1. (1 pt) Write the given second order equation as its equivalent system of first order equations. u =

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1 1 (1 pt) Write the given second order equation as its equivalent system of first order equations u + u + 5u = 0 Use v to represent the velocity function, ie v = u (t) Use v and u for the two functions, rather than u(t) and v(t) (The latter confuses webwork Functions like sin(t) are ok) u = v = Now write the system using matrices: d u u = v v v -v - 5u (1 pt) Write the given second order equation as its equivalent system of first order equations u + 5u + u = 5sin(t), u(1) = 05, u (1) = 1 Use v to represent the velocity function, ie v = u (t) Use v and u for the two functions, rather than u(t) and v(t) (The latter confuses webwork Functions like sin(t) are ok) u = v = Now write the system using matrices: d u u = v v and the initial value for the vector valued function is: u(1) = v(1) v -5v - u + 5sin(t) sin(t) (1 pt) Write the given second order equation as its equivalent system of first order equations t u 5tu + (t )u = 5sin(t) Use v to represent the velocity function, ie v = u (t) Use v and u for the two functions, rather than u(t) and v(t) (The latter confuses webwork Functions like sin(t) are ok) u = v = Now write the system using matrices: d u u = v v + v + 5(1/t)v - (tˆ-)(1/tˆ)u + (1/tˆ)*(-5sin(t)) 0 1

2 1 - (tˆ-)(1/tˆ) --5(1/t) 0 (1/tˆ)*(-5sin(t)) 4 (1 pt) Consider two interconnected tanks as shown in the figure above Tank 1 initial contains 40 L (liters) of water and 170 g of salt, while tank initially contains 60 L of water and 80 g of salt Water containing 50 g/l of salt is poured into tank1 at a rate of 4 L/min while the miture flowing into tank contains a salt concentration of 0 g/l of salt and is flowing at the rate of L/min The two connecting tubes have a flow rate of 55 L/min from tank 1 to tank ; and of 15 L/min from tank back to tank 1 Tank is drained at the rate of 6 L/min You may assume that the solutions in each tank are thoroughly mied so that the concentration of the miture leaving any tank along any of the tubes has the same concentration of salt as the tank as a whole (This is not completely realistic, but as in real physics, we are going to work with the approimate, rather than eact description The real equations of physics are often too complicated to even write down precisely, much less solve) How does the water in each tank change over time? Let p(t) and q(t) be the amount of salt in g at time t in tanks 1 and respectively Write differential equations for p and q (As usual, use the symbols p and q rather than p(t) and q(t)) p = q = Give the initial values: p(0) = q(0) Show the equation that needs to be solved to find a constant solution to the differential equation: p = q A constant solution is obtained if p(t) = for all time t and q(t) = for all time t 50*4 - (55/40)p + (15/60)q 0* - ( 15/60 + 6/60 )q + ( 55/40 )p

3 Generated by the WeBWorK system c WeBWorK Team, Department of Mathematics, University of Rochester

4 1 (1 pt) Calculate the eigenvalues of this matri: If you select the integral curves utility from the main menu, will also be able to plot the integral curves of the associated diffential equations -105 A = smaller eigenvalue = associated eigenvector = (, ) larger eigenvalue = associated, eigenvector = (, ) A All of the solution curves would run away from 0 (Unstable node) B All of the solutions curves would converge towards 0 (Stable node) C The solution curves converge to different points D The solution curves would race towards zero and then veer away towards infinity (Saddle) -81 ( 5, 4 ) 108 ( -, ) D (1 pt) Calculate the eigenvalues of this matri: If you select the integral curves utility from the main menu, will also be able to plot the integral curves of the associated diffential equations 7 A = smaller eigenvalue = associated eigenvector = (, ) larger eigenvalue = associated, eigenvector = (, ) A The solution curves would race towards zero and then veer away towards infinity (Saddle) B All of the solution curves would run away from 0 (Unstable node) C The solution curves converge to different points D All of the solutions curves would converge towards 0 (Stable node) The eigenvalues and eigenvectors are {10,8} and the eigenvectors are ( 1,1 ) and (,1 ) The eigenvalues and eigenvectors are {10,8} and the eigenvectors are ( 1,1 ) and (,1 ) B (1 pt) Match the differential equations and their vector valued function solutions: It will be good practice to multiply at least one solution out fully, to make sure that you know how to do it, but you can get the other answers quickly by process of elimination and just multiply out one row element 1 y (t) = y(t)

