- 1.9.1: Use the theorem of equality of mixed partial derivatives to show th...
- 1.9.2: Show that the expression M(t,y)- j(il~(r,~)/ilt)dj. is a function o...
- 1.9.3: In each of 3-6 find the general solution of the given differential ...
- 1.9.4: In each of 3-6 find the general solution of the given differential ...
- 1.9.5: In each of 3-6 find the general solution of the given differential ...
- 1.9.6: In each of 3-6 find the general solution of the given differential ...
- 1.9.7: In each of 7-1 1, solve the given initial-value problem.
- 1.9.8: In each of 7-1 1, solve the given initial-value problem.
- 1.9.9: In each of 7-1 1, solve the given initial-value problem.
- 1.9.10: In each of 7-1 1, solve the given initial-value problem.
- 1.9.11: In each of 7-1 1, solve the given initial-value problem.
- 1.9.12: In each of 12-14, determine the constant a so that the equation is ...
- 1.9.13: In each of 12-14, determine the constant a so that the equation is ...
- 1.9.14: In each of 12-14, determine the constant a so that the equation is ...
- 1.9.15: Show that every separable equation of the form M(t)+ N(y)dy/dt=O is...
- 1.9.16: Find all functions f(t) suoh that the differential equation y2sint ...
- 1.9.17: Show that if ((aN/ at) - (aM/ ay))/ M = Q (y), then the differentia...
- 1.9.18: The differential equation f (t)(dy /dt) + t2 + y = 0 is known to ha...
- 1.9.19: The differential equation et secy - tany + (&/dt) = 0 has an integr...
- 1.9.20: The Bernoulli differential equation is (dy /dt) + a(t) y = b(t) y "...
Solutions for Chapter 1.9: Exact equations, and why we cannot solve very many differential equations
Full solutions for Differential Equations and Their Applications: An Introduction to Applied Mathematics | 3rd Edition
Solutions for Chapter 1.9: Exact equations, and why we cannot solve very many differential equationsGet Full Solutions
Tv = Av + Vo = linear transformation plus shift.
Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.
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.)
Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).
Column space C (A) =
space of all combinations of the columns of A.
Conjugate Gradient Method.
A sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.
Diagonal matrix D.
dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.
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 IIA-1 yll2 = Y T(AAT)-1 Y = 1 displayed by eigshow; axis lengths ad
Free columns of A.
Columns without pivots; these are combinations of earlier columns.
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!).
Set of n nodes connected pairwise by m edges. A complete graph has all n(n - 1)/2 edges between nodes. A tree has only n - 1 edges and no closed loops.
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.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.
Nullspace matrix N.
The columns of N are the n - r special solutions to As = O.
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 •
Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
Schur complement S, D - C A -} B.
Appears in block elimination on [~ g ].
Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.
Tridiagonal matrix T: tij = 0 if Ii - j I > 1.
T- 1 has rank 1 above and below diagonal.
Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.