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Math 2280 Summer 2006 Test 2 You will have one hour to complete this test. You may not use a calculator, computer, or any other electronic device on this test. Please be sure to show as much of your work as possible, as I will determine your score on a given problem from the work and not necessarily by the answer. Please be neat so that I can grade your answers easily. 1. Transform the given system of differential equations into an equivalent system of first order differential equations. x00 + 3x0 + 4x − 2y = 0 y 00 + 2y 0 − 3x + y = cos t If we make the substitutions x1 = x, x2 = x0 and y1 = y, y2 = y 0 , then the system becomes x01 = x2 x02 = −3x2 − 4x1 + 2y1 . y10 = y2 0 y2 = −2y2 + 3x1 − y1 + cos t 2. Use the Wronskian to show that the following vector functions are linearly independent. −2 2 2 ~x1 (t) = et 2 , ~x2 (t) = et 0 , ~x3 (t) = e5t −2 . 1 1 1 We simply calculate that t 2e −2et 2e5t 0 −2e5t W (t) = 2et t et e e5t = −2et (−2e6t − 2e6t ) + 2e5t (2e2t + 2e2t ) = −2et (−4e6t ) + 2e5t (4e2t ) = 8e7t + 8e7t = 16e7t 6= 0. Thus, the functions are linearly independent. 3. Find a solution of the system below that corresponds to the given initial values. x01 = 9x1 + 5x2 , x02 = −6x1 − 2x2 , x1 (0) = 1, First, we write the system in vector form as 0 x1 9 5 x1 = . −6 −2 x2 x02 1 x2 (0) = 0 Next, we find the eigenvalues of the coefficient matrix. They must satisfy 9−λ 5 = (9−λ)(−2−λ)+30 = −18−9λ+2λ+λ2 +30 = λ2 −7λ+12 = (λ−4)(λ−3). 0 = −6 −2 − λ An eigenvector corresponding to λ = 4 must satisfy (A − 4I)~v1 = ~0, so we row reduce the matrix 5 5 1 1 −1 ∼ ⇒ v1 = −v2 ⇒ ~v1 = . −6 −6 0 0 1 Similarly, for λ = 3, we have 6 5 1 ∼ −6 −5 0 5 6 5 ⇒ v1 = − v2 ⇒ ~v2 = 6 0 −5 6 . This gives us the general solution 4t ~x(t) = c1 e and ~x(0) = c1 −1 1 −1 1 3t + c2 e + c2 −5 6 −5 6 1 0 = , . Therefore, we solve the augmented matrix system .. .. −1 −5 . 1 1 5 . −1 ∼ ∼ 1 .. .. 1 6 . 0 0 1 . 1 0 . 0 .. −6 . . 1 .. 1 This gives us that 4t ~x(t) = −6e −1 1 3t +e −5 6 . 4. Find the general solution to the system below. x01 = 3x1 − x2 , x02 = x1 + 5x2 Again, we write the system in matrix form as 0 x1 3 −1 x1 = . x02 1 5 x2 Then the eigenvalues of the coefficient matrix must satisfy 3−λ −1 0= = (3 − λ)(5 − λ) + 1 = 15 − 3λ − 5λ + λ2 + 1 = λ2 − 8λ + 16 = (λ − 4)(λ − 4). 1 5−λ Therefore, we have a repeated eigenvalue λ = 4. To find a corresponding eigenvector(s), we solve the homogeneous system (A − 4I)~v1 = ~0. −1 −1 1 1 −1 ∼ ⇒ v1 = −v2 ⇒ ~v1 = . 1 1 0 0 1 2 Since we need two linearly independent eigenvectors and there is only one here, we find a generalized eigenvector. Generalized eigenvectors satisfy the equation (A − 4I)~v2 = ~v1 , which becomes the augmented matrix equation .. .. −1 −1 . −1 1 1 . 1 ∼ ⇒ v1 = −v2 + 1 ⇒ ~v2 = 0 . . . 1 1 1 .. 1 0 0 .. 0 This gives the solution ~x(t) = c1 e4t −1 1 + c2 e4t 0 1 +t −1 1 . 5. Find a fundamental matrix for the system 2 −1 0 ~x = ~x −4 2 and then use it to find the solution satisfying the initial value 2 ~x(0) = . −1 The eigenvalues of the coefficient matrix satisfy 2−λ −1 = (2 − λ)2 − 4 ⇒ (2 − λ)2 = 4 ⇒ 2 − λ = ±2 ⇒ λ = 2 ± 2 = 4 or 0. 