Find the general solution of the system y = Ay, where

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1 Math Homework # March, Find the general solution of the system y = Ay, where 5 Answer: The matrix A has characteristic polynomial p(λ = λ + 7λ + = λ + 3(λ +. Hence the eigenvalues are λ = 3and λ =. For λ = 3wehave A λ I = A + 3I =. We see that the eigenspace has dimension and is spanned by v = (, T. Thus we have the solution y (t = e λt v = e 3t (, T. For λ = wehave A λ I = A + I =. We see that the eigenspace has dimension and is spanned by v = (, T. Thus we have the solution y (t = e λt v = e t (, T. Thus, the general solution y(t = C e t + C e 3t Find the solution of the initial value problem for system y = Ay with 5 and y( =. Answer: The system in Exercise 9..3 had general solution y(t = C e t + C e 3t. Thus, if y( = (, T, then ( = C + C = ( The augmented matrix reduces. Therefore, C = and C =, giving particular solution y(t = e t e 3t. ( C C.

2 9..6. Find a fundamental set of real solutions of the system y = Ay, where Answer: The characteristic polynomial is p(λ = λ λ + 8, which has complex roots λ = ± i. For the eigenvalue λ = + i, we have the eigenvector w = ( i, T. The corresponding exponential solution is z(t = e ( +it ( i [ ] = e t [cos t + i sin t] + i ] = e [cos t t sin t + ie t [cos t ( + sin t ]. The real and imaginary parts of z, ] y (t = e [cos t t sin t ] y (t = e [cos t t + sin t are a fundamental set of solutions Find the solution of the initial value problem for system y = Ay with and y( = Answer: A fundamental set of solutions was found in Exercise 9..6, so the solution has the form y(t = C y (t + C y (t, where ] y (t = e [cos t t sin t ] y (t = e [cos t t + sin t At t = wehave ( = y( = C ( = ( C ( + C This system can be readily solved, getting C = and C =. Hence the solution is y(t = y (t y (t. C.

3 Find the general solution of the system y = Ay, where 3 Answer: The matrix A has one eigenvalue, λ =. However, the nullspace of A I = is generated by the single eigenvector, v = (, T, with corresponding solution y (t = e t. To find another solution, we need to find a vector v which satisfies (A Iv = v. Choose w = (,, which is independent of v, and note that (A Iw = = = v. Thus, choose v = w = (, T. Our second solution is y (t = e t (v + tv [ ] = e t + t. Thus, the general solution can be written [( y(t = C e t + C e t ( = e [(C t + C t + t + C ( ] ] Find the solution of the initial value problem for system y = Ay where 3 and y( =. Answer: From Exercise 9..39, If y( = (, T, then y(t = e t [(C + C t ] + C. = C + C. The augmented matrix reduces,, 3

4 and C = and C = 3. Thus, the particular solution is ] y(t = e [( t + 3t + 3 = e t + 3t. + 3t Classify the equilibrium point of each of the system y = Ay where based on the position of (T, D in the trace-determinant plane. Verify your result by creating a phase portrait with your numerical solver. Answer: For ( we have D = det( 6 <. Hence the origin is a saddle point. The following phase plane diagram is typical of a saddle. 5 y x Classify the equilibrium point of each of the system y = Ay where based on the position of (T, D in the trace-determinant plane. Verify your result by creating a phase portrait with your numerical solver. Answer: If 3, 5 8 then the trace is T = and the determinant is D = 3 >. Further, T D = ( (3 = 36 <, so the equilibrium point at the origin is a spiral sink.

5 Further, 3 =, so the motion is counterclockwise. A hand sketch follows. The phase portrait, drawn in a numerical solver, follows. 5 y x Classify the equilibrium point of each of the system y = Ay where 3 based on the position of (T, D in the trace-determinant plane. Verify your result by creating a phase portrait with your numerical solver.

6 Answer: For 3 we have T = tr(, and D = det( 5. The discriminant is T D = 6. Hence the equilibrium point is a spiral source. The phase portrait, drawn in a numerical solver, follows. 5 y x Degenerate nodes. Degenerate nodal sinks are equilibrium points characterized by the fact that all solutions tend to the equilibrium point as t, all tangent to the same line. The solutions of degenerate nodal sources tend to the equilibrium point as t, all tangent to the same line. Exercises discuss degenerate nodes The system y = y 3 has a repeated eigenvalue, λ =, but only one eigenvector, v = (, T. (a Explain why this system lies on the boundary separating nodal sinks from spiral sinks in the trace-determinant plane. Answer: For 3 (b we have T = tr( and D = det(. Since the discriminant T D = the point (T, D lies on the parabola that divides nodal sinks from spiral sinks in the trace-determinant plane. The general solution (see Example.5 in section 9. can be written y(t = e ((C t + C t + C. / Predict the behavior of the solution in the phase plane as t and as t.

