 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 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 (2k...
 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  10th Edition
ISBN: 9780470458310
Solutions for Chapter 6.5: Impulse Functions
Get Full SolutionsElementary Differential Equations and Boundary Value Problems was written by and is associated to the ISBN: 9780470458310. Chapter 6.5: Impulse Functions includes 25 full stepbystep solutions. 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 16352 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: 10.

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

Column space C (A) =
space of all combinations of the columns of A.

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

Cyclic shift
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.

Diagonal matrix D.
dij = 0 if i # j. Blockdiagonal: zero outside square blocks Du.

Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then SI AS = A = eigenvalue matrix.

Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).

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

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!).

GaussJordan method.
Invert A by row operations on [A I] to reach [I AI].

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

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.

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.

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.

Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.

Pseudoinverse A+ (MoorePenrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

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

Similar matrices A and B.
Every B = MI AM has the same eigenvalues as A.

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

Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.