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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

The solution to the initial value problem y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 . The solution to the initial value problem y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 is y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + 7 6 e 3 t + 1 60 e − 4 t _ . Procedure used: 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. Calculation: The given initial value problem is, y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 . The corresponding auxiliary equation is r 2 + r − 12 = 0 . Simplify r 2 + r − 12 = 0 as, r 2 + 4 r − 3 r − 12 = 0 r ( r + 4 ) − 3 ( r + 4 ) = 0 ( r − 3 ) ( r + 4 ) = 0 r = 3 , − 4 That is, the roots are r 1 = 3 and r 2 = − 4 . Thus, the general solution for the homogenous equation is, y ( t ) = c 1 e 3 t + c 2 e − 4 t . The RHS of the differential equation y ″ + y ′ − 12 y = e t + e 2 t − 1 has three terms e t , e 2 t and − 1 . Comparing the term e t with the form C t m e r t , it is observed that m = 0 and r = 1 . Since r = 1 is a 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 = 1 will be, y p ( t ) = t 0 ( A 0 ) e t = A 0 e t Comparing the term e 2 t with the form C t m e r t , it is observed that m = 0 and r = 2 . Since r = 2 is a 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 = 2 will be, y p ( t ) = t 0 ( A 0 ) e 2 t = A 0 e 2 t Comparing the term − 1 with the form C t m e r t , it is observed that m = 0 and r = 0 . Since r = 0 is a 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 e 2 t + B e t + C . Obtain the first and second order derivatives of y p ( t ) = A e 2 t + B e t + C as follows: Differentiate y p ( t ) = A e 2 t + B e t + C with respect to t as, y ′ p ( t ) = 2 A e 2 t + B e t . Differentiate y ′ p ( t ) = 2 A e 2 t + B e t with respect to t as, y ″ p ( t ) = 4 A e 2 t + B e t . Thus, y ″ + y ′ − 12 y for y p will be, y ″ p + y ′ p − 12 y p = ( 4 A e 2 t + B e t ) + ( 2 A e 2 t + B e t ) − 12 ( A e 2 t + B e t + C ) = 4 A e 2 t + B e t + 2 A e 2 t + B e t − 12 A e 2 t − 12 B e t − 12 C = − 6 A e 2 t − 10 B e t − 12 C Setting − 6 A e 2 t − 10 B e t − 12 C = e t + e 2 t − 1 and comparing the coefficients result in − 6 A = 1 , − 10 B = 1 and − 12 C = − 1 . On simplification, A = − 1 6 , B = − 1 10 and C = 1 12 . Thus, the particular solution y p ( t ) = A e 2 t + B e t + C becomes y p ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 . Combine the general solution y ( t ) = c 1 e 3 t + c 2 e − 4 t of the corresponding homogenous equation and the particular solution y p ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 to obtain y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t . Obtain the value of the constants using the initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = 3 as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t , y ( 0 ) = − 1 6 e 2 × 0 − 1 10 e 0 + 1 12 + c 1 e 3 × 0 + c 2 e − 4 × 0 1 = − 1 6 − 1 10 + 1 12 + c 1 + c 2 c 1 + c 2 = 1 + 1 6 + 1 10 − 1 12 c 1 + c 2 = 120 + 20 + 12 − 10 120 Simplify further as, c 1 + c 2 = 142 120 c 1 + c 2 = 71 60 c 1 = 71 60 − c 2 Differentiate y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t with respect to t as, y ′ ( t ) = − 1 6 ( 2 e 2 t ) − 1 10 e t + c 1 ( 3 e 3 t ) + c 2 ( − 4 e − 4 t ) = − 1 3 e 2 t − 1 10 e t + 3 c 1 e 3 t − 4 c 2 e − 4 t Substitute t = 0 and y ′ ( 0 ) = 3 in y ′ ( t ) = − 1 3 e 2 t − 1 10 e t + 3 c 1 e 3 t − 4 c 2 e − 4 t , y ′ ( 0 ) = − 1 3 e 2 × 0 − 1 10 e 0 + 3 c 1 e 3 × 0 − 4 c 2 e − 4 × 0 3 = − 1 3 − 1 10 + 3 c 1 − 4 c 2 3 c 1 − 4 c 2 = 3 + 1 3 + 1 10 3 c 1 − 4 c 2 = 90 + 10 + 3 30 3 c 1 − 4 c 2 = 103 30 Use c 1 = 71 60 − c 2 and simplify as, 3 ( 71 60 − c 2 ) − 4 c 2 = 103 30 71 20 − 3 c 2 − 4 c 2 = 103 30 71 20 − 7 c 2 = 103 30 7 c 2 = 71 20 − 103 30 On further simplification, 7 c 2 = 213 − 206 60 c 2 = 7 60 × 7 c 2 = 1 60 Thus, c 1 = 71 60 − c 2 becomes, c 1 = 71 60 − 1 60 = 70 60 = 7 6 That is, c 1 = 7 6 and c 2 = 1 60 . Thus, the solution y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t becomes, y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + 7 6 e 3 t + 1 60 e − 4 t . Therefore, the solution to the initial value problem y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 is y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + 7 6 e 3 t + 1 60 e − 4 t _ .