7 (c Answer: The general solution can be written y(t = e ((C t + C t + C. / We will write y(t = te t ( C t + C + C t /. Since the factor te t ast we see that y(t as t. Ifwenow look at the other factor, we see that it approaches C (, T as t. Hence we see that y(t and becomes tangent to the half-line through C (, T as t. Next as t, the factor te t, and the other factor again approaches C (, T. Hence we see that as t, y(t, and in the process becomes parallel to the half-line generated by C (, T in the process. Use your numerical solver to sketch the half-line solutions. Then sketch exactly one solution in each region separated by the half-line solutions. Explain how the behavior you see agrees with your findings in part (b. Answer: The following figure shows the half-line solutions and one other in each sector. The solutions clearly exhibit the behavior predicted in part (a. 5 y x Let A be a real matrix with one eigenvalue λ and suppose that the eigenspace of λ has dimension. The matrix in Exercise is an example. According to Theorem. in Section 9., a fundamental set of solutions for the system y = Ay is given by y (t = e λt v and y (t = e λt [v + tv ], where v is a nonzero eigenvector, and v satisfies (A λiv = v. The general solution can be written as y(t = e λt [(C + C tv + C v ],

8 Assume that the eigenvalue λ is negative. (a Describe the exponential solutions. Answer: There is one exponential solution, e λt v. Because λ<, this solution decays to the equilibrium point at the origin along the half line generated by Cv. (b Describe the behavior of the general solution as t. Answer: The general solution is (c (d y(t = e λt [(C + C tv + C v ]. Because λ<, the terms e λt (C + C tv and C e λt v both decay to zero. However, the first term is larger for large values of t. Thus, as t, y(t e λt (C +C tv te λt C v, which implies that the solution approaches zero tangent to the half-line generated by C v. Describe the behavior of the general solution as t. Answer: Because λ<, the terms e λt (C + C tv and C e λt v get infinitely large in magnitude as t. However, the first term is larger in magnitude for negative values of t that are large in magnitude. Thus, as t, y(t e λt (C +C tv te λt C v,. Since t<ast, this implies that the solution eventually parallels the half-line generated by C v. Is the equilibrium point at the origin a degenerate nodal sink or source? Answer: Degenerate nodal sink Redo Exercise under the assumption that the eigenvalue λ is positive. Answer: In general everything moves in the opposite direction in comparison to the situation in Exercise As t the exponential solution tends to along the half-line generated by C v. As t the general solution tends to and becomes parallel to the half-line generated by C v. As t the general solution tends to tangent to the half-line generated by C v. The origin is a degenerate nodal source Where do linear degenerate sources and sinks fit on the trace-determinant plane? Answer: Because the linear degenerate nodal sources and sinks have only one eigenvalue, and because the eigenvalues are given by λ,λ = T ± T D, we must have T D =. Therefore, the degenerate nodal sources and sinks lie on the parabola T D = in the trace-determinant plane. This positioning on the boundary between the nodal sink and sources and the spiral sinks and sources is significant. The solutions attempt to spiral, but they cannot. The presence of the half-line solutions prevents them from spiralling (solutions cannot cross.

9 9..8. Find the general solution of the system y = ( y Answer: For we have A λi = ( ( 3 λ 3 λ 3 λ We can compute the characteristic polynomial p(λ = det(a λi by expanding along the second column to get p(λ = ( λ det 3 λ λ = (λ (λ + 3λ + = (λ (λ + (λ +. Hence the eigenvalues are λ =, λ =, and λ 3 =. For λ = wehave A λ I = A + I = ( 3 3 The nullspace is generated by the vector v = (,, T. For λ = wehave A λ I = A + I = ( The nullspace is generated by the vector v = (,, T. For λ 3 = wehave A λ 3 I = A I = ( The nullspace is generated by the vector v 3 = (,, T.

10 Thus we have three exponential solutions: y (t = e λt v = e t y (t = e λt v = e t y 3 (t = e λ 3t v 3 = e t ( Since the three eigenvalues are distinct, these solutions are linearly independent, and form a fundamental system of solutions. The general solution is y(t = C y (t + C y (t + C 3 y 3 (t Find the solution of the initial value problem for the system in Exercise 9..8 with initial value y( = (,, T. Answer: The solution has the form y(t = C y (t + C y (t + C 3 y 3 (t. where y, y, and y 3 are the fundamental set of solutions found in Exercise Hence we must have ( = y( = C y ( + C y ( + C 3 y 3 ( = C v + C v + C 3 v 3 C = [v, v, v 3 ] C C 3, where v, v, and v 3 are the eigenvectors of A found in Exercise To solve the system we form the augmented matrix ( [v, v, v 3, y(] = This is reduced to the row echelon form ( 3. Backsolving, we find that C 3 =, C = 3, and C =. Hence the solution is y(t = y (t + 3y (t y 3 (t.

