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1 MATH HOMEWORK #7 PART A SOLUTIONS Problem Consider the system x = 5 x. a Express the general solution of the given system of equations in terms of realvalued functions. b Draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as t. Solution. a Letting A be the above matrix, then the characteristic polynomial is deta ri = r 5 r = r + r + 5 = r = r + which has roots r = ±i. Then i 5 i A ri = i i 5 ξ + iξ + i = = = ξ ξ ξ. i 0 0 = ξ = + iξ Thus ξ = + i is an eigenvector with corresponding solution [ x + i = e it = cost sint = + i cost + i 0] cost + i sint. cost + sint sint Taking real and imaginary parts, we find the general solution cost sint x = c cost cost + sint + c. sint b The solutions are periodic, hence are remain bounded as t.

2 Problem Express the general solution of the given system of equations in terms of real-valued functions. x = 0 0 x 0 Solution. Denoting the above matrix by A, then the characteristic polynomial is r 0 deta ri = r 0 r = r 0 r r 0 = r 0 0 r r 0 r + r + Since = r + rr + r + = r 4r 7r 6. r + 4 r + 7 r + 6 = r + r + r + then the eigenvalues are r =, r = i, r = + i. A I = 0 0 = ξ ξ = ξ ξ ξ Thus we have the solution ξ = ξ x = e t

3 Letting r = + i, then A ri = r 0 r 0 r 0 r 0 r 0 0 r r r 0 r r r 0 r r + 0 r + r 0 0 r r r Since r = + i, then = r = + i r + = + i. ξ i ξ = i ξ ξ ξ ξ = ξ i i. Taking ξ =, then we find the eigenvector i i and hence the associated solution x = i i e + it = e t + i cos t + i sin t 0 cos t + sin t = cos t + sin cos t + sin t t cos e t + i cos t sin t t sin e t t Taking real and imaginary parts of this second solution yields the general solution x = c cos t + sin t e t + c cos t + sin t cos e t t cos t + sin t + c cos t sin t sin e t. t

4 Problem Consider the system x α = x. α a Determine the eigenvalues in terms of α. b Find the critical value or values of α where the qualitative nature of the phase portrait for the system changes. c Draw the phase portrait for a value of α slightly below, and for another value slightly above each critical value. Solution. a Letting A be the above matrix, then the characteristic polynomial is deta ri = α r α r = α r + = r αr + α + which has roots r = α ± 4α 4α + = α ± 4 = α ± i = α ± i. b Since α is real, then the eigenvalues are always complex, hence the origin is a spiral point. The only critical value is α = 0: when α < 0, the origin is a stable spiral point, when α = 0, the origin is a center, and when α > 0, the origin is an unstable spiral point. c For α = 0., we compute the following phase portrait and for α = 0., we compute the following phase portait 4

5 Problem Consider the system x α 0 = x. 4 a Determine the eigenvalues in terms of α. b Find the critical value or values of α where the qualitative nature of the phase portrait for the system changes. c Draw the phase portrait for a value of α slightly below, and for another value slightly above each critical value. Solution. a Letting A be the above matrix, then the characteristic polynomial is deta ri = α r 0 4 r = α r 4 r + 0 = r + 4 αr 4α + 0 which has roots r = α 4 ± 4 α + 6α 40 = α 4 ± α + 8α 4. b Solving α + 8α 4 = 0, we find α = 8 ± Then α + 8α 4 is = 8 ± 60 = α 4 ± α 8α α 40 = 8 ± 4 0 > 0 if α < 4 0 < 0 if 4 0 < α < > 0 if < α. = 4 ± 0. Thus for 4 0 < α < the matrix has complex eigenvalues, and since the real part is negative note that , this means the origin is a stable spiral. 5

6 In the other cases, the roots are both real. To determine whether they have the same sign or opposite signs, we first determine when α 4 = ± α + 8α 4: ± α + 8α 4 = α 4 = α + 8α 4 = α 4 = α 8α + 6 = 6α = 40 = α = 5/. Note that 5/ =.5 and When < α < 5/, then both roots are negative, so the origin is a stable node. When 5/ < α, then the roots are of opposite sign, hence the origin is a saddle point. Finally, when α < 4 0, then both roots are negative, so the origin is a stable node. In summary, α < 4 0 = origin is a stable node 4 0 < α < = origin is a stable spiral < α < 5/ = origin is a stable node 5/ < α = origin is a saddle point. c The plots below depict the phase portrait for α =, 5,.4 and 5. 6

