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Solutions for Chapter 2.2: First-Order Differential Equations

Differential Equations with Boundary-Value Problems, | 8th Edition | ISBN: 9781111827069 | Authors: Dennis G. Zill, Warren S. Wright

Full solutions for Differential Equations with Boundary-Value Problems, | 8th Edition

ISBN: 9781111827069

Differential Equations with Boundary-Value Problems, | 8th Edition | ISBN: 9781111827069 | Authors: Dennis G. Zill, Warren S. Wright

Solutions for Chapter 2.2: First-Order Differential Equations

Solutions for Chapter 2.2
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Textbook: Differential Equations with Boundary-Value Problems,
Edition: 8
Author: Dennis G. Zill, Warren S. Wright
ISBN: 9781111827069

Since 60 problems in chapter 2.2: First-Order Differential Equations have been answered, more than 20488 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Differential Equations with Boundary-Value Problems,, edition: 8. Chapter 2.2: First-Order Differential Equations includes 60 full step-by-step solutions. Differential Equations with Boundary-Value Problems, was written by and is associated to the ISBN: 9781111827069.

Key Math Terms and definitions covered in this textbook
  • Augmented matrix [A b].

    Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.

  • Cofactor Cij.

    Remove row i and column j; multiply the determinant by (-I)i + j •

  • Elimination matrix = Elementary matrix Eij.

    The identity matrix with an extra -eij in the i, j entry (i #- j). Then Eij A subtracts eij times row j of A from row i.

  • Hermitian matrix A H = AT = A.

    Complex analog a j i = aU of a symmetric matrix.

  • Hilbert matrix hilb(n).

    Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.

  • Jordan form 1 = M- 1 AM.

    If A has s independent eigenvectors, its "generalized" eigenvector matrix M gives 1 = diag(lt, ... , 1s). The block his Akh +Nk where Nk has 1 's on diagonall. Each block has one eigenvalue Ak and one eigenvector.

  • Krylov subspace Kj(A, b).

    The subspace spanned by b, Ab, ... , Aj-Ib. Numerical methods approximate A -I b by x j with residual b - Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.

  • Least squares solution X.

    The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.

  • Left inverse A+.

    If A has full column rank n, then A+ = (AT A)-I AT has A+ A = In.

  • Left nullspace N (AT).

    Nullspace of AT = "left nullspace" of A because y T A = OT.

  • Lucas numbers

    Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = 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.

  • Nilpotent matrix N.

    Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.

  • Orthogonal matrix Q.

    Square matrix with orthonormal columns, so QT = Q-l. Preserves length and angles, IIQxll = IIxll and (QX)T(Qy) = xTy. AlllAI = 1, with orthogonal eigenvectors. Examples: Rotation, reflection, permutation.

  • Orthogonal subspaces.

    Every v in V is orthogonal to every w in W.

  • Pascal matrix

    Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).

  • Pivot columns of A.

    Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

  • Pivot.

    The diagonal entry (first nonzero) at the time when a row is used in elimination.

  • Reflection matrix (Householder) Q = I -2uuT.

    Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.

  • Simplex method for linear programming.

    The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

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