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Solutions for Chapter 3.3: Complex Roots of the Characteristic Equation

Elementary Differential Equations and Boundary Value Problems | 11th Edition | ISBN: 9781119256007 | Authors: Boyce, Diprima, Meade

Full solutions for Elementary Differential Equations and Boundary Value Problems | 11th Edition

ISBN: 9781119256007

Elementary Differential Equations and Boundary Value Problems | 11th Edition | ISBN: 9781119256007 | Authors: Boyce, Diprima, Meade

Solutions for Chapter 3.3: Complex Roots of the Characteristic Equation

Solutions for Chapter 3.3
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Textbook: Elementary Differential Equations and Boundary Value Problems
Edition: 11
Author: Boyce, Diprima, Meade
ISBN: 9781119256007

Chapter 3.3: Complex Roots of the Characteristic Equation includes 36 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Since 36 problems in chapter 3.3: Complex Roots of the Characteristic Equation have been answered, more than 12619 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Elementary Differential Equations and Boundary Value Problems, edition: 11. Elementary Differential Equations and Boundary Value Problems was written by and is associated to the ISBN: 9781119256007.

Key Math Terms and definitions covered in this textbook
  • Affine transformation

    Tv = Av + Vo = linear transformation plus shift.

  • Companion matrix.

    Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).

  • Complex conjugate

    z = a - ib for any complex number z = a + ib. Then zz = Iz12.

  • Cramer's Rule for Ax = b.

    B j has b replacing column j of A; x j = det B j I det A

  • Diagonal matrix D.

    dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.

  • Eigenvalue A and eigenvector x.

    Ax = AX with x#-O so det(A - AI) = o.

  • Factorization

    A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.

  • Free variable Xi.

    Column i has no pivot in elimination. We can give the n - r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

  • Linearly dependent VI, ... , Vn.

    A combination other than all Ci = 0 gives L Ci Vi = O.

  • Markov matrix M.

    All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

  • Normal equation AT Ax = ATb.

    Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b - Ax) = o.

  • Normal matrix.

    If N NT = NT N, then N has orthonormal (complex) eigenvectors.

  • Partial pivoting.

    In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

  • Projection matrix P onto subspace S.

    Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.

  • Rank r (A)

    = number of pivots = dimension of column space = dimension of row space.

  • Semidefinite matrix A.

    (Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

  • Similar matrices A and B.

    Every B = M-I AM has the same eigenvalues as A.

  • Skew-symmetric matrix K.

    The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.

  • Vector addition.

    v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.

  • Vector v in Rn.

    Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.

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