 2.2.41: In each of 1 through 6, find the general solution, using the method...
 2.2.65: This problem gauges the relative effects of initial position and ve...
 2.2.82: In each of 1 through 10, find the general solution x 2 y+ 2x y 6y =...
 2.2.1: In each of 1 through 5, verify that y1 and y2 are solutions of the ...
 2.2.17: In each of 1 through 10, write the general solution. y y 6y = 0 2
 2.2.42: y 4y+ 3y = 2 cos(x + 3) 3
 2.2.66: Repeat the experiment of 1, except now use the critically damped, u...
 2.2.83: x 2 y+ 3x y+ y = 0 3
 2.2.2: y 16y = 0; y(0) = 12, y (0) = 3 y1(x) = e4x , y2(x) = e4x 3
 2.2.18: y 2y+ 10y = 0 3
 2.2.43: y+ 9y = 12 sec(3x)
 2.2.67: 4 through 9 explore the effects of changing the initial position or...
 2.2.84: x 2 y+ x y+ 4y = 0 4
 2.2.3: y+ 3y+ 2y = 0; y(0) = 3, y (0) = 1 y1(x) = e2x , y2(x) = ex 4.
 2.2.19: y+ 6y+ 9y = 0 4
 2.2.44: y 2y 3y = 2 sin2 (x) 5
 2.2.68: y+ 4y+ 2y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and ...
 2.2.85: x 2 y+ x y 4y = 0 5
 2.2.4: y 6y+ 13y = 0; y(0) = 1, y (0) = 1 y1(x) = e3x cos(2x), y2(x) = e3x...
 2.2.20: y 3y= 0 5
 2.2.45: y 3y+ 2y = cos(ex ) 6
 2.2.69: y+ 4y+ 4y = 0; y(0) = A, y (0) = 0; A has values 1, 3, 6, 10,4 and ...
 2.2.86: x 2 y+ x y 16y = 0 6
 2.2.5: y 2y+ 2y = 0; y(0) = 6, y (0) = 1 y1(x) = ex cos(x), y2(x) = ex sin...
 2.2.21: y+ 10y+ 26y = 0 6
 2.2.46: y 5y+ 6y = 8 sin2 (4x) I
 2.2.70: y+ 4y+ 4y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and ...
 2.2.87: x 2 y+ 3x y+ 10y = 0 7
 2.2.6: y+ 36y = x 1, Yp(x) = (x 1)/36
 2.2.22: y+ 6y 40y = 0 7
 2.2.47: In each of 7 through 16, find the general solution, using the metho...
 2.2.71: y+ 2y+ 5y = 0; y(0) = A, y (0) = 0; A has values 1, 3, 6, 10,4 and ...
 2.2.88: x 2 y+ 6x y+ 6y = 0 8
 2.2.7: y 16y = 4x 2 ; Yp(x) = x 2 /4 + 1/2
 2.2.23: y+ 3y+ 18y = 0 8
 2.2.48: y y 6y = 8e2x 9
 2.2.72: y+ 2y+ 5y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and ...
 2.2.89: x 2 y 5x y+ 58y = 0 9
 2.2.8: y+ 3y+ 2y = 15; Yp(x) = 15/2 9
 2.2.24: y+ 16y+ 64y = 0 9
 2.2.49: y 2y+ 10y = 20x 2 + 2x 8 1
 2.2.73: y+ 2y+ 5y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and ...
 2.2.90: x 2 y+ 25x y+ 144y = 0 1
 2.2.9: y 6y+ 13y = ex ; Yp(x) = 8ex 1
 2.2.25: y 14y+ 49y = 0 1
 2.2.50: y 4y+ 5y = 21e2x 1
 2.2.74: An object having a mass of 1 gram is attached to the lower end of a...
 2.2.91: x 2 y 11x y+ 35y = 0 I
 2.2.10: y 2y+ 2y = 5x 2 ; Yp(x) = 5x 2 /2 5x 4 1
 2.2.26: y 6y+ 7y = 0 I
 2.2.51: y 6y+ 8y = 3ex 1
 2.2.75: How many times can the mass pass through the equilibrium point in o...
 2.2.92: In each of 11 through 16, solve the initial value problem. x 2 y+ 5...
 2.2.11: Here is a sketch of a proof of Theorem 2.2. Fill in the details. De...
 2.2.27: In each of 11 through 20, solve the initial value problem. y+ 3y= 0...
 2.2.52: y+ 6y+ 9y = 9 cos(3x) 1
 2.2.76: How many times can the mass pass through equilibrium in critical da...
 2.2.93: x 2 y x y= 0; y(2) = 5, y (2) = 8 13
 2.2.12: Let y1(x) = x 2 and y2(x) = x 3 . Show that W(x) = x 4 . Now W(0) =...
 2.2.28: y+ 2y 3y = 0; y(0) = 6, y (0) = 2 13
 2.2.29: y 2y+ y = 0; y(1) = y (1) = 0 14
 2.2.53: y 3y+ 2y = 10 sin(x) 1
 2.2.77: In underdamped, unforced motion, what effect does the damping const...
 2.2.94: x 2 y 3x y+ 4y = 0; y(1) = 4, y (1) = 5 Co
 2.2.13: Show that y1(x) = x and y2(x) = x 2 are linearly independent soluti...
 2.2.30: y 4y+ 4y = 0; y(0) = 3, y (0) = 5 15
 2.2.54: y 4y= 8x 2 + 2e3x 1
 2.2.78: Suppose y(0) = y (0) = 0. Determine the maximum displacement of the...
 2.2.95: x 2 y+ 25x y+ 144y = 0; y(1) = 4, y (1) = 0 15
 2.2.14: Suppose y1 and y2 are solutions of equation (2.2) on (a, b) and tha...
 2.2.31: y+ y 12y = 0; y(2) = 2, y (2) = 1 16
 2.2.55: y 4y+ 13y = 3e2x 5e3x 1
