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MATH 215/255 Fall 2014 Assignment 8 due 11/19 §3.4, §3.7, §3.5, §3.9 Solutions to selected exercises can be found in [Lebl], starting from page 303. • 3.4.6: a) Find the general solution of x01 = 2x1 , x02 = 3x2 using the eigenvalue method (first write the system in the form ~x0 = A~x). b) Solve the system by solving each equation and verify you get the same general solution. x1 (t) 2 0 Answer. a). Let ~x(t) = and A = , then ~x0 = A~x. Let’s find the x2 (t) 0 3 2−λ 0 eigenvalues of A, that is, we should solve det (A − λI2 ) = det = 0, 0 3−λ then λ1 = 2, and λ2 = 3. For λ1 = 2, let’s solve A~x = 2~x, that is, 0 0 x1 0 . = 0 1 x2 0 Then x1 x2 = x1 1 1 is an eigenvalue for 2. , that is, 0 0 For λ2 = 3, let’s solve A~x = 3~x, that is, 0 x1 −1 0 . = 0 x2 0 0 Then x1 x2 = x2 0 0 is an eigenvalue for 3. , that is, 1 1 Therefore, the general solution to ~x0 = A~x is: 2t ~x(t) = C1 e 1 0 3t + C2 e 0 1 = C1 e2t C2 e3t . b). Since x01 = 2x1 , then x1 = C1 e2t . Since x02 = 3x2 , then x2 = C2 e3t . Hence the general soluton to ~x0 = A~x is: ~x(t) = C1 e2t C2 e3t . • 3.4.7: Find the general solution of x01 = 3x1 + x2 , x02 = 2x1 + 4x2 using the eigenvalue method. 3 1 Answer. Let ~x(t) = and A = , then ~x0 = A~x. Let’s find the 2 4 3−λ 1 eigenvalues of A, that is, we should solve det (A − λI2 ) = det = 0, 2 4−λ then (λ − 3)(λ − 4) − 2 = 0, that is, λ2 − 7λ + 10 = 0, which implies that x1 (t) x2 (t) λ1 = 2, and λ2 = 5. For λ1 = 2, let’s solve A~x = 2~x, that is, 1 1 x1 0 = . 2 2 x2 0 1 1 x1 , that is, is an eigenvalue for 2. Then = x1 −1 −1 x2 For λ2 = 5, let’s solve A~x = 5~x, that is, −2 1 x1 0 = . 2 −1 0 x2 x1 1 1 is an eigenvalue for 3. Then , that is, = x1 2 x2 2 Therefore, the general solution to ~x0 = A~x is: C1 e2t + C2 e5t 1 1 = + C2 e5t ~x(t) = C1 e2t . −C1 e2t + 2C2 e5t 2 −1 • 3.4.8: Find the general solution of x01 = x1 − 2x2 , x02 = 2x1 + x2 using the eigenvalue method. Do not use complex exponentials in your solution. 1 −2 x1 (t) , then ~x0 = A~x. Let’s find the and A = Answer. Let ~x(t) = 2 1 x2 (t) 1 − λ −2 = 0, eigenvalues of A, that is, we should solve det (A − λI2 ) = det 2 1−λ then (λ − 1)2 + 4 = 0, which implies that λ1 = 1 + 2i, and λ2 = 1 − 2i. For λ1 = 1 + 2i, let’s solve A~x = (1 + 2i)~x, that is, −2i −2 x1 0 = . 2 −2i x2 0 x1 1 1 Then = x1 , that is, is an eigenvalue for 1 + 2i. Then a complex x2 −i −i solution can be (1+2i)t 1 e λ1 t e = −i −ie(1+2i)t t e cos(2t) + iet sin(2t) = et sin(2t) − iet cos(2t) t e cos(2t) et sin(2t) = + i et sin(2t) −et cos(2t) 2 Therefore, the general solution to ~x0 = A~x is: ~x(t) = C1 et cos(2t) et sin(2t) + C2 et sin(2t) −et cos(2t) = C1 et cos(2t) + C2 et sin(2t) C1 et sin(2t) − C2 et cos(2t) . 5 −3 . Find the general solution of ~x0 = A~x. 3 −1 • 3.7.2: Let A = Answer. Let’s find the eigenvalues of A, that is, we should solve det (λI2 − A) = λ−5 3 det = 0, then (λ − 5)(λ + 1) + 9 = 0, that is, λ2 − 4λ + 4 = 0, which −3 λ + 1 implies that λ1 = λ2 = 2. It’s easy to see that 2I2 − A 6= 0, that is, λ = 2 is a defective 1 eigenvalue of A. For λ = 2, let’s solve A~x = 2~x, that is, x1 0 −3 3 . = 0 −3 3 x2 Then x1 x2 = x1 1 1 is an eigenvalue for 2. Then a solution can be , that is, 1 1 1 . e2t 1 1 , that is, e2t~v is a solution to ~x0 = A~x. Let ~y (t) = e2t (t~v + w) ~ = Let ~v = 1 te2t~v + e2t w ~ for some w ~ be such that ~y be a solution to ~x0 = A~x, then ~y 0 = e2t~v + 2te2t~v + 2e2t w ~ = A~y = te2t A~v + e2t Aw. ~ Since A~v = 2~v , then e2t~v + 2e2t w ~ = e2t Aw, ~ that is, (2I2 − A)w ~ = −~v , which implies that −3 3 w1 −1 = . −3 3 w2 −1 So −3w1 + 3w2 = −1, we can take w1 = 0 and w2 = − 31 , then 1 0 2t 2t 2t 2t ~y (t) = te ~v + e w ~ = te +e . 