Explanation Given: The differential equation is y ″ ′ − 12 y ″ + 27 y ′ + 40 y = 0 . The initial value conditions are y ( 0 ) = − 3 y ′ ( 0 ) = − 6 y ″ ( 0 ) = − 12 . Approach: The auxiliary equation of a differential equation a n y ( n ) + a n − 1 y ( n − 1 ) + ⋅ ⋅ ⋅ + a 1 y ( 1 ) + a 0 is a n r n + a n − 1 r n − 1 + ⋅ ⋅ ⋅ + a 1 r 1 + a 0 ... ( 1 ) . Here, y ( n ) denotes the n th derivative of y . If x = a is a zero of the polynomial a n x n + a n − 1 x n − 1 + ⋅ ⋅ ⋅ + a 1 x 1 + a 0 , then ( x − a ) is a factor of the polynomial ... ( 2 ) . If the auxiliary equation has roots r 1 = a + b i , r 2 = a − b i then the solution of the differential equation is y = A e r 1 t + B e r 2 t ... ( 3 ) If the auxiliary equation has same root r 1 = r 2 = r , then the solution of the differential equation is y = A e r t + t B e r t Calculation: The differential equation is y ″ ′ − 12 y ″ + 27 y ′ + 40 y = 0 . The auxiliary equation of the differential equation is r 3 − 12 r 2 + 27 r + 40 = 0 Let p ( r ) = r 3 − 12 r 2 + 27 r + 40 Put p ( r ) = 0 r 3 − 12 r 2 + 27 r + 40 = 0 Calculate p ( − 1 ) Substitute − 1 for r in p ( r ) = r 3 − 12 r 2 + 27 r + 40 . p ( − 1 ) = ( − 1 ) 3 − 12 × ( − 1 ) 2 + 27 × ( − 1 ) + 40 = − 1 − 12 − 27 + 40 = 0 This implies ( r + 1 ) is a factor of p ( r ) . Divide p ( r ) by ( r + 1 ) . ( r + 1 ) r 2 − 13 r + 40 r 3 − 12 r 2 + 27 r + 40 − r 3 − r 2 _ − 13 r 2 + 27 r + 40 + 13 r 2 + 13 r _ 40 r + 40 − 40 r − 40 _ 0 _ Now, D ividend = Divisor × Quotient + Remainder . Substitute r 3 − 12 r 2 + 27 r + 40 for Dividend , ( r + 1 ) for Divisor , r 2 − 13 r + 40 for Quotient and 0 for remainder in the above equation. r 3 − 12 r 2 + 27 r + 40 = ( r + 1 ) ( r 2 − 13 r + 40 ) + 0 = ( r + 1 ) ( r 2 − 8 r − 5 r + 40 ) = ( r + 1 ) ( r ( r − 8 ) − 5 ( r − 8 ) ) = ( r + 1 ) ( r − 5 ) ( r − 8 ) The zeroes of p ( r ) are − 1 , 5 , 8 . y ( t ) = C 1 e − t + C 2 e 5 t + C 3 e 8 t ( 4 ) Now, y ( 0 ) = − 3 Substitute 0 for t , − 3 for y in the equation ( 4 ) C 1 + C 2 + C 3 = − 3

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

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

Publisher: Addison Wesley

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Chapter 4.RP, Problem 33RP

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(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).

Let us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 = 0. What level of type II error would you recommend here? Round your answer to four decimal places (e.g. 98.7654). Use a = 0.05. β = i What sample size would be required? Assume the sample sizes are to be equal.…

= Consider the hypothesis test Ho: μ₁ = μ₂ against H₁ μ₁ μ2. Suppose that sample sizes are n₁ = 15 and n₂ = 15, that x1 = 4.7 and X2 = 7.8 and that s² = 4 and s² = 6.26. Assume that o and that the data are drawn from normal distributions. Use απ 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) for a true difference in means of 3? (c) Assuming equal sample sizes, what sample size should be used to obtain ẞ = 0.05 if the true difference in means is - 2? Assume that α = 0.05. (a) The null hypothesis is 98.7654). rejected. The P-value is 0.0008 (b) The power is 0.94 . Round your answer to four decimal places (e.g. Round your answer to two decimal places (e.g. 98.76). (c) n₁ = n2 = 1 . Round your answer to the nearest integer.

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Fundamentals of Differential Equations [With CDROM] - 7th Edition

Ch. 4.1 - Verify that for b=0 and Fext(t)=0, equation (3)...Ch. 4.1 - If Fext(t)=0, equation (3) becomes my+by+ky=0. For...Ch. 4.1 - Show that if Fext(t)=0, m=1, k=9, and b=6, then...Ch. 4.1 - Prob. 4E Ch. 4.1 - Verify that the exponentially damped sinusoid...Ch. 4.1 - An external force F(t)=2cos2t is applied to a...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 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Undamped oscillators that are driven at resonance...

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The solution to the differential equation

Solution Summary: The author explains the solution of the differential equation: y(t)=-e-t

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The solution to the differential equation