Let be a solution of y+ py+ qy = 0 on an open interval I.

In each of Problems 1 through 5, verify that y1 and y2 are solutions of the homogeneous differential equation, calculate the Wronskian of these solutions, write the general solution, and solve the initial value problem y+ 36y = 0; y(0) = 5, y (0) = 2 y1(x) = sin(6x), y2(x) = cos(6x) 2

y 16y = 0; y(0) = 12, y (0) = 3 y1(x) = e4x , y2(x) = e4x 3

y+ 3y+ 2y = 0; y(0) = 3, y (0) = 1 y1(x) = e2x , y2(x) = ex 4.

y 6y+ 13y = 0; y(0) = 1, y (0) = 1 y1(x) = e3x cos(2x), y2(x) = e3x sin(2x) 5.

y 2y+ 2y = 0; y(0) = 6, y (0) = 1 y1(x) = ex cos(x), y2(x) = ex sin(x) In

y+ 36y = x 1, Yp(x) = (x 1)/36

y 16y = 4x 2 ; Yp(x) = x 2 /4 + 1/2

y+ 3y+ 2y = 15; Yp(x) = 15/2 9

y 6y+ 13y = ex ; Yp(x) = 8ex 1

y 2y+ 2y = 5x 2 ; Yp(x) = 5x 2 /2 5x 4 1

Here is a sketch of a proof of Theorem 2.2. Fill in the details. Denote W(y1, y2) = W for convenience. For conclusion (1), use the fact that y1 and y2 are solutions of equation (2.2) to write y 1 + py 1 + qy1 = 0 y 2 + py 2 + qy2 = 0. Multiply the first equation by y2 and the second by y1 and add. Use the resulting equation to show that W + pW = 0. Solve Solve this linear equation to verify the conclusion of part (1). To prove conclusion (2), show first that, if y2(x) = ky1(x) for all x in I, then W(x) = 0. A similar conclusion holds if y1(x) = ky2(x). Thus, linear dependence implies vanishing of the Wronskian. Conversely, suppose W(x) = 0 on I. Suppose first that y2(x) does not vanish on I. Differentiate y1/y2 to show that y2 2 d dx y1 y2  = W(x) = 0 on I. This means that y1/y2 has a zero derivative on I, hence y1/y2 = c, so y1 = cy2 and these solutions are linearly dependent. A technical argument, which we omit, covers the case that y2(x) can vanish at points of I.

Let y1(x) = x 2 and y2(x) = x 3 . Show that W(x) = x 4 . Now W(0) = 0, but W(x) = 0 if x = 0. Why does this not violate Theorem 2.3 conclusion (1)?

Show that y1(x) = x and y2(x) = x 2 are linearly independent solutions of x 2 y 2x y+ 2y = 0 on (1, 1) but that W(0) = 0. Why does this not contradict Theorem 2.3 conclusion (1)? 1

Suppose y1 and y2 are solutions of equation (2.2) on (a, b) and that p and q are continuous. Suppose y1 and y2 both have a relative extremum at some point between a and b. Show that y1 and y2 are linearly dependent.

Let be a solution of y+ py+ qy = 0 on an open interval I. Suppose (x0)=0 for some x0 in this interval. Suppose (x) is not identically zero. Prove that (x0) = 0. 16.

Let y1 and y2 be distinct solutions of equation (2.2) on an open interval I. Let x0 be in I, and suppose y1(x0) = y2(x0) = 0. Prove that y1 and y2 are linearly dependent on I. Thus, linearly independent solutions cannot share a common zero.

