 6.5.1: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.2: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.3: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.4: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.5: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.6: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.7: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.8: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.9: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.10: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.11: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.12: In each of 1 through 12: (a) Find the solution of the given initial...
 6.5.13: Consider again the system in Example 1 of this section, in which an...
 6.5.14: Consider the initial value problem y + y + y = (t 1), y(0) = 0, y (...
 6.5.15: Consider the initial value problem y + y + y = k(t 1), y(0) = 0, y ...
 6.5.16: Consider the initial value problem y + y = fk(t), y(0) = 0, y (0) =...
 6.5.17: 17 through 22 deal with the effect of a sequence of impulses on an ...
 6.5.18: 17 through 22 deal with the effect of a sequence of impulses on an ...
 6.5.19: 17 through 22 deal with the effect of a sequence of impulses on an ...
 6.5.20: 17 through 22 deal with the effect of a sequence of impulses on an ...
 6.5.21: 17 through 22 deal with the effect of a sequence of impulses on an ...
 6.5.22: 17 through 22 deal with the effect of a sequence of impulses on an ...
 6.5.23: The position of a certain lightly damped oscillator satisfies the i...
 6.5.24: Proceed as in for the oscillator satisfying y + 0.1y + y = 15 k=1 [...
 6.5.25: (a) By the method of variation of parameters, show that the solutio...
Solutions for Chapter 6.5: Impulse Functions
Full solutions for Elementary Differential Equations and Boundary Value Problems  9th Edition
ISBN: 9780470383346
Solutions for Chapter 6.5: Impulse Functions
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. Since 25 problems in chapter 6.5: Impulse Functions have been answered, more than 12722 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: Elementary Differential Equations and Boundary Value Problems, edition: 9. Elementary Differential Equations and Boundary Value Problems was written by and is associated to the ISBN: 9780470383346. Chapter 6.5: Impulse Functions includes 25 full stepbystep solutions.

CayleyHamilton Theorem.
peA) = det(A  AI) has peA) = zero matrix.

Change of basis matrix M.
The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)

Circulant matrix C.
Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn  l . Cx = convolution c * x. Eigenvectors in F.

Commuting matrices AB = BA.
If diagonalizable, they share n eigenvectors.

Covariance matrix:E.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x  x) (x  x) T is positive (semi)definite; :E is diagonal if the Xi are independent.

Cross product u xv in R3:
Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].

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

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.

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.

Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

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, ... , AjIb. 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.

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.

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

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

Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Spectrum of A = the set of eigenvalues {A I, ... , An}.
Spectral radius = max of IAi I.

Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B II·

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