 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: 3. Consider again the system in Example 1 of this section, in which...
 6.5.14: Consider the initial value problemy + y + y = (t 1), y(0) = 0, y(0)...
 6.5.15: Consider the initial value problemy + y + y = k(t 1), y(0) = 0, y(0...
 6.5.16: Consider the initial value problemy + y = fk(t), y(0) = 0, y(0) = 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 satisfyingy + 0.1y + y = 15k=1[t (...
 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  10th Edition
ISBN: 9780470458327
Solutions for Chapter 6.5: Impulse Functions
Get Full SolutionsChapter 6.5: Impulse Functions includes 25 full stepbystep solutions. Elementary Differential Equations was written by and is associated to the ISBN: 9780470458327. This textbook survival guide was created for the textbook: Elementary Differential Equations, edition: 10. This 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 11639 students have viewed full stepbystep solutions from this chapter.

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

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

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

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

Elimination.
A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.

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

Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.

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.

Matrix multiplication AB.
The i, j entry of AB is (row i of A)·(column j of B) = L aikbkj. By columns: Column j of AB = A times column j of B. By rows: row i of A multiplies B. Columns times rows: AB = sum of (column k)(row k). All these equivalent definitions come from the rule that A B times x equals A times B x .

Multiplication Ax
= Xl (column 1) + ... + xn(column n) = combination of columns.

Network.
A directed graph that has constants Cl, ... , Cm associated with the edges.

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.

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.

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

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

Solvable system Ax = b.
The right side b is in the column space of A.

Tridiagonal matrix T: tij = 0 if Ii  j I > 1.
T 1 has rank 1 above and below diagonal.

Unitary matrix UH = U T = UI.
Orthonormal columns (complex analog of Q).

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