11 9..5. Find the general solution of y = Answer: In system y = Ay, where ( y 3 8 ( 7 3 3, 3 8 we have A λi = ( 7 λ 3 3 λ. 3 8 λ We can compute the characteristic polynomial by expanding along the third column. We get p(λ = det(a λi = ( λ det 7 λ 3 3 λ = (λ + (λ + λ + 5. Hence we have one real eigenvalue λ =, and the quadratic λ + λ + 5 has complex roots λ = + i, and λ = i. For the eigenvalue λ =, we look for a vector in the nullspace (eigenspace of A λ I = A + I = ( The eigenspace is generated by v = (,, T. Thus, one solution is y (t = e t. For the eigenvalue λ = + i, we look for a vector in the nullspace (eigenspace of A λ I = A + ( ii = ( 5 i 3 5 i. 3 8 i The eigenspace is generated by ( 5 + i,, 3 i T. Thus, we have the complex conjugate solutions 5 + i z(t = e ( +it 3 i and 5 i z(t = e ( it. 3 + i

12 Using Euler s formula, we find the real and imaginary parts of the solution z(t. 5 + i z(t = e t e it 3 i Thus we have the solutions = e t (cos t + i sin t = e t ( 5 cos t sin t cos t 3 cos t + sin t [( 5 3 ] + i ( cos t 5 sin t + ie t sin t cos t + 3 sin t 5 cos t sin t y (t = Re(z(t = e t cos t 3 cos t + sin t cos t 5 sin t y 3 (t = Im(z(t = e t sin t cos t + 3 sin t and The general solution is y(t = C y (t + C y (t + C 3 y 3 (t Use a computer to help you find a fundamental set of solutions to the system y = Ay, where ( Answer: Using MATLAB we proceed as follows. [-7 -; - 8;38-8]; ee = eig(a ee = + i - i -

13 We see that A has two complex conjugate eigenvalues ± i and one real eigenvalue. Let s look at the real eigenvalue first. v3 = null(a - ee(3*eye(3, r v3 = -/ v3( = v3 = Thus we have the real solution y 3 (t = e t. Next we look at the complex eigenvalue + i w = null(a - ee(*eye(3, r w = -/ + /i 3/ + 3/i w=*w w = - + i 3 + 3i Thus we have the eigenvector w = ( + i, 3 + 3i, T. We also have the complex valued solution. + i z(t = e (+it 3 + 3i. Expanding using Euler s formula, we get [ ] z(t = e t [cos t + i sin t] 3 + i 3 ] = e [cos t t 3 sin t 3 + ie t [cos t ( 3 + sin t ] 3

14 Since the real and imaginary parts of z(t are solutions we get two real solutions ] y (t = Re(z(t = e [cos t t 3 sin t 3 and ] y (t = Im(z(t = e [cos t t 3 + sin t 3. The functions y, y, and y 3 form a fundamental set of solutions Use a computer to help you find a fundamental set of solutions to the system y = Ay, where Answer: Using MATLAB we proceed as follows. A=[ -5;-6- - ;3 - -5; ]; format rat ee = eig(a Thus A has eigenvalues and, with repeated three times. Let s look at the eigenspace for. null(a-ee(*eye(, r ans = - - Thus the eigenspace has dimension and is spanned by the vector (,,, T. Hence we get the solution y (t = e t.

15 Next we look at the eigenspace for. null(a-ee(*eye(, r ans = -/3 -/3 5/3 V=3*ans V = Thus the eigenspace has dimension 3 and it is spanned by the indicated column vectors. We get three linearly independent solutions. 5 y (t = e t 3, y 3 (t = e t, and y 3 (t = e t. 3 The functions y, y, y 3, and y form a fundamental set of solutions Use a computer to help you find the solution to the system y = Ay, with the given matrix A in Exercise and the initial value y( = (,, 5 T. Answer: In Exercise we found the fundamental set of solutions ] y (t = Re(z(t = e [cos t t 3 sin t 3, ] y (t = Im(z(t = e [cos t t 3 + sin t 3, and y 3 (t = e t. Our solution has the form y(t = C y (t + C y (t + C 3 y 3 (t. At t = wehave = y( = C 3 + C 3 + C 3 5 C = 3 3 C. C 3

16 We can use MATLAB to solve this system of equations V=[- ;3 3 ; ]; y = [- 5] ; C = V\y C = - Thus our solution is y(t = y (t y (t + y 3 (t Use a computer to help you find the solution to the system y = Ay, with the given matrix A in Exercise 9..3 and the initial value y( = (, 5,, T. Answer: In Exercise 9..3 we found the fundamental set of solutions y (t = e t, y 3 (t = e t 3 y (t = e t 3, and y (t = e t, 5. 3 Our solution has the form y(t = C y (t + C y (t + C 3 y 3 (t + C y (t. At t = we have = y( = C + C + C 3 + C C 3 C =. 3 C 3 3 C We can use MATLAB to solve this system as follows. V=[ ; 3 ; - 3 ; 3]; y=[- 5 ] ; C = V\y C =

17 Thus, C = C = C 3 = C =, leading to y(t = y (t + y (t + y 3 (t + y (t e t e = t + 3e t e t + 3e t. e t + 3e t