1 − 13 Therefore, the general solution to ~x0 = A~x is: 2t ~x(t) = C1 e 1 1 + C2 te 2t 1 1 + C2 e 3 2t 0 − 31 = C1 e2t + C2 te2t C1 e2t + C2 te2t − C32 e2t . a a • 3.7.104: Let A = , where a, b and c are unknowns. Suppose that 5 is a b c 1 doubled eigenvalue of defect 1, and suppose that is the eigenvector. Find A and 0 show that there is only one solution. λ − a −a Answer. Let’s solve det (λI2 − A) = det = 0, that is, (λ − a)(λ − −b λ − c c) − ab = 0, which implies that λ2 − (a + c)λ + ac − ab = 0. Since 5 is a doubled eigenvalue of defect 1, that is, λ = 5 is the only solution to λ2 − (a + c)λ + ac − ab = 0, then and ac − ab = 25. a + c = 10, Since 1 0 (1) is the eigenvalue for λ = 5, that is, 5 − a −a −b 5 − c 1 0 = 0 0 Then we have 5 − a = 0, − b = 0. and (2) By (1) and (2), we have a = 5, b = 0, and c = 5. Therefore, we know that A= 5 5 0 5 . • 3.5.101: Describe the behavior of the following systems without solving: a) x0 = x + y, y 0 = x − y. b) x01 = x1 + x2 , x02 = 2x2 . c) x01 = −2x2 , x02 = 2x1 . d) x0 = x + 3y, y 0 = −2x − 4y. e) x0 = x − 4y, y 0 = −4x + y. Answer. a) The coefficient matrix is 1 1 A= , 1 −1 (3) 2 which √ has characteristic polynomial λ − 2 = 0. Therefore the eigenvalues are λ = ± 2. Since they are both real and have opposite signs, the behaviour is saddle . 4 b) The coefficient matrix is A= 1 1 , 0 2 (4) which has characteristic polynomial (λ − 1)(λ − 2) = 0. Therefore the eigenvalues are λ = 1, 2. Since they are both real and positive the behaviour is source (unstable) . c) The coefficient matrix is 0 −2 A= , 2 0 (5) which has characteristic polynomial λ2 +4 = 0. Therefore the eigenvalues are λ = ±2i. Since they are both purely imaginary, the behaviour is center point (ellipses) . d) The coefficient matrix is 1 3 A= , −2 −4 (6) which has characteristic polynomial λ2 + 3λ + 2 = 0. Therefore the eigenvalues are λ = −1, −2. Since they are both real and negative, the behaviour is sink (stable) . e) The coefficient matrix is 1 −4 , A= −4 1 (7) which has characteristic polynomial λ2 − 2λ − 15 = 0. Therefore the eigenvalues are λ = −3, 5. Since they are both real and of opposite sign, the behaviour is saddle . • 3.5.102: Suppose that ~x0 = A~x where A is a 2 × 2 matrix with eigenvalues 2 ± i. Describe the behavior. Answer. Since the eigenvalues are complex with positive real parts, the equilibrium at the origin is a spiral source . All nonzero solutions spiral around and move away from the origin. • 3.9.4: Find a particular solution to x0 = x + 2y + 2t, y 0 = 3x + 2y − 4, a) using undetermined coefficients, b) using variation of parameters. Answer. a) Since the inhomogeneous term is 2 0 t + , 0 −4 we try ~xp (t) = t~a + ~b, where ~a = a1 ~ b ,b = 1 a2 b2 are to be determined. Substituting ~xp (t) = t~a + ~b into the system gives 5 a1 a1 + 2a2 b1 + 2b2 2t =t + + . a2 3a1 + 2a2 3b1 + 2b2 −4 Equating the coefficients of t on both sides of the equation, we obtain a1 + 2a2 = −2 3a1 + 2a2 = 0. Equating the constants on both sides of the equation, we obtain b1 + 2b2 = a1 3b1 + 2b2 = a2 + 4. Solving the first two equations gives a1 = 1 and a2 = −3/2. Solving the next two equations, we obtain b1 = 3/4 and b2 = 1/8. Therefore a particular solution is t + 43 ~xp (t) = − 32 t + . 