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

(a) The solution to the differential equation y ″ + 2 y ′ + 4 y = t + cos 2 t . The solution to the differential equation y ″ + 2 y ′ + 4 y = t + cos 2 t is, y ( t ) = 1 4 sin 2 t + t 4 − 1 8 _ . 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 ″ + 2 y ′ + 4 y = t + cos 2 t . Given that, y 1 ( t ) = 1 4 sin 2 t is a solution to y ″ + 2 y ′ + 4 y = cos 2 t and y 2 ( t ) = t 4 − 1 8 is a solution to y ″ + 2 y ′ + 4 y = t . By the superposition principle, the solution to y ″ + 2 y ′ + 4 y = t + cos 2 t is obtained as follows: Here, k 1 = 1 and k 2 = 1 . Thus, y ( t ) = k 1 y 1 + k 2 y 2 is a solution to y ″ + 2 y ′ + 4 y = t + cos 2 t . That is, the solution will be, y ( t ) = k 1 y 1 + k 2 y 2 = y 1 + y 2 = 1 4 sin 2 t + t 4 − 1 8 Therefore, the solution to the differential equation y ″ + 2 y ′ + 4 y = t + cos 2 t is, y ( t ) = 1 4 sin 2 t + t 4 − 1 8 _ . (b) The solution to the differential equation y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t . The solution to the differential equation y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t is, y ( t ) = − 3 4 sin 2 t + t 2 − 1 4 _ . The given differential equation is y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t . Given that, y 1 ( t ) = 1 4 sin 2 t is a solution to y ″ + 2 y ′ + 4 y = cos 2 t and y 2 ( t ) = t 4 − 1 8 is a solution to y ″ + 2 y ′ + 4 y = t . By the superposition principle, the solution to y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t is obtained as follows: Here, k 1 = − 3 and k 2 = 2 . Thus, y ( t ) = k 1 y 1 + k 2 y 2 is a solution to y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t . That is, the solution will be, y ( t ) = k 1 y 1 + k 2 y 2 = − 3 y 1 + 2 y 2 = − 3 ( 1 4 sin 2 t ) + 2 ( t 4 − 1 8 ) = − 3 4 sin 2 t + t 2 − 1 4 Therefore, the solution to the differential equation y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t is, y ( t ) = − 3 4 sin 2 t + t 2 − 1 4 _ . (c) The solution to the differential equation y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t . The solution to the differential equation y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t is, y ( t ) = − 3 sin 2 t + 11 t 4 − 11 8 _ . The given differential equation is y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t . Given that, y 1 ( t ) = 1 4 sin 2 t is a solution to y ″ + 2 y ′ + 4 y = cos 2 t and y 2 ( t ) = t 4 − 1 8 is a solution to y ″ + 2 y ′ + 4 y = t . By the superposition principle, the solution to y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t is obtained as follows: Here, k 1 = − 12 and k 2 = 11 . Thus, y ( t ) = k 1 y 1 + k 2 y 2 is a solution to y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t . That is, the solution will be, y ( t ) = k 1 y 1 + k 2 y 2 = − 12 y 1 + 11 y 2 = − 12 ( 1 4 sin 2 t ) + 11 ( t 4 − 1 8 ) = − 3 sin 2 t + 11 t 4 − 11 8 Therefore, the solution to the differential equation y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t is, y ( t ) = − 3 sin 2 t + 11 t 4 − 11 8 _ .

(a) The solution to the differential equation y ″ + 2 y ′ + 4 y = t + cos 2 t . The solution to the differential equation y ″ + 2 y ′ + 4 y = t + cos 2 t is, y ( t ) = 1 4 sin 2 t + t 4 − 1 8 _ . 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 ″ + 2 y ′ + 4 y = t + cos 2 t . Given that, y 1 ( t ) = 1 4 sin 2 t is a solution to y ″ + 2 y ′ + 4 y = cos 2 t and y 2 ( t ) = t 4 − 1 8 is a solution to y ″ + 2 y ′ + 4 y = t . By the superposition principle, the solution to y ″ + 2 y ′ + 4 y = t + cos 2 t is obtained as follows: Here, k 1 = 1 and k 2 = 1 . Thus, y ( t ) = k 1 y 1 + k 2 y 2 is a solution to y ″ + 2 y ′ + 4 y = t + cos 2 t . That is, the solution will be, y ( t ) = k 1 y 1 + k 2 y 2 = y 1 + y 2 = 1 4 sin 2 t + t 4 − 1 8 Therefore, the solution to the differential equation y ″ + 2 y ′ + 4 y = t + cos 2 t is, y ( t ) = 1 4 sin 2 t + t 4 − 1 8 _ . (b) The solution to the differential equation y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t . The solution to the differential equation y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t is, y ( t ) = − 3 4 sin 2 t + t 2 − 1 4 _ . The given differential equation is y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t . Given that, y 1 ( t ) = 1 4 sin 2 t is a solution to y ″ + 2 y ′ + 4 y = cos 2 t and y 2 ( t ) = t 4 − 1 8 is a solution to y ″ + 2 y ′ + 4 y = t . By the superposition principle, the solution to y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t is obtained as follows: Here, k 1 = − 3 and k 2 = 2 . Thus, y ( t ) = k 1 y 1 + k 2 y 2 is a solution to y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t . That is, the solution will be, y ( t ) = k 1 y 1 + k 2 y 2 = − 3 y 1 + 2 y 2 = − 3 ( 1 4 sin 2 t ) + 2 ( t 4 − 1 8 ) = − 3 4 sin 2 t + t 2 − 1 4 Therefore, the solution to the differential equation y ″ + 2 y ′ + 4 y = 2 t − 3 cos 2 t is, y ( t ) = − 3 4 sin 2 t + t 2 − 1 4 _ . (c) The solution to the differential equation y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t . The solution to the differential equation y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t is, y ( t ) = − 3 sin 2 t + 11 t 4 − 11 8 _ . The given differential equation is y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t . Given that, y 1 ( t ) = 1 4 sin 2 t is a solution to y ″ + 2 y ′ + 4 y = cos 2 t and y 2 ( t ) = t 4 − 1 8 is a solution to y ″ + 2 y ′ + 4 y = t . By the superposition principle, the solution to y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t is obtained as follows: Here, k 1 = − 12 and k 2 = 11 . Thus, y ( t ) = k 1 y 1 + k 2 y 2 is a solution to y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t . That is, the solution will be, y ( t ) = k 1 y 1 + k 2 y 2 = − 12 y 1 + 11 y 2 = − 12 ( 1 4 sin 2 t ) + 11 ( t 4 − 1 8 ) = − 3 sin 2 t + 11 t 4 − 11 8 Therefore, the solution to the differential equation y ″ + 2 y ′ + 4 y = 11 t − 12 cos 2 t is, y ( t ) = − 3 sin 2 t + 11 t 4 − 11 8 _ .

Solution Summary: Theorem used: The function k_1y2 is a 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

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. 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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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