ) = c 1 e − t + c 2 e t , the two independent solutions are e − t and e t . Let y 1 ( t ) = e − t and y 2 ( t ) = e t . Then, y ′ 1 ( t ) = − e − t and y ′ 2 ( t ) = e t . Set the particular solution y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) as y p ( t ) = v 1 ( t ) e − t + v 2 ( t ) e t . That is, a = 1 , y 1 ( t ) = e − t , y 2 ( t ) = e t , y ′ 1 ( t ) = − e − t , y ′ 2 ( t ) = e t and f = 2 t + 4 . The equation v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 1 ( t ) = − ( 2 t + 4 ) e t 1 [ e − t ( e t ) − ( − e − t ) e t ] = − ( 2 t + 4 ) e t [ 1 + 1 ] = − ( t + 2 ) e t The equation v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 2 ( t ) = ( 2 t + 4 ) e − t 1 [ e − t ( e t ) − ( − e − t ) e t ] = ( 2 t + 4 ) e − t [ 1 + 1 ] = ( t + 2 ) e − t Integrating v ′ 2 ( t ) = ( t + 2 ) e − t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ ( t + 2 ) e − t d t = ( t + 2 ) ∫ e − t d t − ∫ − e − t d t = − ( t + 2 ) e − t − e − t + C 1 = − ( t + 3 ) e − t + C 1 Integrating v ′ 1 ( t ) = − ( t + 2 ) e t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) d t = ∫ − ( t + 2 ) e t d t = − ( t + 2 ) ∫ e t d t − ∫ − e t d t = − ( t + 2 ) e t + e t + C 2 = − ( t + 1 ) e t + C 2 Since particular solution is to be obtained, take both C 1 and C 2 zero for simplicity. That is, v 1 ( t ) = − ( t + 1 ) e t and v 2 ( t ) = − ( t + 3 ) e − t . Thus, the particular solution y p ( t ) = v 1 ( t ) e − t + v 2 ( t ) e t becomes, y p ( t ) = − ( t + 1 ) e t e − t − ( t + 3 ) e − t e t = − ( t + 1 ) − ( t + 3 ) = − 2 t − 4 Therefore, the particular solution to the differential equation y ″ − y = 2 t + 4 is y p ( t ) = − 2 t − 4 _ . With the number of steps to obtain a particular solution, the method of undetermined coefficients gives quick result than the method of variation of parameters.

Solution Summary: The author explains the particular solution to the differential equation y′′-y=2t+4 using undetermined coefficients and variation of parameters.

Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

9th Edition

ISBN: 9780134768748

Author: Nagle

Publisher: PEARSON

See similar textbooks

Concept explainers

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Videos

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 4.6, Problem 8E

Chapter 4.6, Problem 10E

Students have asked these similar questions

Hi can anyone help me with getting point of Symmetry for Rayleigh equation limit cycles and stability. Thqnx you