5 y (t) = y(t) y (t) = y(t) A B C y(t) = y(t) = 0 e 5t e t y(t) = - 1 A B C e 45t 4 (1 pt) Solve the system d = with the initial value (0) = 5 (t) = -1*eˆ(-*t)*1 + *eˆ(*t)* -1*eˆ(-*t)*1 + *eˆ(*t)* 5 (1 pt) Solve the system d = with the initial value (0) = -0 (t) = -5*eˆ(8*t)*4 + -5*-1-5*eˆ(8*t)* + -5*1 6 (1 pt) Multiplying the differential equation d f + a f (t) = g(t), where a is a constant and g(t) is a smooth function, by e at, gives e at d f + eat a f (t) = e at g(t), d ( e at f (t) ) = e at g(t),

6 e at f (t) = e at g(t), f (t) = e at e at g(t) Use this to solve the initial value problem d 1 - =, 0 with (0) =, - ie find first (t) and then 1 (t) 1 (t) =, (t) = -*-/( - 1) * eˆ(*t) + ( - -*-/( - 1)) * eˆ(1*t) - * eˆ(*t) Generated by the WeBWorK system c WeBWorK Team, Department of Mathematics, University of Rochester

7 1 (1 pt) Consider the system of differential equations d = y, dy = y For this system, the smaller eigenvalue is and the larger eigenvalue is A All of the solution curves would run away from 0 (Unstable node) B The solution curves converge to different points C The solution curves would race towards zero and then veer away towards infinity (Saddle) D All of the solutions curves would converge towards 0 (Stable node) The solution to the above differential equation with initial values (0) =, y(0) = 6 is (t) =, y(t) = D *075*ep(-9*t)+4*075*ep(-09*t) *( )*ep(-9*t)+4*( )*ep(-09*t) (1 pt) Consider the interaction of two species of animals in a habitat We are told that the change of the populations (t) and y(t) can be modeled by the equations d = y, dy = 05 06y For this system, the smaller eigenvalue is and the larger eigenvalue is A All of the solution curves would run away from 0 (Unstable node) B The solution curves converge to different points C All of the solutions curves would converge towards 0 (Stable node) D The solution curves would race towards zero and then veer away towards infinity (Saddle) The solution to the above differential equation with initial values (0) = 6, y(0) = is (t) =, y(t) = Symbiosis D -0*15*ep(-11*t)+4*15*ep(09*t) -0*(-11-04)*ep(-11*t)+4*(09-04)*ep(09*t) 1

8 (1 pt) Consider the interaction of two species of animals in a habitat We are told that the change of the populations (t) and y(t) can be modeled by the equations d = 05 08y, dy = y For this system, the smaller eigenvalue is and the larger eigenvalue is A All of the solution curves would run away from 0 (Unstable node) B The solution curves would race towards zero and then veer away towards infinity (Saddle) C All of the solutions curves would converge towards 0 (Stable node) D The solution curves converge to different points The solution to the above differential equation with initial values (0) = 9, y(0) = 4 is (t) =, y(t) = 0 1 Competition A -1*-08*ep(0*t)+175*-08*ep(1*t) -1*(0-05)*ep(0*t)+175*(1-05)*ep(1*t) 4 (1 pt) Consider the interaction of two species of animals in a habitat We are told that the change of the populations (t) and y(t) can be modeled by the equations d = y, dy = 15 06y For this system, the smaller eigenvalue is and the larger eigenvalue is A The solution curves converge to different points B All of the solution curves would run away from 0 (Unstable node) C All of the solutions curves would converge towards 0 (Stable node) D The solution curves would race towards zero and then veer away towards infinity (Saddle) The solution to the above differential equation with initial values (0) = 6, y(0) = is (t) =, y(t) = Symbiosis D -0*1*ep(-11*t)+6*1*ep(19*t)