The solution to the initial value problem y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 . The solution to the initial value problem y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 is y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + 7 6 e 3 t + 1 60 e − 4 t _ . Procedure used: 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. Calculation: The given initial value problem is, y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 . The corresponding auxiliary equation is r 2 + r − 12 = 0 . Simplify r 2 + r − 12 = 0 as, r 2 + 4 r − 3 r − 12 = 0 r ( r + 4 ) − 3 ( r + 4 ) = 0 ( r − 3 ) ( r + 4 ) = 0 r = 3 , − 4 That is, the roots are r 1 = 3 and r 2 = − 4 . Thus, the general solution for the homogenous equation is, y ( t ) = c 1 e 3 t + c 2 e − 4 t . The RHS of the differential equation y ″ + y ′ − 12 y = e t + e 2 t − 1 has three terms e t , e 2 t and − 1 . Comparing the term e t with the form C t m e r t , it is observed that m = 0 and r = 1 . Since r = 1 is a 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 = 1 will be, y p ( t ) = t 0 ( A 0 ) e t = A 0 e t Comparing the term e 2 t with the form C t m e r t , it is observed that m = 0 and r = 2 . Since r = 2 is a 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 = 2 will be, y p ( t ) = t 0 ( A 0 ) e 2 t = A 0 e 2 t Comparing the term − 1 with the form C t m e r t , it is observed that m = 0 and r = 0 . Since r = 0 is a 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 e 2 t + B e t + C . Obtain the first and second order derivatives of y p ( t ) = A e 2 t + B e t + C as follows: Differentiate y p ( t ) = A e 2 t + B e t + C with respect to t as, y ′ p ( t ) = 2 A e 2 t + B e t . Differentiate y ′ p ( t ) = 2 A e 2 t + B e t with respect to t as, y ″ p ( t ) = 4 A e 2 t + B e t . Thus, y ″ + y ′ − 12 y for y p will be, y ″ p + y ′ p − 12 y p = ( 4 A e 2 t + B e t ) + ( 2 A e 2 t + B e t ) − 12 ( A e 2 t + B e t + C ) = 4 A e 2 t + B e t + 2 A e 2 t + B e t − 12 A e 2 t − 12 B e t − 12 C = − 6 A e 2 t − 10 B e t − 12 C Setting − 6 A e 2 t − 10 B e t − 12 C = e t + e 2 t − 1 and comparing the coefficients result in − 6 A = 1 , − 10 B = 1 and − 12 C = − 1 . On simplification, A = − 1 6 , B = − 1 10 and C = 1 12 . Thus, the particular solution y p ( t ) = A e 2 t + B e t + C becomes y p ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 . Combine the general solution y ( t ) = c 1 e 3 t + c 2 e − 4 t of the corresponding homogenous equation and the particular solution y p ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 to obtain y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t . Obtain the value of the constants using the initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = 3 as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t , y ( 0 ) = − 1 6 e 2 × 0 − 1 10 e 0 + 1 12 + c 1 e 3 × 0 + c 2 e − 4 × 0 1 = − 1 6 − 1 10 + 1 12 + c 1 + c 2 c 1 + c 2 = 1 + 1 6 + 1 10 − 1 12 c 1 + c 2 = 120 + 20 + 12 − 10 120 Simplify further as, c 1 + c 2 = 142 120 c 1 + c 2 = 71 60 c 1 = 71 60 − c 2 Differentiate y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t with respect to t as, y ′ ( t ) = − 1 6 ( 2 e 2 t ) − 1 10 e t + c 1 ( 3 e 3 t ) + c 2 ( − 4 e − 4 t ) = − 1 3 e 2 t − 1 10 e t + 3 c 1 e 3 t − 4 c 2 e − 4 t Substitute t = 0 and y ′ ( 0 ) = 3 in y ′ ( t ) = − 1 3 e 2 t − 1 10 e t + 3 c 1 e 3 t − 4 c 2 e − 4 t , y ′ ( 0 ) = − 1 3 e 2 × 0 − 1 10 e 0 + 3 c 1 e 3 × 0 − 4 c 2 e − 4 × 0 3 = − 1 3 − 1 10 + 3 c 1 − 4 c 2 3 c 1 − 4 c 2 = 3 + 1 3 + 1 10 3 c 1 − 4 c 2 = 90 + 10 + 3 30 3 c 1 − 4 c 2 = 103 30 Use c 1 = 71 60 − c 2 and simplify as, 3 ( 71 60 − c 2 ) − 4 c 2 = 103 30 71 20 − 3 c 2 − 4 c 2 = 103 30 71 20 − 7 c 2 = 103 30 7 c 2 = 71 20 − 103 30 On further simplification, 7 c 2 = 213 − 206 60 c 2 = 7 60 × 7 c 2 = 1 60 Thus, c 1 = 71 60 − c 2 becomes, c 1 = 71 60 − 1 60 = 70 60 = 7 6 That is, c 1 = 7 6 and c 2 = 1 60 . Thus, the solution y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + c 1 e 3 t + c 2 e − 4 t becomes, y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + 7 6 e 3 t + 1 60 e − 4 t . Therefore, the solution to the initial value problem y ″ + y ′ − 12 y = e t + e 2 t − 1 , y ( 0 ) = 1 , y ′ ( 0 ) = 3 is y ( t ) = − 1 6 e 2 t − 1 10 e t + 1 12 + 7 6 e 3 t + 1 60 e − 4 t _ .

Solution Summary: The author explains the solution to the initial value problem y′′+xprime, where m is a nonnegative integer.

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4 Solutions

Fundamentals of Differential Equations (9th Edition)

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:...

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Intro to the Laplace Transform & Three Examples; Author: Dr. Trefor Bazett;https://www.youtube.com/watch?v=KqokoYr_h1A;License: Standard YouTube License, CC-BY