Prove it pass to pass

proof heb (a+b)" - {("r) a". b-r

Chapter 4 Solutions

Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 3E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 5E Ch. 4.2 - Prob. 6E Ch. 4.2 - Prob. 7E Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Prob. 10E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 15E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - Prob. 19E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Prob. 24E Ch. 4.2 - Prob. 25E Ch. 4.2 - Prob. 26E Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - Prob. 31E Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Using the definition in Problem 35, prove that if...Ch. 4.2 - Prob. 37E Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.2 - In Problems 3741, find three linearly independent...Ch. 4.2 - Prob. 42E Ch. 4.2 - Prob. 43E Ch. 4.2 - Solve the initial value problem: y 2y y + 2y =...Ch. 4.2 - Prob. 45E Ch. 4.2 - Prob. 46E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - Prob. 5E Ch. 4.3 - Prob. 6E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 920, find a general solution. 9. y 8y...Ch. 4.3 - In Problems 920, find a general solution. 10. y +...Ch. 4.3 - Prob. 11E Ch. 4.3 - Prob. 12E Ch. 4.3 - In Problems 920, find a general solution. 13. y ...Ch. 4.3 - Prob. 14E Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - In Problems 920, find a general solution. 17. y y...Ch. 4.3 - In Problems 920, find a general solution. 18. 2y +...Ch. 4.3 - Prob. 19E Ch. 4.3 - In Problems 920, find a general solution. 20. y y...Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 25E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 27E Ch. 4.3 - Prob. 28E Ch. 4.3 - Prob. 29E Ch. 4.3 - Prob. 30E Ch. 4.3 - Prob. 31E Ch. 4.3 - Prob. 32E Ch. 4.3 - Prob. 33E Ch. 4.3 - Prob. 34E Ch. 4.3 - Swinging Door. The motion of a swinging door with...Ch. 4.3 - Prob. 36E Ch. 4.3 - Prob. 37E Ch. 4.3 - Prove the sum of angles formula for the sine...Ch. 4.4 - In Problems 18, decide whether or not the method...Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Prob. 5E Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 16E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 18E Ch. 4.4 - Prob. 19E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Prob. 28E Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - Prob. 31E Ch. 4.4 - Prob. 32E Ch. 4.4 - Prob. 33E Ch. 4.4 - Prob. 34E Ch. 4.4 - Prob. 35E Ch. 4.4 - In Problems 3336, use the method of undetermined...Ch. 4.5 - Given that y1(t) = cos t is a solution to y y + y...Ch. 4.5 - Prob. 2E Ch. 4.5 - Prob. 3E Ch. 4.5 - Prob. 4E Ch. 4.5 - Prob. 5E Ch. 4.5 - Prob. 6E Ch. 4.5 - Prob. 7E Ch. 4.5 - In Problems 38, a nonhomogeneous equation and a...Ch. 4.5 - Prob. 9E Ch. 4.5 - Prob. 10E Ch. 4.5 - Prob. 11E Ch. 4.5 - Prob. 12E Ch. 4.5 - Prob. 13E Ch. 4.5 - In Problems 916 decide whether the method of...Ch. 4.5 - Prob. 15E Ch. 4.5 - Prob. 16E Ch. 4.5 - Prob. 17E Ch. 4.5 - Prob. 18E Ch. 4.5 - Prob. 19E Ch. 4.5 - In Problems 1722, find a general solution to the...Ch. 4.5 - Prob. 21E Ch. 4.5 - Prob. 22E Ch. 4.5 - Prob. 23E Ch. 4.5 - Prob. 24E Ch. 4.5 - Prob. 25E Ch. 4.5 - Prob. 26E Ch. 4.5 - Prob. 27E Ch. 4.5 - Prob. 28E Ch. 4.5 - Prob. 29E Ch. 4.5 - Prob. 30E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 33E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 35E Ch. 4.5 - Prob. 36E Ch. 4.5 - Prob. 37E Ch. 4.5 - Prob. 38E Ch. 4.5 - In Problems 3740, find a particular solution to...Ch. 4.5 - Prob. 40E Ch. 4.5 - Prob. 41E Ch. 4.5 - Prob. 42E Ch. 4.5 - Prob. 43E Ch. 4.5 - Prob. 44E Ch. 4.5 - Prob. 46E Ch. 4.5 - Prob. 47E Ch. 4.5 - Prob. 48E Ch. 4.6 - Prob. 1E Ch. 4.6 - In Exercises 12, rectangles have been drawn to...Ch. 4.6 - Prob. 3E Ch. 4.6 - Prob. 4E Ch. 4.6 - Prob. 5E Ch. 4.6 - Prob. 6E Ch. 4.6 - Prob. 7E Ch. 4.6 - Prob. 8E Ch. 4.6 - Prob. 9E Ch. 4.6 - Prob. 10E Ch. 4.6 - Prob. 11E Ch. 4.6 - Prob. 12E Ch. 4.6 - Prob. 13E Ch. 4.6 - Prob. 14E Ch. 4.6 - Prob. 15E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - Prob. 17E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - For f(x) = 3x + 2, evaluate the Riemann sums: (a)...Ch. 4.6 - For g(x) = 4x 1, evaluate the Riemann sums: (a)...Ch. 4.6 - Prob. 21E Ch. 4.6 - Use Table 5.9 to evaluate the Riemann sums: Table...Ch. 4.6 - The graph of f(t) is in Figure 5.40. Which of the...Ch. 4.6 - Prob. 24E Ch. 4.6 - Prob. 25E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 3E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 5E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 7E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 9E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 11E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 13E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - Prob. 17E Ch. 4.7 - Prob. 18E Ch. 4.7 - Prob. 19E Ch. 4.7 - Prob. 20E Ch. 4.7 - Prob. 21E Ch. 4.7 - Prob. 22E Ch. 4.7 - Prob. 23E Ch. 4.7 - Solve the following CauchyEuler equations by using...Ch. 4.7 - Prob. 25E Ch. 4.7 - Prob. 26E Ch. 4.7 - Prob. 27E Ch. 4.7 - Prob. 28E Ch. 4.7 - Prove that if y1 and y2 are linearly independent...Ch. 4.7 - Prob. 30E Ch. 4.7 - Prob. 31E Ch. 4.7 - Prob. 32E Ch. 4.7 - Prob. 33E Ch. 4.7 - Prob. 34E Ch. 4.7 - Prob. 35E Ch. 4.7 - Prob. 36E Ch. 4.7 - Prob. 37E Ch. 4.7 - Prob. 38E Ch. 4.7 - Prob. 39E Ch. 4.7 - Prob. 40E Ch. 4.7 - Prob. 41E Ch. 4.7 - In Problems 41 through 44, a differential equation...Ch. 4.7 - Prob. 43E Ch. 4.7 - Prob. 44E Ch. 4.7 - Prob. 45E Ch. 4.7 - Prob. 46E Ch. 4.7 - Prob. 47E Ch. 4.7 - Prob. 48E Ch. 4.7 - Prob. 49E Ch. 4.7 - Prob. 50E Ch. 4.7 - Prob. 51E Ch. 4.7 - Prob. 52E Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Try to predict the qualitative features of the...Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Use reduction of order to derive the solution...Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.9 - Prob. 1E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 3E Ch. 4.9 - Prob. 4E Ch. 4.9 - Prob. 5E Ch. 4.9 - Prob. 6E Ch. 4.9 - Prob. 7E Ch. 4.9 - Prob. 8E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 10E Ch. 4.9 - Prob. 11E Ch. 4.9 - Prob. 12E Ch. 4.9 - Prob. 13E Ch. 4.9 - Prob. 14E Ch. 4.9 - Prob. 15E Ch. 4.9 - Prob. 16E Ch. 4.9 - Prob. 17E Ch. 4.9 - Prob. 18E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 2E Ch. 4.10 - Prob. 3E Ch. 4.10 - Prob. 4E Ch. 4.10 - Prob. 5E Ch. 4.10 - Prob. 6E Ch. 4.10 - Prob. 7E Ch. 4.10 - Prob. 8E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 10E Ch. 4.10 - Prob. 11E Ch. 4.10 - Prob. 12E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 14E Ch. 4.10 - Prob. 15E Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 11RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 15RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 25RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP Ch. 4 - Prob. 29RP Ch. 4 - Prob. 30RP Ch. 4 - Prob. 31RP Ch. 4 - Prob. 32RP Ch. 4 - Prob. 33RP Ch. 4 - Prob. 34RP Ch. 4 - Prob. 35RP Ch. 4 - Prob. 36RP Ch. 4 - Prob. 37RP Ch. 4 - Prob. 38RP Ch. 4 - A 32-lb weight is attached to a vertical spring,...Ch. 4 - Prob. 1TWE Ch. 4 - Prob. 2TWE Ch. 4 - Prob. 3TWE Ch. 4 - For students with a background in linear algebra:...