0= −4 2−λ To find an eigenvector corresponding to λ = 4, we calculate 1 1 −2 −1 1 12 ⇒ v1 = − v2 ⇒ ~v1 = . ∼ −2 −4 −2 0 0 2 Corresponding to λ = 0, we have 1 2 −1 1 1 − 12 ⇒ v1 = v2 ⇒ ~v2 = . ∼ 2 −4 2 0 0 2 This gives us fundamental matrix Φ(t) = with e4t −2e4t Φ(0) = 1 −2 1 2 To find Φ−1 (0), we solve the augmented matrix system .. .. 1 1 . 1 0 ∼ 1 1 . 1 0 ∼ 1 0 . . 0 1 −2 2 .. 0 1 0 4 .. 2 1 3 1 2 , . .. . .. . 1 2 − 14 1 2 1 4 ⇒ Φ−1 (0) = 1 2 1 2 − 14 1 4 . Finally, we have that ~x(t) = Φ(t)Φ−1 (0)~x0 = 1 1 4t 1 e4t 1 − 14 − 41 e4t + 14 2 2 2 2e + 2 = 1 1 4t 1 1 −2e4t 2 −1 −1 −e4t + 1 2 4 2e + 2 5 4t 3 e4t + 1 + 41 e4t − 14 4e + 4 = . = 1 1 5 −2e4t + 2 − 2 e4t − 2 − 2 e4t + 32 6. Use the method of undetermined coefficients to find a particular solution to the nonhomogeneous problem x0 = x + 2y + 3, y 0 = 2x + y − 2. As a matrix equation, the system is 0 x 1 = y0 2 2 1 x y 3 −2 + . The eigenvalues of the coefficient matrix satisfy 1−λ 2 = (1 − λ)2 − 4 ⇒ (1 − λ)2 = 4 ⇒ 1 − λ = ±2 ⇒ λ = 1 ± 2 = 3 or − 1. 0= 2 1−λ Thus, we take as our guess ~xp (t) = a1 a2 b1 b2 + t. Differentiating with respect to t, we have ~x0p (t) = Also, we have that 1 2 a1 1 ~ A~xp + f (t) = + 2 1 a2 2 2 1 b1 b2 b1 b2 t+ . 3 −2 = a1 + 2a2 + 3 2a1 + a2 − 2 + b1 + 2b2 2b1 + b2 t. Equating the ”coefficients” of the various powers of t gives us a1 + a2 + 3 = b1 b + 2b2 = 0 and 1 . 2a1 + a2 − 2 = b2 2b1 + b2 = 0 Since the coefficient matrix is nonsingular, the second set of equations implies that b1 = b2 = 0. In this case, the first set of equations corresponds to the augmented matrix equation .. .. .. 7 3 . 1 2 . −3 ∼ 1 2 . −3 ∼ 1 0 . .. .. .. 0 1 . −8 2 1 . 2 0 −3 . 8 3 This gives us that ~xp (t) = 4 7 3 − 83 . 7. Without completely solving the system below, determine the type (node, center, etc.) of the fixed point at the origin, and say whether it is asymptotically stable, stable, or unstable. dx = y, dt As a matrix equation, the system is x0 y0 = dy = −5x − 4y dt 0 1 −5 −4 x y . The eigenvalues of the coefficient matrix satisfy −λ 1 = −λ(−4 − λ) + 5 = λ2 + 4λ + 5 0= −5 −4 − λ The quadratic formula tells us that the eigenvalues are p √ −4 ± 16 − 4(1)(5) −4 ± −4 = = −2 ± i. λ= 2 2 Since the coefficient matrix has complex conjugate eigenvalues with negative real part, the fixed point at the origin is a stable spiral, which is asymptotically stable. 8. Given the nonlinear system dx = x20 + y, dt dy = − sin 5x − y 8 − 4y, dt find the linearization about the fixed point at the origin and classify the fixed point of the linearization as in problem 7. The linearization of the system will have coefficient matrix J(0, 0), where 20x19 1 J(x, y) = . −5 cos 5x −8y 7 − 4 This gives that J(0, 0) = 0 1 −5 −4 . Since this is that same matrix as in problem 7, we have that the linearization has a stable spiral fixed point at the origin, and this is of course an asymptotically stable fixed point. 5