# sol10 - Chapter 10 Exercises 10.1 1 y = f(t = 2 3 t2 1 e y...

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315 Chapter 10 Exercises 10.1 1. y = f ( t ) = 3 2 e t 2 1 2 ; y = 3 te t 2 y 2 ty = 3 te t 2 2 t 3 2 e t 2 1 2 = t 2. y = f ( t ) = t 2 1 2 y = 2 t ( y ) 2 4 y = (2 t ) 2 4 t 2 1 2 = 2 3. y = f ( t ) = ( e t + 1) 1 y = e t ( e t + 1) 2 y + y 2 = e t ( e t + 1) 2 + 1 ( e t + 1) 2 = 1 e t + 1 = y ; y (0) = 1 1 + 1 = 1 2 4. y = f ( t ) = 5 e 2 t y = 10 e 2 t y = 20 e 2 t y 3 y + 2 y = 20 e 2 t 3(10 e 2 t ) + 2( 5 e 2 t ) = 0 y = 5 e 0 = 5, y = 10 e 0 = 10 5. The highest derivative is the second derivative so the differential equation is of second order. y , y ( t ) = t y ( t ) = 1; y ( t ) = 0 (1 t 2 ) y 2 t y + 2 y = t 2 )(0) 2 t (1) + 2( t ) = 0 6. The highest derivative is the second derivative so the differential equation is of second order. y , y ( t ) = 1 2 (3 t 2 1); y ( t ) = 3 t y ( t ) = 3 t 2 ) y 2 t y + 6 y = t 2 )(3) 2 t t ) + 6 1 2 t 2 = 0 7. Yes. Given y = f ( t ) = 3, y = 0 = 6 2(3). 8. Yes. Given y = f ( t ) = –4, for all t . y = 0 = t 2 ( 4 + 4) 9. y = 5 is a constant solution. y = t 2 y 5 t 2 = t 2 ( y 5); dy dt = 0, and t 2 ( y 5) = 0 at y = 5. 10. y = 4 y ( y 7) ; y = 0 or y = 7 are constant solutions. dy dt = 0, and 4 y ( y 7) = y = 0 and y = 7. 11. f (0) = y (0) = 4. Since , we must have in particular y = 2 y 3 y = 2 y 3 = 2(4) 3 = 5 . 12. f (0) = y (0) = 0. Since y = e t + y , we must have in particular . y = e 0 + y (0) = 1 + 0 = 1 13. y = .2(160 y ) . When y = 60, y = .2(160 60) = .2(100) = 20 feet per second per second. 14. y = .0004 y (1000 y ) . When y = 1 2 (1000) = 500, y = .0004(500)(1000 500) = 100 . So the fish population is growing at a rate of 100 fish per month at the time there are 500 fish in the lake. 15. y = .05 y 10, 000 . a. When t = 1, y = 150,000. y = .05(150,000) – 10,000 = –2500 So the balance is decreasing at a rate of \$2500 per year after 1 year. b. y = .05( y 200, 000) c. The rate of change of the savings account balance is proportional to the difference between the balance at the end of t years and \$200,000. 16. y = .04 y + 2000 a. When t = 1, y = 10,000. y = .04(10,000) + 2000 = 2400 The balance is increasing at a rate of \$2400 per year after 1 year. b. y = .04( y + 50, 000) c. The rate of change of the savings account balance is proportional to the sum of the balance at the end of t years and \$50,000.

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Chapter 10: Differential Equations ISM: Calculus & Its Applications, 11e 316 17. The number of people who have heard the news broadcast after t hours is increasing at a rate that is proportional to the difference between that number and 200,000. At the beginning of the broadcast there are 10 people tuned in. 18. The size of the paramecium population after t days is increasing at a rate that is proportional to the difference between that number and 500. There are 20 paramecia at the beginning of the count. 19. , k > 0 y = k ( C y ) 20. = k (20 – y ), k > 0 y 21. y = k ( P b y ), y (0) = P 0 22. a. dy dt = t y ; dy dt t = 0 y =− 1 = 0 ( 1) = 1.
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sol10 - Chapter 10 Exercises 10.1 1 y = f(t = 2 3 t2 1 e y...

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