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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

(a) To find: The mass y ( t ) of radioactive isotope R A 2 for t ≥ 0 and k = 20 / sec if initially y ( 0 ) = 40 kg ; The term in the solution that eventually dominates. The mass y ( t ) of R A 2 for t ≥ 0 and k = 20 / sec is y ( t ) = 5 e − 10 t + 35 e − 20 t _ . The term 5 e − 10 t _ , is the term in the solution y ( t ) = 5 e − 10 t + 35 e − 20 t that eventually dominates. From Example 2, “A rock contains two radioactive isotopes, namely, R A 1 and R A 2 and the rate of change in R A 2 is d y d t = rate of creation − rate of decay .” Assume that the rate at which R A 1 decays into R A 2 is 50 e − 10 t kg / sec . That is, the rate of creation is 50 e − 10 t . The decay constant is k = 20 / sec . That is, the rate of decay is k = 20 y . Thus, the rate of change in R A 2 becomes d y d t = 50 e − 10 t − 20 y d y d t + 20 y = 50 e − 10 t , y ( 0 ) = 40 kg Solve the differential equation by the method of linear equation as follows. Step 1: Rewrite the differential equation of the standard form d y d t + P ( t ) y = Q ( t ) as follows. d y d t + 20 y = 50 e − 10 t ... ( 1 ) Here, P ( t ) = 20 and Q ( t ) = 50 e − 10 t . Step 2: Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 20 ) d t ] = e 20 t Step 3: Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) y ) = μ ( t ) Q ( t ) . e 20 t ( d y d t + 20 y ) = 50 e − 10 t e 20 t e 20 t d y d t + 20 e 20 t y = 50 e 10 t d d t ( e 20 t y ) = 50 e 10 t Step 4: To obtain the general solution y ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e 20 t y ) = 50 e 10 t and solve for y ( t ) by dividing μ ( t ) . e 20 t y = 50 ∫ e 10 t d t e 20 t y = 50 10 e 10 t + C e 20 t y = 5 e 10 t + C y = 5 e 10 t e − 20 t + C e − 20 t y = 5 e − 10 t + C e − 20 t Step 5: Apply the initial condition t = 0 and y ( 0 ) = 40 in the above result to find the constant C . y = 5 e − 10 t + C e − 20 t 40 = 5 e 0 + C e 0 40 = 5 + C C = 35 Substitute C = 35 in the solution y = 5 e − 10 t + C e − 20 t and simplify further. y ( t ) = 5 e − 10 t + 35 e − 20 t Thus, the mass of R A 2 for t ≥ 0 and k = 20 / sec is y ( t ) = 5 e − 10 t + 35 e − 20 t _ . From the above equation, take y 1 ( t ) = 5 e − 10 t and y 2 ( t ) = 35 e − 20 t . Use online graphing calculator to sketch the curves y 1 ( t ) = 5 e − 10 t and y 2 ( t ) = 35 e − 20 t as shown in Figure 1. From Figure 1, it is observed that initially the curve y 1 ( t ) = 5 e − 10 t has smaller magnitude than y 2 ( t ) = 35 e − 20 t . But it dominates y 2 ( t ) = 35 e − 20 t as t increases. Therefore, the term 5 e − 10 t _ , is the term in the solution y ( t ) = 5 e − 10 t + 35 e − 20 t that eventually dominates. (b) To find: The mass y ( t ) of radioactive isotope R A 2 for t ≥ 0 and k = 10 / sec if initially y ( 0 ) = 40 kg . The mass y ( t ) of R A 2 for t ≥ 0 and k = 10 / sec is y ( t ) = 10 e − 10 t ( 5 t + 4 ) _ . From Example 2, “A rock contains two radioactive isotopes, namely, R A 1 and R A 2 and the rate of change in R A 2 is d y d t = rate of creation − rate of decay .” Assume that the rate at which R A 1 decays into R A 2 is 50 e − 10 t kg / sec . That is, the rate of creation is 50 e − 10 t . The decay constant is k = 10 / sec . That is, the rate of decay is k = 20 y . Thus, the rate of change in R A 2 becomes, d y d t = 50 e − 10 t − 10 y d y d t + 10 y = 50 e − 10 t , y ( 0 ) = 40 kg Solve the differential equation by the method of linear equation as follows. Step 1: Rewrite the differential equation of the standard form d y d t + P ( t ) y = Q ( t ) as follows. d y d t + 10 y = 50 e − 10 t ... ( 1 ) Here, P ( t ) = 10 and Q ( t ) = 50 e − 10 t . Step 2: Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 10 ) d t ] = e 10 t Step 3: Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) y ) = μ ( t ) Q ( t ) . e 10 t ( d y d t + 10 y ) = 50 e − 10 t e 10 t e 10 t d y d t + 10 e 10 t y = 50 e 0 d d t ( e 10 t y ) = 50 Step 4: To obtain the general solution y ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e 10 t y ) = 50 and solve for y ( t ) by dividing μ ( t ) . e 10 t y = ∫ 50 d t e 10 t y = 50 t + C y ( t ) = 50 t e − 10 t + C e − 10 t Step 5: Apply the initial condition t = 0 and y ( 0 ) = 40 in the above result to find the constant C . 40 = 50 e − 10 ( 0 ) ( 0 ) + C e 0 40 = 0 + C C = 40 Substitute C = 40 in the solution y ( t ) = 50 e − 10 t t + C e − 10 t and simplify further. y ( t ) = 50 e − 10 t t + 40 e − 10 t y ( t ) = 10 e − 10 t ( 5 t + 4 ) Thus, the mass y ( t ) of R A 2 for t ≥ 0 and k = 10 / sec is y ( t ) = 10 e − 10 t ( 5 t + 4 ) _ .

(a) To find: The mass y ( t ) of radioactive isotope R A 2 for t ≥ 0 and k = 20 / sec if initially y ( 0 ) = 40 kg ; The term in the solution that eventually dominates. The mass y ( t ) of R A 2 for t ≥ 0 and k = 20 / sec is y ( t ) = 5 e − 10 t + 35 e − 20 t _ . The term 5 e − 10 t _ , is the term in the solution y ( t ) = 5 e − 10 t + 35 e − 20 t that eventually dominates. From Example 2, “A rock contains two radioactive isotopes, namely, R A 1 and R A 2 and the rate of change in R A 2 is d y d t = rate of creation − rate of decay .” Assume that the rate at which R A 1 decays into R A 2 is 50 e − 10 t kg / sec . That is, the rate of creation is 50 e − 10 t . The decay constant is k = 20 / sec . That is, the rate of decay is k = 20 y . Thus, the rate of change in R A 2 becomes d y d t = 50 e − 10 t − 20 y d y d t + 20 y = 50 e − 10 t , y ( 0 ) = 40 kg Solve the differential equation by the method of linear equation as follows. Step 1: Rewrite the differential equation of the standard form d y d t + P ( t ) y = Q ( t ) as follows. d y d t + 20 y = 50 e − 10 t ... ( 1 ) Here, P ( t ) = 20 and Q ( t ) = 50 e − 10 t . Step 2: Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 20 ) d t ] = e 20 t Step 3: Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) y ) = μ ( t ) Q ( t ) . e 20 t ( d y d t + 20 y ) = 50 e − 10 t e 20 t e 20 t d y d t + 20 e 20 t y = 50 e 10 t d d t ( e 20 t y ) = 50 e 10 t Step 4: To obtain the general solution y ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e 20 t y ) = 50 e 10 t and solve for y ( t ) by dividing μ ( t ) . e 20 t y = 50 ∫ e 10 t d t e 20 t y = 50 10 e 10 t + C e 20 t y = 5 e 10 t + C y = 5 e 10 t e − 20 t + C e − 20 t y = 5 e − 10 t + C e − 20 t Step 5: Apply the initial condition t = 0 and y ( 0 ) = 40 in the above result to find the constant C . y = 5 e − 10 t + C e − 20 t 40 = 5 e 0 + C e 0 40 = 5 + C C = 35 Substitute C = 35 in the solution y = 5 e − 10 t + C e − 20 t and simplify further. y ( t ) = 5 e − 10 t + 35 e − 20 t Thus, the mass of R A 2 for t ≥ 0 and k = 20 / sec is y ( t ) = 5 e − 10 t + 35 e − 20 t _ . From the above equation, take y 1 ( t ) = 5 e − 10 t and y 2 ( t ) = 35 e − 20 t . Use online graphing calculator to sketch the curves y 1 ( t ) = 5 e − 10 t and y 2 ( t ) = 35 e − 20 t as shown in Figure 1. From Figure 1, it is observed that initially the curve y 1 ( t ) = 5 e − 10 t has smaller magnitude than y 2 ( t ) = 35 e − 20 t . But it dominates y 2 ( t ) = 35 e − 20 t as t increases. Therefore, the term 5 e − 10 t _ , is the term in the solution y ( t ) = 5 e − 10 t + 35 e − 20 t that eventually dominates. (b) To find: The mass y ( t ) of radioactive isotope R A 2 for t ≥ 0 and k = 10 / sec if initially y ( 0 ) = 40 kg . The mass y ( t ) of R A 2 for t ≥ 0 and k = 10 / sec is y ( t ) = 10 e − 10 t ( 5 t + 4 ) _ . From Example 2, “A rock contains two radioactive isotopes, namely, R A 1 and R A 2 and the rate of change in R A 2 is d y d t = rate of creation − rate of decay .” Assume that the rate at which R A 1 decays into R A 2 is 50 e − 10 t kg / sec . That is, the rate of creation is 50 e − 10 t . The decay constant is k = 10 / sec . That is, the rate of decay is k = 20 y . Thus, the rate of change in R A 2 becomes, d y d t = 50 e − 10 t − 10 y d y d t + 10 y = 50 e − 10 t , y ( 0 ) = 40 kg Solve the differential equation by the method of linear equation as follows. Step 1: Rewrite the differential equation of the standard form d y d t + P ( t ) y = Q ( t ) as follows. d y d t + 10 y = 50 e − 10 t ... ( 1 ) Here, P ( t ) = 10 and Q ( t ) = 50 e − 10 t . Step 2: Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 10 ) d t ] = e 10 t Step 3: Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) y ) = μ ( t ) Q ( t ) . e 10 t ( d y d t + 10 y ) = 50 e − 10 t e 10 t e 10 t d y d t + 10 e 10 t y = 50 e 0 d d t ( e 10 t y ) = 50 Step 4: To obtain the general solution y ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e 10 t y ) = 50 and solve for y ( t ) by dividing μ ( t ) . e 10 t y = ∫ 50 d t e 10 t y = 50 t + C y ( t ) = 50 t e − 10 t + C e − 10 t Step 5: Apply the initial condition t = 0 and y ( 0 ) = 40 in the above result to find the constant C . 40 = 50 e − 10 ( 0 ) ( 0 ) + C e 0 40 = 0 + C C = 40 Substitute C = 40 in the solution y ( t ) = 50 e − 10 t t + C e − 10 t and simplify further. y ( t ) = 50 e − 10 t t + 40 e − 10 t y ( t ) = 10 e − 10 t ( 5 t + 4 ) Thus, the mass y ( t ) of R A 2 for t ≥ 0 and k = 10 / sec is y ( t ) = 10 e − 10 t ( 5 t + 4 ) _ .