In each of Problems 1 through 10, write the general solution. y y 6y = 0 2

y 2y+ 10y = 0 3

y+ 6y+ 9y = 0 4

y 3y= 0 5

y+ 10y+ 26y = 0 6

y+ 6y 40y = 0 7

y+ 3y+ 18y = 0 8

y+ 16y+ 64y = 0 9

y 14y+ 49y = 0 1

y 6y+ 7y = 0 I

In each of Problems 11 through 20, solve the initial value problem. y+ 3y= 0; y(0) = 3, y (0) = 6 12

y+ 2y 3y = 0; y(0) = 6, y (0) = 2 13

y 2y+ y = 0; y(1) = y (1) = 0 14

y 4y+ 4y = 0; y(0) = 3, y (0) = 5 15

y+ y 12y = 0; y(2) = 2, y (2) = 1 16

y 2y 5y = 0; y(0) = 0, y (0) = 3 17

y 2y+ y = 0; y(1) = 12, y (1) = 5 18

y 5y+ 12y = 0; y(2) = 0, y (2) = 4 19

y y+ 4y = 0; y(2) = 1, y (2) = 3 20

y+ y y = 0; y(4) = 7, y (4) = 1 21

This problem illustrates how small changes in the coefficients of a differential equation may cause dramatic changes in the solution. (a) Find the general solution (x) of y 2y + 2 y = 0 with = 0. (b) Find the general solution (x) of y 2y + (2 2 )y = 0 with a positive constant. (c) S (c) Show that, as  0, the solution in part (b) does not approach the solution in part (a), even though the differential equation in part (b) would appear to more closely resemble that of part (a) as is chosen smaller.

(a) Find the solution of the initial value problem y 2y + 2 y = 0; y(0) = c, y (0) = d with = 0. (b) Find the solution of y 2y + (2 2 )y = 0; y(0) = c, y (0) = d. (c) Is it true that (x)(x) as 0? 23. Sup

Suppose is a solution of y+ ay + by = 0; y(x0) = A, y (x0) = B with a, b, A, and B as given numbers and a and b positive. Show that lim x (x) = 0. 24

Use power series expansions to derive Eulers formula. Hint: Write ex = n=0 1 n! x n , sin(x) = n=0 (1)n (2n + 1)! x 2n+1 , and cos(x) = n=0 (1)n (2n)! x 2n . Let x = i with real, and use the fact that i 2n = (1) n and i 2n+1 = (1) n i. for every positive integer n

In each of Problems 1 through 6, find the general solution, using the method of variation of parameters for a particular solution. y+ y = tan(x)

y 4y+ 3y = 2 cos(x + 3) 3

y+ 9y = 12 sec(3x)

y 2y 3y = 2 sin2 (x) 5

y 3y+ 2y = cos(ex ) 6

y 5y+ 6y = 8 sin2 (4x) I

In each of Problems 7 through 16, find the general solution, using the method of undetermined coefficients for a particular solution. y y 2y = 2x 2 + 5 8

y y 6y = 8e2x 9

y 2y+ 10y = 20x 2 + 2x 8 1

y 4y+ 5y = 21e2x 1

y 6y+ 8y = 3ex 1

y+ 6y+ 9y = 9 cos(3x) 1

y 3y+ 2y = 10 sin(x) 1

y 4y= 8x 2 + 2e3x 1

y 4y+ 13y = 3e2x 5e3x 1

y 2y+ y = 3x + 25 sin(3x) I

In each of Problems 17 through 24, solve the initial value problem.y 4y = 7e2x + x; y(0) = 1, y (0) = 3 1

y+ 4y= 8 + 34 cos(x); y(0) = 3, y (0) = 2 19

y+ 8y+ 12y = ex + 7; y(0) = 1, y (0) = 0 20

y 3y= 2e2x sin(x); y(0) = 1, y (0) = 2 21

y 2y 8y = 10ex + 8e2x ; y(0) = 1, y (0) = 4 22

y y+ y = 1; y(1) = 4, y (1) = 2 23

y y = 5 sin2 (x); y(0) = 2, y (0) = 4 2

y+ y = tan(x); y(0) = 4, y (0) = 3 2

This problem gauges the relative effects of initial position and velocity on the motion in the unforced, overdamped case. Solve the initial value problems y+ 4y + 2y = 0; y(0) = 5, y (0) = 0 and y+ 4y + 2y = 0; y(0) = 0, y (0) = 5. Graph the solutions on the same set of axes. 2. Rep

Repeat the experiment of Problem 1, except now use the critically damped, unforced equation y+ 4y + 4y = 0.

