The general solution for this problem is x t c 1 e t 4 cos2 t 2 sin2 t c 2 e t

# The general solution for this problem is x t c 1 e t

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The general solution for this problem is x ( t ) = c 1 e - t 4 cos(2 t ) 2 sin(2 t ) + c 2 e - t 4 sin(2 t ) - 2 cos(2 t ) + - 2 2 . We solve the IVP, so x (0) = c 1 4 0 + c 2 0 - 2 + - 2 2 = 5 2 or 4 c 1 - 2 c 2 = 7 0 It follows that c 1 = 7 4 and c 2 = 0, so x ( t ) = 7 e - t 4 4 cos(2 t ) 2 sin(2 t ) + - 2 2 . 4. From the information, we write the rate of change in amounts , A 1 ( t ) and A 2 ( t ). The rate of change in amount is concentra- tion times flow rate. dA i dt = amount enter - amount leave. For Tank 1, the amount entering is f 1 q 1 and f 3 c 2 , while the amount leaving is f 4 c 1 . Simi- lar expressions give the equation for Tank 2. The concentration equations follow by simply dividing the amount equations by the appropriate volumes. With the data the amount equations are dA 1 dt = 0 . 3 · 8 + 0 . 4 · c 2 - 0 . 7 · c 1 , dA 2 dt = 0 . 2 · 15 + 0 . 2 · c 1 - 0 . 4 · c 2 .

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Dividing by the volumes gives the concentration equations dc 1 dt = - 7 2000 c 1 + 1 500 c 2 + 3 250 , dc 2 dt = 1 500 c 1 - 1 250 c 2 + 3 100 . The equilibria are found first by solving: 7 2000 c 1 e - 1 500 c 2 e = 3 250 , - 1 500 c 1 e + 1 250 c 2 e = 3 100 . It follows that c 1 e = 10 . 8 g/l, while c 2 e = 12 . 9 g/l. This gives the equilibrium solution, which will be the asymptotic limit for the concentrations. We make a change of variables z 1 ( t ) = c 1 ( t ) - 10 . 8 and z 2 ( t ) = c 2 ( t ) - 12 . 9. The homogeneous equation, ˙ z = Az , is ˙ z 1 ˙ z 2 = - 0 . 0035 0 . 002 0 . 002 - 0 . 004 z 1 z 2 . The characteristic equation is det - 0 . 0035 - λ 0 . 002 0 . 002 - 0 . 004 - λ = λ 2 + 0 . 0075 λ + 0 . 00001 = 0 , which gives λ 1 = - 0 . 0057656 with its associated eigenvector v 1 = 1 - 1 . 13278 and λ 2 = - 0 . 0017344 with its associated eigenvector v 2 = 1 . 13278 1 . This results in the general solution x 1 ( t ) x 2 ( t ) = c 1 1 - 1 . 13278 e - 0 . 0057656 t + c 2 1 . 13278 1 e - 0 . 0017344 t + 10 . 8 12 . 9 , which is a stable node . The solution to the IVP satisfies 1 1 . 13278 - 1 . 13278 1 c 1 c 2 = 2 - 10 . 8 3 - 12 . 9 = - 8 . 8 - 9 . 9 , which gives c 1 = 1 . 05752 and c 2 = - 8 . 70206. It follows that the unique solution to the IVP is x 1 ( t ) x 2 ( t ) = 1 . 05752 - 1 . 19794 e - 0 . 0057656 t + - 9 . 85754 - 8 . 70206 e - 0 . 0017344 t + 10 . 8 12 . 9 . This solution clearly shows that the trajectory converges asymptotically to the equilibrium solution as expected. 5. The predator-prey model is given by: dH dt = 0 . 1 H - 0 . 0005 H 2 - 0 . 016 HP = f 1 ( H, P ) , dP dt = 0 . 005 HP - 0 . 2 P = f 2 ( H, P ) .
The equilibria satisfy: H e (0 . 1 - 0 . 0005 H e - 0 . 016 P e ) = 0 , P e (0 . 005 H e - 0 . 2) = 0 . One equilibrium is ( H e , P e ) = (0 , 0), the extinction equilibrium. When P e = 0, then there is another equilibrium at H e = 200, the carrying capacity. Finally, there is a third equilibrium with 0 . 005 H e - 0 . 2 = 0 and 0 . 1 - 0 . 0005 H e - 0 . 016 P e = 0. The first equation gives H e = 40, which gives P e = 5 in the second equation. Thus, there is a coexistence equilibrium, ( H e , P e ) = (40 , 5). As we did in class, we use Taylor’s theorem to linearize this system about the equilibria (finding the Jacobian matrix ). If h ( t ) = H ( t ) - H e and p ( t ) = P ( t ) - P e , then the linearized system can be written: ˙ h ˙ p = ∂f 1 ( H e ,P e ) ∂h ∂f 1 ( H e ,P e ) ∂p ∂f 2 ( H e ,P e ) ∂h ∂f 2 ( H e ,P e ) ∂p h p = 0 . 1 - 0 . 001 H e - 0 . 016 P e - 0 . 016 H e 0 . 005 P e 0 . 005 H e - 0 . 2 ! h p . The linear system about ( H e , P e ) = (0 , 0) is ˙ h ˙ p = 0 . 1 0 0 - 0 . 2 h p , which has eigenvalues λ 1 = - 0 . 2 with associated eigenvector v 1 = 0 1 and λ 2 = 0 . 1 with associated eigenvector v 2 = 1 0 . This is a saddle node at the extinction equilibrium. The general linear solution is given by h ( t ) p ( t ) = c 1 0 1 e - 0 . 2 t + c 2 1 0 e 0 . 1 t . Thus, if there are both predators and prey, then the solution moves away from extinction (unstable).

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