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

Whether the method of undetermined coefficients along with the superposition principle can be applied to obtain a particular solution to the differential equation y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 t . No _ , the method of undetermined coefficients cannot be applied to obtain a particular solution to the differential equation y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 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. A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. Theorem used: Superposition principle: If y 1 is a solution to the differential equation a y ″ + b y ′ + c y = f 1 ( t ) and y 2 is a solution to a y ″ + b y ′ + c y = f 2 ( t ) , then for any constants k 1 and k 2 , the function k 1 y 1 + k 2 y 2 is a solution to the differential equation a y ″ + b y ′ + c y = k 1 f 1 ( t ) + k 2 f 2 ( t ) . Calculation: The given differential equation is, y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 t . Note that, the RHS of the differential equation has three terms 4 sin 3 t , − e 3 t t 2 and 1 t . The term 1 t can be written as t − 1 , which has negative power. That is, t − 1 does not satisfy the required form C t m e r t , where m is a nonnegative integer. Thus, the method of undetermined coefficients cannot be used. Therefore, the method of undetermined coefficients cannot be applied to obtain a particular solution to the differential equation y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 t .

Whether the method of undetermined coefficients along with the superposition principle can be applied to obtain a particular solution to the differential equation y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 t . No _ , the method of undetermined coefficients cannot be applied to obtain a particular solution to the differential equation y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 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. A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. Theorem used: Superposition principle: If y 1 is a solution to the differential equation a y ″ + b y ′ + c y = f 1 ( t ) and y 2 is a solution to a y ″ + b y ′ + c y = f 2 ( t ) , then for any constants k 1 and k 2 , the function k 1 y 1 + k 2 y 2 is a solution to the differential equation a y ″ + b y ′ + c y = k 1 f 1 ( t ) + k 2 f 2 ( t ) . Calculation: The given differential equation is, y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 t . Note that, the RHS of the differential equation has three terms 4 sin 3 t , − e 3 t t 2 and 1 t . The term 1 t can be written as t − 1 , which has negative power. That is, t − 1 does not satisfy the required form C t m e r t , where m is a nonnegative integer. Thus, the method of undetermined coefficients cannot be used. Therefore, the method of undetermined coefficients cannot be applied to obtain a particular solution to the differential equation y ″ − 6 y ′ − 4 y = 4 sin 3 t − e 3 t t 2 + 1 t .

Solution Summary: The author explains that the method of undetermined coefficients along with the superposition principle can be applied to obtain a particular solution to the differential equation.

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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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

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