Knowledge Booster

Background pattern image

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Intermediate Algebra
Algebra
ISBN:9780998625720
Author:Lynn Marecek
Publisher:OpenStax College
Elementary Algebra
Algebra
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:OpenStax - Rice University
PREALGEBRA
Algebra
ISBN:9781938168994
Author:OpenStax
Publisher:OpenStax
Algebra: Structure And Method, Book 1
Algebra
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:McDougal Littell
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt

Text book image

Intermediate Algebra

Algebra

ISBN:9780998625720

Author:Lynn Marecek

Publisher:OpenStax College

Text book image

Elementary Algebra

Algebra

ISBN:9780998625713

Author:Lynn Marecek, MaryAnne Anthony-Smith

Publisher:OpenStax - Rice University

Text book image

Algebra

ISBN:9781938168994

Author:OpenStax

Publisher:OpenStax

Text book image

Algebra: Structure And Method, Book 1

Algebra

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:McDougal Littell

Text book image

Holt Mcdougal Larson Pre-algebra: Student Edition...

Algebra

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Text book image

Big Ideas Math A Bridge To Success Algebra 1: Stu...

Algebra

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:Houghton Mifflin Harcourt

SEE MORE TEXTBOOKS

What is a Linear Equation in One Variable?; Author: Don't Memorise;https://www.youtube.com/watch?v=lDOYdBgtnjY;License: Standard YouTube License, CC-BY

Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY