j The DE dy dt y y 3 e t is a Bernoullis equation where we make the

J the de dy dt y y 3 e t is a bernoullis equation

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j. The DE dy dt + y = y 3 e t is a Bernoulli’s equation, where we make the substitution u = y 1 - 3 = y - 2 , so du dt = - 2 y - 3 dy dt . Multiplying the above equation by - 2 y - 3 , we obtain the linear DE in u ( t ) - 2 y - 3 dy dt - 2 y - 2 = - 2 e t or du dt - 2 u = - 2 e t . This has the integrating factor μ ( t ) = e - 2 t , so d dt e - 2 t u ( t ) = - 2 e - t or e - 2 t u ( t ) = 2 e - t + C. It follows that 1 y 2 ( t ) = u ( t ) = 2 e t + Ce 2 t , so 1 = 2 + C or C = - 1 . Thus, y ( t ) = 1 2 e t - e 2 t .
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2. a. The solution to the white lead problem is P ( t ) = 10 e - kt , where t = 0 represents 1970. From the data at 1975, we have 8 . 5 = 10 e - 5 k or e 5 k = 10 / 8 . 5 = 1 . 17647. Thus, k = 0 . 032504 yr - 1 . To find the half-life, we compute 5 = 10 e - kt , so t = ln(2) /k = 21 . 33 yr is the half-life of lead-210. b. The differential equation can be written P 0 = - k ( P - r/k ), so we make the substitution z ( t ) = P ( t ) - r/k . This leaves the initial value problem z 0 = - kz, z (0) = P (0) - r/k = 10 - r/k, which has the solution z ( t ) = ( P (0) - r/k ) e - kt = P ( t ) - r/k . Thus, the solution is P ( t ) = 10 - r k e - kt + r k = 2 . 3086 e - kt + 7 . 6914 , where k = 0 . 032504. In the limit, lim t →∞ P ( t ) = 7 . 6914 disintegrations per minute of 210 Pb . 3. a. The differential equation describing the temperature of the tea satisfies H 0 = - k ( H - 21) , H (0) = 85 and H (5) = 81 . Make the substitution z ( t ) = H ( t ) - 21, which gives the differential equation z 0 = - kz, z (0) = H (0) - 21 = 64 . The solution becomes z ( t ) = 64 e - kt = H ( t ) - 21 or H ( t ) = 64 e - kt + 21 . To find k , we solve H (5) = 81 = 64 e - 5 k + 21 or e 5 k = 64 / 60 = 1 . 0667. Thus, k = 0 . 012908 min - 1 . The water was at boiling point when 64 e - kt + 21 = 100 or e - kt = 79 / 64. It follows that t = - ln(79 / 64) /k = - 16 . 3 min. This means that the talk went 16.3 min over its scheduled ending. b. To obtain a temperature of at least 93 C, then we need to find the time that satisfies H ( t ) = 93 = 64 e - kt + 21, so e - kt = 72 / 64 = 1 . 125. Solving for t gives t = - ln(72 / 64) /k = - 9 . 125 min. It follows that you must arrive at the hot water within 16 . 3 - 9 . 1 = 7 . 2 min of the scheduled end of the talks. 4. a. Substituting the parameters into the differential equation gives c 0 = 1 10 6 (22000 - 2000 c ) = - 0 . 002( c - 11) . We make the substitution z ( t ) = c ( t ) - 11, which gives the initial value problem z 0 = - 0 . 002 z with z (0) = c (0) - 11 = - 11. The solution of this differential equation is z ( t ) = - 11 e - 0 . 002 t = c ( t ) - 11, so c ( t ) = 11 - 11 e - 0 . 002 t .
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b. Solve the equation c ( t ) = 11 - 11 e - 0 . 002 t = 5, so e 0 . 002 t = 11 / 6 or t = 500 ln(11 / 6) = 303 . 1 days. The limiting concentration lim t →∞ c ( t ) = 11 . The graph is below. 0 200 400 600 800 1000 0 2 4 6 8 10 12 t c ( t ) Problem 4 5. The differential equation is separable, so write Z T - 1 2 dT = k Z dt or 2 T 1 2 ( t ) = kt + C. It follows that T ( t ) = kt + C 2 2 . The initial condition T (0) = 1 implies C = 2, so T ( t ) = kt 2 + 1 2 . Since T (4) = 4 k 2 + 1 2 = 25, 2 k + 1 = 5 or k = 2. Thus, the solution for the spread of the disease in this orchard is T ( t ) = ( t + 1) 2 . When t = 10, T (10) = 121. 6. The differential equation with the information in the problem is given by: dH dt = - k ( H - 25) , H (0) = 35 , where t = 0 is 7 AM. We make the change of variables z ( t ) = H ( t ) - 25, so z (0) = 10. The problem now becomes dz dt = - kz, z (0) = 10 , which has the solution z ( t ) = 10 e - kt or H ( t ) = 25 + 10 e - kt .
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From the information at 9 AM, we see H (2) = 33 . 5 = 25 + 10 e - 2 k or e 2 k = 10 8 . 5 or k = ln 10 8 . 5 2 = 0 . 081259 .
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