1 8 b) The coefficient matrix and inhomogeneous term for the system ~x0 = P ~x + f~ is 2t 1 2 ~ . , f= P = −4 3 2 2 1 and P has eigenvalues λ = −1 and 4 with corresponding eigenvectors 3 −1 respectively. Thus we can construct a fundamental matrix solution as −t e 2e4t X(t) = . −e−t 3e4t Then −1 X(t) 1 = 3t 5e 3e4t −2e4t e−t e−t 1 = 5 3et −2et e−4t e−4t . Let ~xp (t) = X(t)~u(t) be a particular solution to ~x0 + P ~x = f~. Then 1 1 3et −2et 2t 6tet + 8et 0 −1 ~ ~u (t) = X(t) f = = . −4 5 e−4t e−4t 5 2te−4t − 4e−4t Then Z ~u(t) = 0 ~u (t) dt = Z 1 5 6tet + 8et 2te−4t − 4e−4t 1 dt = 5 6tet + 2et 1 −4t − 2 te + 78 e−4t . So a particular solution is −t 1 t + 34 e 2e4t 6tet + 2et ~xp (t) = X(t)~u(t) = = . − 12 te−4t + 87 e−4t − 32 t + 81 5 −e−t 3e4t 6 • 3.9.101: Find a particular solution to x0 = 5x + 4y + t, y 0 = x + 8y − t. a) using undetermined coefficients. b) using variation of parameters. Answer. a) Since the inhomogeneous term contains only t’s we try a particular solution of the form ~xp = t~a + ~b. Substituting into the system gives a1 5a1 + 4a2 5b1 + 4b2 t =t + + . a2 a1 + 8a2 b1 + 8b2 −t Equating powers of t in both entries we get the systems 5a1 + 4a2 = −1 a1 + 8a2 = 1 and 5b1 + 4b2 = a1 b1 + 8b2 = a2 . Solving the first system gives a1 = −1/3 and a2 = 1/6. Solving the second system gives b1 = −5/54 and b2 = 7/216. Therefore the particular solution is 5 − 31 t − 54 ~xp = 1 . 7 6 t + 216 b) The corresponding matrix and inhomogeneous vector for the problem ~x0 = P ~x + f~ is t 5 4 ~ . , f= P = −t 1 8 1 −4 and . Then we P has eigenvalues λ = 9, 4 with corresponding eigenvectors 1 1 can take fundamental matrix to be: 9t e −4e4t X(t) = . e9t e4t Then X(t)−1 = 1 5e13t e4t 4e4t −e9t e9t = 1 5 e−9t 4e−9t −e−4t e−4t . Let ~xp (t) = X(t)~u(t) be a particular solution to ~x0 = P ~x + f~, then −9t 1 1 −3te−9t t e 4e−9t 0 −1 ~ u (t) = X(t) f = = −t 5 −e−4t e−4t 5 −2te−4t 7 Then Z ~u(t) = Z 0 ~u (t) dt = 1 5 −3te−9t −2te−4t 1 dt = 5 1 −9t 1 −9t + 27 e 3 te 1 −4t 1 −4t te + e 2 8 So a particular solution can be: 1 ~x(t) = X(t)~u(t) = 5 e9t −4e4t e9t e4t 1 −9t 1 −9t + 27 e 3 te 1 −4t 1 −4t te + e 2 8 5 − 13 t − 54 = 1 . 7 6 t + 216 • 3.9.102: Find a particular solution to x0 = y + et , y 0 = x + et . a) using undetermined coefficients. b) using variation of parameters. Answer. a) The corresponding matrix is 0 1 , A= 1 0 which has characteristic equation λ2 − 1 = 0. Hence the eigenvalues are λ = ±1. Observe that the eigenvalue λ = 1 matches the power of the exponent that corresponds to the non-homogeneous part. Also, this eigenvalue has a corresponding eigenvector 1 . The particular solution is therefore of the form tet~a + et~b as is easily verified ~v = 1 by substituting into the equation and keeping in mind that A~v = ~v ~x0p = et~a + tet~a + et~b = A~xp + et~v = tet A~a + et A~b + et~v Then ~a + ~b = A~b + ~v , and ~a = A~a. Then we can take ~a = ~v = 1 , 1 and ~b = 0. Hence the particular solution is ~xp = tet 1 . 1 b) Matrix λ1 = 1, λ2 = −1 and the corresponding eigenvectors A has eigenvalues 1 1 ~v1 = , ~v2 = . Thus, the fundamental matrix is 1 −1 X(t) = et e−t et −e−t 8 . Its inverse is X(t) −1 1 = −2 −e−t −e−t −et et 1 = 2 e−t e−t et −et . Let ~xp (t) = X(t)~u(t) be a particular solution to ~x0 = A~x + f~, then t Z Z −t 1 e e−t e t −1 ~ u(t) = X(t) f dt = dt = , t t t e −e e 0 2 and finally ~x(t) = X(t)~u(t) = et e−t et −e−t 9 t 0 t te . = tet