 5.2.1: dx dt = 2 x y dy dt = y x2 (a) x0 = 2, y0 = 1 (b) x0 = 0, y0 = 1 (c...
 5.2.2: dx dt = 2 x y dy dt = y x (a) x0 = 1, y0 = 1 (b) x0 = 2, y0 = 1 (...
 5.2.3: dx dt = x(x 1) dy dt = x2 y (a) x0 = 1, y0 = 0 (b) x0 = 0.8, y0 = 0...
 5.2.4: The VolterraLotka system for two competitive species is dx dt = x(...
 5.2.5: dx dt = x(x 3y + 150) dy dt = y(2x y + 100)
 5.2.6: dx dt = x(10 x y) dy dt = y(30 2x y)
 5.2.7: dx dt = x(100 x 2y) dy dt = y(150 x 6y)
 5.2.8: dx dt = x(x y + 100) dy dt = y(x2 y2 + 2500)
 5.2.9: dx dt = x(x y + 40) dy dt = y(x2 y2 + 2500)
 5.2.10: dx dt = x(4x y + 160) dy dt = y(x2 y2 + 2500)
 5.2.11: dx dt = x(8x 6y + 480) dy dt = y(x2 y2 + 2500)
 5.2.12: dx dt = x(2 x y) dy dt = y(y x2)
 5.2.13: dx dt = x(2 x y) dy dt = y(y x)
 5.2.14: dx dt = x(x 1) dy dt = y(x2 y)
 5.2.15: Some species live in a cooperative mannereach species helping the o...
 5.2.16: da dt = ab 2 db dt = ab 2
 5.2.17: da dt = 2 ab 2 db dt = 3 2 ab 2
 5.2.18: da dt = 2 ab 2 a2 3 db dt = 3 2 ab 2
 5.2.19: da dt = 2 ab 2 + b2 6 db dt = 3 2 ab 2 b2 3
 5.2.20: da dt = 2 ab 2 ab2 3 db dt = 3 2 ab 2 2ab2 3
 5.2.21: Sketch the nullclines and find the direction of the vector field al...
 5.2.22: Show that there is at least one solution in each of the second and ...
 5.2.23: Find the linearized system near the equilibrium points (0, 0) and (...
Solutions for Chapter 5.2: QUALITATIVE ANALYSIS
Full solutions for Differential Equations 00  4th Edition
ISBN: 9780495561989
Solutions for Chapter 5.2: QUALITATIVE ANALYSIS
Get Full SolutionsThis textbook survival guide was created for the textbook: Differential Equations 00, edition: 4. This expansive textbook survival guide covers the following chapters and their solutions. Since 23 problems in chapter 5.2: QUALITATIVE ANALYSIS have been answered, more than 16933 students have viewed full stepbystep solutions from this chapter. Chapter 5.2: QUALITATIVE ANALYSIS includes 23 full stepbystep solutions. Differential Equations 00 was written by and is associated to the ISBN: 9780495561989.

Adjacency matrix of a graph.
Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected). Adjacency matrix of a graph. Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected).

Characteristic equation det(A  AI) = O.
The n roots are the eigenvalues of A.

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

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

Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

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.

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.

Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA1 yll2 = Y T(AAT)1 Y = 1 displayed by eigshow; axis lengths ad

Fast Fourier Transform (FFT).
A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn1c can be computed with ne/2 multiplications. Revolutionary.

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!).

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.

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.

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

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.

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

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

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

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

Toeplitz matrix.
Constant down each diagonal = timeinvariant (shiftinvariant) filter.