 9.3.1: In each of 1 through 4, verify that (0, 0) is a critical point, sho...
 9.3.2: In each of 1 through 4, verify that (0, 0) is a critical point, sho...
 9.3.3: In each of 1 through 4, verify that (0, 0) is a critical point, sho...
 9.3.4: In each of 1 through 4, verify that (0, 0) is a critical point, sho...
 9.3.5: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.6: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.7: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.8: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.9: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.10: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.11: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.12: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.13: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.14: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.15: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.16: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.17: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.18: In each of 5 through 18:(a) Determine all critical points of the gi...
 9.3.19: Consider the autonomous systemdx/dt = y, dy/dt = x + 2x3.(a) Show t...
 9.3.20: Consider the autonomous systemdx/dt = x, dy/dt = 2y + x3.(a) Show t...
 9.3.21: The equation of motion of an undamped pendulum is d2/dt2 + 2 sin = ...
 9.3.22: (a) By solving the equation for dy/dx, show that the equation of th...
 9.3.23: The motion of a certain undamped pendulum is described by the equat...
 9.3.24: Consider again the pendulum equations (see 23)dx/dt = y, dy/dt = 4 ...
 9.3.25: This problem extends to a damped pendulum. The equations of motion ...
 9.3.26: Theorem 9.3.2 provides no information about the stability of a crit...
 9.3.27: In this problem we show how small changes in the coefficients of a ...
 9.3.28: In this problem we show how small changes in the coefficients of a ...
 9.3.29: In this problem we derive a formula for the natural period of an un...
 9.3.30: A generalization of the damped pendulum equation discussed in the t...
Solutions for Chapter 9.3: Locally Linear Systems
Full solutions for Elementary Differential Equations and Boundary Value Problems  10th Edition
ISBN: 9780470458310
Solutions for Chapter 9.3: Locally Linear Systems
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Elementary Differential Equations and Boundary Value Problems, edition: 10. Elementary Differential Equations and Boundary Value Problems was written by and is associated to the ISBN: 9780470458310. Chapter 9.3: Locally Linear Systems includes 30 full stepbystep solutions. Since 30 problems in chapter 9.3: Locally Linear Systems have been answered, more than 16331 students have viewed full stepbystep solutions from this chapter.

Cholesky factorization
A = CTC = (L.J]))(L.J]))T for positive definite A.

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 nl  An).

Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).

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

Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and

Eigenvalue A and eigenvector x.
Ax = AX with x#O so det(A  AI) = o.

Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

Independent vectors VI, .. " vk.
No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.

Inverse matrix AI.
Square matrix with AI A = I and AAl = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B1 AI and (AI)T. Cofactor formula (Al)ij = Cji! detA.

Iterative method.
A sequence of steps intended to approach the desired solution.

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

lAII = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n  1, volume of box = I det( A) I.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.

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

Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q 1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

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.

Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.

Reflection matrix (Householder) Q = I 2uuT.
Unit vector u is reflected to Qu = u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q1 = Q.

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