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APPM 2360: Section exam 3 7:00pm – 8:30pm, November 28, 2012. ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your student ID number, (3) recitation section (4) your instructor’s name, and (5) a grading table. Text books, class notes, and calculators are NOT permitted. A one-page crib sheet is allowed. 1 0 1 Problem 1: (20 points) Given the matrix A = 0 1 2 , −2 0 3 (a) (14 points) Find the eigenvalues and eigenvectors of matrix A. (b) (6 points) Find the eigenspaces for matrix A and give their dimension. Solution: (a) det(A − λI)v = (λ − 1)(−λ2 + 4λ − 5) = 0 leads to the eigenvalues λ1 = 2 + ı, λ2 = 2 − ı ~0 yields the corresponding eigenvectors ~v1 = and det(A − λI)v = 1 λ3 ı = 1, and solving 1 ı 0 2 − 2 2 + 2 1 − ı , ~v2 = 1 + ı and ~v3 = 1 . 0 1 1 (b) Eλ1 = span {~v1 } with dim(Eλ1 ) = 1 Eλ2 = span {~v2 } with dim(Eλ2 ) = 1 Eλ3 = span {~v3 } with dim(Eλ3 ) = 1 Problem 2: (20 points) Consider the second order differential equations 5y 00 − 10 y 0 + 15 y = 0 (a) (10 points) Construct the general solution. (b) (4 points) Construct the particular solution with initial condition y(0) = −2, y 0 (0) = 1. (c) (6 points) Letting x1 (t) = y(t) and x2 (t) = y 0 (t), construct the linear system of two first x1 0 order differential equations for x1 (t) and x2 (t), i.e., ~x = A~x where ~x = . (Do not x2 solve the linear system.) Solution: √ (a) Characteristic equation: r2 − 2r +√ 3 = 0. A pair√of complex conjugate roots r1,2 = 1 ± 2. General solution: y(t) = et (c1 cos( 2t) + c2 sin( 2t)). √ √ √ √ (b) Note that y 0 (t) = et ((c1 + 2c2 ) cos( 2t)+(c2 − 2c1 ) sin( 2t)). Applying initial condition, a linear system is obtained: c1 + So c1 = −2 and c2 = √3 . 2 √ c1 = −2 2c2 = 1. √ The particular solution is y(t) = et (−2 cos( 2t) + (c) x01 = x2 x02 = −3x1 + 2x2 . √3 2 √ sin( 2t)). Problem 3: (20 points) (a) (12 points) Use the Method of Undetermined Coefficients to write down the general form of the particular solution, yp , for the differential equations given below but do not solve: (i) y 00 + 6y 0 + 9y = e−3t (ii) y 00 + 6y 0 + 9y = t2 sin(6t) (iii) y 00 + 6y 0 + 9y = te2t cos(t) (iv) y 00 + 9y = −6y 0 + e−2t + t2 e−2t (b) (8 points) Now use the Method of Undetermined Coefficients to find a particular solution of the equation, y 00 + 6y 0 + 9y = e−3t Solution: (a)(i) yp = At2 e−3t , since y = e−t and y = te−3t are solutions of the corresponding homogeneous equation. (ii) yp = (At2 + Bt + C) sin(6t) + (Dt2 + Et + F ) cos(6t) (iii) yp = (At + B)e2t sin(t) + (Ct + D)e2t cos(t) (iii) yp = (At2 + Bt + C)e−2t (b) If yp = At2 e−3t , then yp0 = 2Ate−3t − 3At2 e−3t , and yp00 = 2Ae−3t − 12Ate−3t + 9Ate−3t , now 1 substituting yp into the equation y 00 + 6y 0 + 9y = e−3t yields A = 1/2 and so yp = − t2 e−3t . 2 PLEASE TURN OVER Problem 4: (20 points) Consider the following nonhomogeneous Euler equation with initial conditions t2 x00 + 3tx0 − 3x = t x(1) = 0, x0 (1) = 0 (a) (6 points) Find the general homogenous solution by assuming the form xh (t) = tr . (b) (9 points) Find the general solution to the nonhomogenous problem. (c) (5 points) Find the solution to the initial-value problem. Solution: (a) The characteristic equation is given by r(r − 1) + 3r − 3 = r2 + 2r − 3 = 0. Therefore r1 = 1, r2 = −3. The general homogenous solution is given by xh (t) = c1 t + c2 t−3 (b) Solve the normalize problem 3 3 1 x00 + x0 − 2 x = t t t by variation of parameters where xp (t) = v1 (t)t + v2 (t)t−3 . Therefore 0 0 t t−3 v1 = v20 et 1 −3t−4 The Wronksian for the homogenous solutions is −4t−3 . Using Cramer’s rule: 0 t 0 t−3 −1 t 1 t−1 −3t−4 1 t3 0 0 v1 = = , v = = − 2 −4t−3 4t −4t−3 4 Therefore 1 1 v1 (t) = ln(t), v2 (t) = − t4 4 16 Hence 1 1 1 1 xp (t) = ln(t)t − t4 t−3 = t ln(t) − t 4 16 4 16 The general solution is thus given by t xg (t) = c1 t + c2 t−3 + (4 ln(t) − 1) 16 (c) Solve the initial value problem gives 1 1 1 c1 16 = 3 1 −3 c2 − 16 Therefore 1 1 1 16 3 1 −3 − 16 1 1 1 1 1 16 ∼ ∼ 0 −4 − 14 0 1 1 0 0 ∼ 1 0 1 16 Hence x(t) = 1 16 1 16 1 −3 t t + (4 ln(t) − 1) 16 16 Problem 5: (20 points) Give a short answer to the following statements. ∼ (a) (4 points) Let A be a 4 × 4 real matrix. Two distinct complex eigenvalues λ1 and λ2 of A are known. Compute the remaining eigenvalues. (b) (4 points) Let A be a n×n real matrix. If λ1 is an eigenvalue of A with algebraic multiplicity 2, then rank (A − λ1 I) = n − 2. (TRUE/FALSE) (c) (4 points) Any solution y(t) to y 00 + 5y 0 + 6y = 0 tends to zero as t → ∞. (TRUE/FALSE) (d) (4 points) Let A be a 3 × 3 real matrix with real eigenvalues. The eigenvectors of A form a basis for R3 . (TRUE/FALSE) (e) (4 points) Let A be a n × n real matrix with λ1 one of its eigenvalues. Then (A − λ1 I) is an invertible matrix. (TRUE/FALSE) Solution: (a) λ3 = λ̄1 and λ4 = λ̄2 (b) FALSE (c) TRUE (Roots of the characteristic equation are −2 amd −3) (d) FALSE (Linearly independent eigenvectors might not be obtained for repeated eigenvalues) (e) FALSE (det (A − λ1 I) has to be zero to obtain a non-trivial eigenvector)