 2.2.79: Consider overdamped forced motion governed by y+ 6y + 2y = 4 cos(3t...
 2.2.96: x 2 y 9x y+ 24y = 0; y(1) = 1, y (1) = 10 16
 2.2.15: Let be a solution of y+ py+ qy = 0 on an open interval I. Suppose (...
 2.2.32: y 2y 5y = 0; y(0) = 0, y (0) = 3 17
 2.2.56: y 2y+ y = 3x + 25 sin(3x) I
 2.2.80: Carry out the program of for the critically damped, forced system g...
 2.2.97: x 2 y+ x y 4y = 0; y(1) = 7, y (1) = 3 17
 2.2.16: Let y1 and y2 be distinct solutions of equation (2.2) on an open in...
 2.2.33: y 2y+ y = 0; y(1) = 12, y (1) = 5 18
 2.2.57: In each of 17 through 24, solve the initial value problem.y 4y = 7e...
 2.2.81: Carry out the program of for the underdamped, forced system governe...
 2.2.98: Here is another approach to solving an Euler equation. For x > 0, s...
 2.2.34: y 5y+ 12y = 0; y(2) = 0, y (2) = 4 19
 2.2.58: y+ 4y= 8 + 34 cos(x); y(0) = 3, y (0) = 2 19
 2.2.99: Outline a procedure for solving the Euler equation for x < 0. Hint:...
 2.2.35: y y+ 4y = 0; y(2) = 1, y (2) = 3 20
 2.2.59: y+ 8y+ 12y = ex + 7; y(0) = 1, y (0) = 0 20
 2.2.36: y+ y y = 0; y(4) = 7, y (4) = 1 21
 2.2.60: y 3y= 2e2x sin(x); y(0) = 1, y (0) = 2 21
 2.2.37: This problem illustrates how small changes in the coefficients of a...
 2.2.61: y 2y 8y = 10ex + 8e2x ; y(0) = 1, y (0) = 4 22
 2.2.38: (a) Find the solution of the initial value problem y 2y + 2 y = 0; ...
 2.2.62: y y+ y = 1; y(1) = 4, y (1) = 2 23
 2.2.39: Suppose is a solution of y+ ay + by = 0; y(x0) = A, y (x0) = B with...
 2.2.63: y y = 5 sin2 (x); y(0) = 2, y (0) = 4 2
 2.2.40: Use power series expansions to derive Eulers formula. Hint: Write e...
 2.2.64: y+ y = tan(x); y(0) = 4, y (0) = 3 2
Solutions for Chapter 2: Linear SecondOrder Equations
Full solutions for Advanced Engineering Mathematics  7th Edition
ISBN: 9781111427412
Solutions for Chapter 2: Linear SecondOrder Equations
Get Full SolutionsThis textbook survival guide was created for the textbook: Advanced Engineering Mathematics, edition: 7. Since 99 problems in chapter 2: Linear SecondOrder Equations have been answered, more than 7773 students have viewed full stepbystep solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 2: Linear SecondOrder Equations includes 99 full stepbystep solutions. Advanced Engineering Mathematics was written by and is associated to the ISBN: 9781111427412.

Affine transformation
Tv = Av + Vo = linear transformation plus shift.

Back substitution.
Upper triangular systems are solved in reverse order Xn to Xl.

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

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

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.

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.

Graph G.
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.

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.

Iterative method.
A sequence of steps intended to approach the desired solution.

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

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

Nullspace N (A)
= All solutions to Ax = O. Dimension n  r = (# columns)  rank.

Orthogonal subspaces.
Every v in V is orthogonal to every w in W.

Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.

Permutation matrix P.
There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or 1) based on the number of row exchanges to reach I.

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

Reduced row echelon form R = rref(A).
Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.

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

Singular matrix A.
A square matrix that has no inverse: det(A) = o.

Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.