Solution Summary: The author explains that the mass of radioactive isotope RA_2fortge 0 and k=20/mathrmsec

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 716, solve the equation. 7. xdydx=1y3 Ch. 2.2 - In Problems 716, solve the equation. 8. dxdt=3xt2 Ch. 2.2 - In Problems 716, solve the equation. 9....Ch. 2.2 - In Problems 716, solve the equation. 10....

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In Chapter 1 we indicated...Ch. 2.2 - As stated in this section, the separation of...Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Mixing. Suppose a brine containing 0.3 kilogram...Ch. 2.2 - Newtons Law of Cooling. According to Newtons law...Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Compound Interest. If P(t) is the amount of...Ch. 2.2 - Free Fall. In Section 2.1, we discussed a model...Ch. 2.2 - Grand Prix Race. Driver A had been leading...Ch. 2.2 - Prob. 40E Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - Radioactive Decay. In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. 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A geometric problem...Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.5 - Prob. 1E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Verify that when the linear differential equation...Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 8E Ch. 2.6 - Use the method discussed under Homogeneous...Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Use the method discussed under Bernoulli Equations...Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Use the method discussed under Equations with...Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - In Problems 3340, solve the equation given in: 36....Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Show that equation (13) reduces to an equation of...Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2 - In Problems 130, solve the equation. 1....Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - In Problems 130, solve the equation. 6. 2xy3 dx ...Ch. 2 - In Problems 130, solve the equation. 7. t3y2 dt +...Ch. 2 - Prob. 8RP Ch. 2 - In Problems 130, solve the equation. 9. (x2 + y2)...Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP Ch. 2 - Prob. 23RP Ch. 2 - In Problems 130, solve the equation. 24. (y/x +...Ch. 2 - Prob. 25RP Ch. 2 - Prob. 26RP Ch. 2 - Prob. 27RP Ch. 2 - Prob. 28RP Ch. 2 - Prob. 29RP Ch. 2 - Prob. 30RP Ch. 2 - Prob. 31RP Ch. 2 - Prob. 32RP Ch. 2 - Prob. 33RP Ch. 2 - Prob. 34RP Ch. 2 - Prob. 35RP Ch. 2 - Prob. 36RP Ch. 2 - Prob. 37RP Ch. 2 - Prob. 38RP Ch. 2 - Prob. 39RP Ch. 2 - Prob. 40RP Ch. 2 - Prob. 41RP

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