7 Problem A mass m on a spring with constant k satisfies the differential equation see.7 mu + ku = 0, where ut is the displacement at time t of the mass from its equilibrium position. a Let x = u, x = u, and show that the resulting system is x 0 = x. k/m 0 b Find the eigenvalues of the matrix for the system in part a. c Sketch several trajectories of the system. Choose one of your trajectories, and sketch the corresponding graphs of x versus t and x versus t. Sketch both graphs on one set of axes. d What is the relation between the eigenvalues of the coefficient matrix and the natural frequency of the spring-mass system? Solution. a In standard form, the differential equation is u + k m u = 0. Letting x = u and x = u, this becomes x + k m x = 0 = x = k m x so we have the system x = x x = k m x or in matrix form x = 0 x. k/m 0 b Denoting the matrix above by A, then the characteristic polynomial is deta ri = r k/m r = r + k/m k which has roots r = ±i m. c Since the eigenvalues are pure imaginary, then the origin is a center. For plotting purposes, we let k = and m =. 7

8 d The general solution to the second-order equation is ut = c cos k k m t + c sin m t k which has natural frequency. We observe that this is exactly the magnitude m k of the eigenvalues ±i m, i.e., k m = k ±i m. Problem Consider the system x = 4 x. 8 4 a Draw a direction field and sketch a few trajectories. b Describe how the solutions behave as t. c Find the general solution of the system of equations. Solution. a 8

9 b All points on the line x = x are equilibria. Solutions that pass through points x off this line are all unbounded: solutions with trajectories above this line x have x, x and those below have x, x as t. c The characteristic polynomial is deta ri = 4 r 8 4 r = 4 r4 + r + 6 = r = r which has a double root r = 0. Then 4 A 0I = A = = ξ = ξ so we find one eigenvector with corresponding solution x =. ξ = ξ ξ To find a generalized eigenvector, we solve 4 4 / = ξ = ξ / ξ ξ = = 0 = ξ ξ ξ / +. / 0 We find the generalized eigenvector with corresponding solution / x 0 = t +. / Thus the general solution is [ x = c x + c x = c + c t + 9 ] 0. /

10 Problem Consider the system x = 5/ x. 5/ a Draw a direction field and sketch a few trajectories. b Describe how the solutions behave as t. c Find the general solution of the system of equations. Solution. a b All trajectories eventually converge to the origin as t. c The characteristic polynomial is r deta ri = 5 5 r = r + r = r + r = r + r + 4 = r + / which has a double root r = /. We row reduce 5/ 5/ 5/ 5/ 0 0 = ξ = ξ and find eigenvector with corresponding solution x = e t/. 0

11 We row reduce 5/ 5/ 5/ 5/ 5/ 5/ / = ξ ξ = /5 = ξ = ξ + /5 ξ ξ 0 = = = ξ ξ ξ + /5 + /5 0 and find generalized eigenvector with corresponding solution /5 x = te t/ + 0 e /5 t/. Thus the general solution is [ x = c x + c x = c e t/ + c te t/ + ] 0 e /5 t/. Problem Consider again the electric circuit in Problem 6 of 7.6. This circuit is described by the system of differential equations d I 0 = L I dt V C. V RC a Show that the eigenvalues are real and equal if L = 4R C. b Suppose that R = Ω, C = F and L = 4 H. Suppose also that I0 = A and V0 = V. Find It and Vt. Solution. a The characteristic polynomial is r deta ri = L C = r r + + RC r RC LC = r + RC r + LC which has roots r = RC ± 4 R C LC = RC ± L 4R C LR C The roots are equal if and only if the discriminant is 0, which occurs if and only if L 4R C = 0, i.e., L = 4R C. 0 /4 b With the above values, the matrix A is. Then A has characteristic polynomial. r + r + 4 = r + /

12 with double root r = /. We row reduce / /4 / / = ξ / / 0 0 = ξ = ξ = ξ ξ ξ = = = ξ ξ ξ and obtain the eigenvector with corresponding solution x = e t/. We row reduce / /4 / / / / = ξ + ξ = = ξ = 4 ξ ξ ξ 0 = = = ξ ξ 4 ξ and find the generalized eigenvector with corresponding solution 4 x = te t/ 0 + e 4 t/. Thus the general solution is [ ] I = c V x + c x = c e t/ + c te t/ 0 + e 4 t/. Imposing the initial conditions, we find I0 0 c = = c V0 + c = 4 c + 4c hence c = and 4c = 4 so c =. Thus the solution of the IVP is I = e V t/ + te t/ 0 + e 4 t/ = e t/ + te t/.

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