9 -0*(-11-14)*ep(-11*t)+6*(19-14)*ep(19*t) 5 (1 pt) Consider the interaction of two species of animals in a habitat We are told that the change of the populations (t) and y(t) can be modeled by the equations d = 16 05y, dy = 5 + 6y For this system, the smaller eigenvalue is and the larger eigenvalue is A All of the solution curves would run away from 0 (Unstable node) B All of the solutions curves would converge towards 0 (Stable node) C The solution curves converge to different points D The solution curves would race towards zero and then veer away towards infinity (Saddle) The solution to the above differential equation with initial values (0) = 9, y(0) = 5 is (t) =, y(t) = Competition A *-05*ep(11*t) - 1*-05*ep(41*t) *(11-16)*ep(11*t) - 1*(41-16)*ep(41*t) Generated by the WeBWorK system c WeBWorK Team, Department of Mathematics, University of Rochester

10 1 (1 pt) Solve the system d 6 = =? 1 Describe the trajectory *(- (-/1)*sin(*t) + cos(*t)) + (4/1)*(sin(*t)) *(-(-*-/1)*sin(*t)-1*sin(*t)) + 4*(cos(*t) + (-/1)*sin(*t)) Ellipse clockwise Predator-prey (1 pt) Solve the system d -5 - = =? 1 Describe the trajectory eˆ(-5*t) * (8*cos(*t) - 5*sin(*t)) eˆ(-5*t) * (8*sin(*t) + 5*cos(*t)) Spiral inward counterclockwise Predator-prey (1 pt) Solve the system d - = 6-6 =? 1 Describe the trajectory 6*(- (1/-1)*sin(*t) + cos(*t)) + (/-1)*(sin(*t)) 6*(-(1*1/-1)*sin(*t) + 1*sin(*t)) + *(cos(*t) + (1/-1)*sin(*t)) Ellipse counterclockwise Predator-prey 1

11 4 (1 pt) Solve the system d = =? 1 Describe the trajectory eˆ(-05*t) * (4*cos(*t) - 9*sin(*t)) eˆ(-05*t) * (4*sin(*t) + 9*cos(*t)) Spiral inward counterclockwise Predator-prey 5 (1 pt) Solve the system d - = 7 5 =? 1 Describe the trajectory eˆ(*t) * (7*cos(*t) - 5*sin(*t)) eˆ(*t) * (7*sin(*t) + 5*cos(*t)) Spiral outward counterclockwise Predator-prey Generated by the WeBWorK system c WeBWorK Team, Department of Mathematics, University of Rochester

12 1 (1 pt) Solve the system d 1-4 = 4-7 = A All of the solution curves would run away from 0 (Unstable node) B The solution curves would race towards zero and then veer away towards infinity (Saddle) C The solution curves converge to different points D All of the solutions curves would converge towards 0 (Stable node) d = eˆ(-*t) * (+4*t) eˆ(-*t) * (+4*t) D (1 pt) Solve the system = A All of the solution curves would run away from 0 (Unstable node) B The solution curves would race towards zero and then veer away towards infinity (Saddle) C All of the solutions curves would converge towards 0 (Stable node) D The solution curves converge to different points d = ep(t/) * (+15*t) -ep(t/) * (+15*t) A (1 pt) Solve the system = 1

13 A The solution curves converge to different points B The solution curves would race towards zero and then veer away towards infinity (Saddle) C All of the solution curves would run away from 0 (Unstable node) D All of the solutions curves would converge towards 0 (Stable node) ep(-t) * (-6*t) -ep(-t) * (1+6*t) D 4 (1 pt) Solve the system d 9 = -1-4 = +4*t 4-14*t Generated by the WeBWorK system c WeBWorK Team, Department of Mathematics, University of Rochester

14 1 (1 pt) Find the equilibrium solution for 1(t) = (t) = (0) = 1; (0) = 8 Equilibrium: e 1 =, e =? 1 Describe the trajectory 10 6 Ellipse counterclockwise Predator-prey (1 pt) Find the equilibrium solution for 1(t) = (t) = (0) = 11; (0) = 4 Equilibrium: e 1 =, e =? 1 Describe the trajectory 10 6 Spiral inward counterclockwise Predator-prey 1

15 (1 pt) Find the equilibrium solution for 1(t) = (t) = (0) = 0; (0) = 4 Equilibrium: e 1 =, e = A The solution curves converge to different points B All of the solutions curves would converge towards the equilibrium point (Stable node) C The solution curves would race towards the equilibrium point and then veer away towards infinity (Saddle) D All of the solution curves would run away from the equilibrium point (Unstable node) 11 7 C Generated by the WeBWorK system c WeBWorK Team, Department of Mathematics, University of Rochester

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