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13 Nov 2019

# question 23 please

Step 3 Multiply both sides of the differential equation by t and get [ty] = dt Step 4 Integrating both sides and solving for y, we obtain C1C- 2 Int C To satisfy the initial condition, we must have 21n(1)+C-C [In(1)-ì´. Hence, the solution of the initial-value problem is 2Int 1 Figure 3 Solution of the initial-value problem fy + ty 2, y(1)= 1, t > 0 t > 0 Among all the solution curves that are shown in Fig. 3, th problem is the curve that goes through the point (1, 1) e solution of the > Now Try Exercise In the following section, we present several interesting applications of fre linear differential equations. Check Your Understanding 10.3 1. Using an integrating factor, solve yty 1e 2. Find an integrating factor for the differential equi EXERCISES 10.3 in Exercises 1-6, find an integrating factor for each equation Taket0. In Exercises 21-28, solve the initial-value problem. 22. ty, + y = In t, y(e) = 0, t > 0 Int-1 23. y+ 24, y, 2(10-y), y(0) = 1 10-9e-2t 10+10f In Exercises 7-20, solve the given equation using an integrating factor. Taket>0. 26. ty'-y=-1, y(1) = 1, t>01 27, y, + 2y cos(2t)-2 cos(2t), y(5) = 0 y 28, ty' + y-sin t, y()-0, t > 0-1 cost Technology Exercises 29. Consider the initial-value -1- - 9. 10Â· y' = 2(20-y) 20 + C'e-2t 12. e +1 11ã y' = .5(35-y) 13. y, + 10+1-0y=åº14.3-h-e2t tr" + Cee 15. (1 + tly, + y--I 17. 6y'+ty t 19,y, + y = 2-et y'=-th+10, (a) Is the solution increasing or decreasing whie y(0)=50. 16, y, = e-"(y + 1) _1+ 18. ety' + y = 1 1 + Ce", C'e-e-t 20 , + y = 1 [Hint: Compute y'(0),1 i + Ce-i(t+1)3/2 (b) Find the solution and plot it for OStS t k indicates answers that are in the back of the book. 19. y = 2-ne, + Ce-r 29. (a) y'(0)--40, y(t) is decreasing at t = 0

question 23 please

Step 3 Multiply both sides of the differential equation by t and get [ty] = dt Step 4 Integrating both sides and solving for y, we obtain C1C- 2 Int C To satisfy the initial condition, we must have 21n(1)+C-C [In(1)-ì´. Hence, the solution of the initial-value problem is 2Int 1 Figure 3 Solution of the initial-value problem fy + ty 2, y(1)= 1, t > 0 t > 0 Among all the solution curves that are shown in Fig. 3, th problem is the curve that goes through the point (1, 1) e solution of the > Now Try Exercise In the following section, we present several interesting applications of fre linear differential equations. Check Your Understanding 10.3 1. Using an integrating factor, solve yty 1e 2. Find an integrating factor for the differential equi EXERCISES 10.3 in Exercises 1-6, find an integrating factor for each equation Taket0. In Exercises 21-28, solve the initial-value problem. 22. ty, + y = In t, y(e) = 0, t > 0 Int-1 23. y+ 24, y, 2(10-y), y(0) = 1 10-9e-2t 10+10f In Exercises 7-20, solve the given equation using an integrating factor. Taket>0. 26. ty'-y=-1, y(1) = 1, t>01 27, y, + 2y cos(2t)-2 cos(2t), y(5) = 0 y 28, ty' + y-sin t, y()-0, t > 0-1 cost Technology Exercises 29. Consider the initial-value -1- - 9. 10Â· y' = 2(20-y) 20 + C'e-2t 12. e +1 11ã y' = .5(35-y) 13. y, + 10+1-0y=åº14.3-h-e2t tr" + Cee 15. (1 + tly, + y--I 17. 6y'+ty t 19,y, + y = 2-et y'=-th+10, (a) Is the solution increasing or decreasing whie y(0)=50. 16, y, = e-"(y + 1) _1+ 18. ety' + y = 1 1 + Ce", C'e-e-t 20 , + y = 1 [Hint: Compute y'(0),1 i + Ce-i(t+1)3/2 (b) Find the solution and plot it for OStS t k indicates answers that are in the back of the book. 19. y = 2-ne, + Ce-r 29. (a) y'(0)--40, y(t) is decreasing at t = 0

Irving HeathcoteLv2

27 Mar 2019