Problems 4 through 9 explore the effects of changing the initial position or initial velocity on the motion of the object. In each, use the same set of axes to graph the solution of the initial value problem for the given values of A and observe the effect that these changes cause in the solution.y+ 4y+ 2y = 0; y(0) = A, y (0) = 0; A has values 1, 3, 6, 10,4 and 7. 5

y+ 4y+ 2y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and 7. 6.

y+ 4y+ 4y = 0; y(0) = A, y (0) = 0; A has values 1, 3, 6, 10,4 and 7. 7

y+ 4y+ 4y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and 7. 8.

y+ 2y+ 5y = 0; y(0) = A, y (0) = 0; A has values 1, 3, 6, 10,4 and 7. 9.

y+ 2y+ 5y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and 7. 10

y+ 2y+ 5y = 0; y(0) = 0, y (0) = A; A has values 1, 3, 6, 10,4 and 7. 10

An object having a mass of 1 gram is attached to the lower end of a spring having a modulus of 29 dynes per centimeter. The mass in turn is adhered to a dashpot that imposes a damping force of 10v dynes, where v(t) is the velocity of the mass at time t in centimeters per second. Determine the motion of the mass if it is pulled down 3 centimeters from equilibrium and then struck upward with a blow sufficient to impart a velocity of 1 centimeter per second. Graph the solution. Solve the problem when the initial velocity is (in turn) 2, 4, 7, and 12 centimeters per second. Graph these solutions on the same axes to visualize the influence of the initial velocity on the motion.

How many times can the mass pass through the equilibrium point in overdamped motion? What condition can be placed on the initial displacement to ensure that it never passes through equilibrium?

How many times can the mass pass through equilibrium in critical damping? What condition can be placed on y(0) to ensure that the mass never passes through the equilibrium point? How does the initial velocity influence whether the mass passes through the equilibrium point?

In underdamped, unforced motion, what effect does the damping constant have on the frequency of the oscillations?

Suppose y(0) = y (0) = 0. Determine the maximum displacement of the mass in critically damped, unforced motion. Show that the time at which this maximum occurs is independent of the initial displacement.

Consider overdamped forced motion governed by y+ 6y + 2y = 4 cos(3t). (a) Find the solution satisfying y(0) = 6, y= 0. (b) Find the solution satisfying y(0) = 0, y (0) = 6. Graph these solutions on the same set of axes to compare the effects of initial displacement and velocity on the motion. Cop

Carry out the program of Problem 15 for the critically damped, forced system governed by y+ 4y + 4y = 4 cos(3t). 17

Carry out the program of Problem 15 for the underdamped, forced system governed by y+ y + 3y = 4 cos(3t). 2

In each of Problems 1 through 10, find the general solution x 2 y+ 2x y 6y = 0 2

x 2 y+ 3x y+ y = 0 3

x 2 y+ x y+ 4y = 0 4

x 2 y+ x y 4y = 0 5

x 2 y+ x y 16y = 0 6

x 2 y+ 3x y+ 10y = 0 7

x 2 y+ 6x y+ 6y = 0 8

x 2 y 5x y+ 58y = 0 9

x 2 y+ 25x y+ 144y = 0 1

x 2 y 11x y+ 35y = 0 I

In each of Problems 11 through 16, solve the initial value problem. x 2 y+ 5x y 21y = 0; y(2) = 1, y (2) = 0 12

x 2 y x y= 0; y(2) = 5, y (2) = 8 13

x 2 y 3x y+ 4y = 0; y(1) = 4, y (1) = 5 Co

x 2 y+ 25x y+ 144y = 0; y(1) = 4, y (1) = 0 15

x 2 y 9x y+ 24y = 0; y(1) = 1, y (1) = 10 16

x 2 y+ x y 4y = 0; y(1) = 7, y (1) = 3 17

Here is another approach to solving an Euler equation. For x > 0, substitute y = xr into the differential equation to obtain a quadratic equation for r. Roots of this quadratic equation yield solutions y = xr . Use this approach to solve the Euler equations of Examples 2.22, 2.22, and 2.23.

Outline a procedure for solving the Euler equation for x < 0. Hint: Let t = ln |x| in this case.