G(s) = 5 s2+12 4s s2+8

In each of Problems 1 through 5, use Table 3.1 to determine the Laplace transform of the function. f (t) = 3t cos(2t)

g(t) = e4t sin(8t)

h(t) = 14t sin(7t)

w(t) = cos(3t) cos(7t)

k(t) = 5t 2 e4t + sin(3t)

In each of Problems 6 through 10, use Table 3.1 to determine the inverse Laplace transform of the function.R(s) = 7 s29

Q(s) = s s2+64

G(s) = 5 s2+12 4s s2+8

P(s) = 1 s+42 1 (s+3)4

F(s) = 5s (s2+1)2

Show that L[ f ](s) =  n=0 (n+1)T nT est f (t) dt.

Show that (n+1)T nT est f (t) dt = ensT T 0 est f (t) dt.

From Problems 11 and 12, show that L[ f ](s) =   n=0 ensTT 0 est f (t) dt.

Recall the geometric series  n=0 r n = 1 1 r for |r| < 1. With this and the result of Problem 13, show that L[ f ](s) = 1 1 esT T 0 est f (t) dt. In each of Problems 15 through 22, a periodic function is given (sometimes by a graph). Use the result of Problem 14 to compute its Laplace transform.

f has period of 6, and f (t) =  5 for 0 < t 3, 0 for 3 < t 6

f (t) = |E sin(t)| with E and positive numbers.

f has the graph of Figure 3.1.

f has the graph of Figure 3.2.

f has the graph of Figure 3.3.

f has the graph of Figure 3.4.

f has the graph of Figure 3.5

f has the graph of Figure 3.6

In each of Problems 1 through 10, use the Laplace transform to solve the initial value problem. y+ 4y = 1; y(0) = 3

y 9y = t; y(0) = 5

y+ 4y = cos(t); y(0) = 0

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

y 2y = 1 t; y(0) = 4

y+ y = 1; y(0) = 6, y (0) = 0

y 4y+ 4y = cos(t); y(0) = 1, y (0) = 1

y+ 9y = t 2 ; y(0) = y (0) = 0

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

y 5y+ 6y = et ; y(0) = 0, y (0) = 2

Prove Theorem 3.1. Hint: Write L[ f  ](s) = 0 est f  (t) dt and integrate by parts.

Derive equation (3.3). Hint: Integrate by parts twice.

In each of Problems 1 through 15, find the Laplace transform of the function(t 3 3t + 2)e2t

e3t (t 2)

f (t) =  1 for 0 t < 7 cos(t) for t 7

e4t (t cos(t))

f (t) =  t for 0 t < 3 1 3t for t 3

f (t) =  2t sin(t) for 0 t < 0 for t

et (1 t 2 + sin(t))

f (t) =  t 2 for 0 t < 2 1 t 3t 2 for t 2

f (t) =  cos(t) for 0 t < 2 2 sin(t) for t 2

f (t) = 4 for 0 t < 1 0 for 1 t < 3

tet cos(3t)

et (1 cosh(t))

f (t) =  t 2 for 0 t < 16 1 for t 16

f (t) =  1 cos(2t) for 0 t < 3 0 for t 3

e5t (t 4 + 2t 2 + t)

In each of Problems 16 through 25, find the inverse Laplace transform. 1 s2 + 4s + 12

1 s2 + 4s + 12

1 s2 4s + 5

e5s /s3

e2s s2 + 9

3 s + 2 e4s

s2 + 6s + 7

s 4 s2 8s + 10

s + 2 s2 + 6s + 1

1 s 5 es

1 s(s2 + 16) e

In each of Problems 27 through 32, solve the initial value problem. y+ 4y = f (t); y(0) = 1, y (0) = 0, with f (t) =  0 for 0 t < 4 3 for t 4

y 2y 3y = f (t); y(0) = 1, y (0) = 0, with f (t) =  0 for 0 t < 4 12 for t 4

y 8y = g(t); y(0) = y (0) = y(0) = 0, with g(t) =  0 for 0 t < 6 2 for t 6

y+ 5y+ 6y = f (t); y(0) = y (0) = 0, with f (t) =  2 for 0 t < 3 0 for t 3

y y+ 4y 4y = 0; y(0) = y (0) = 0, y(0) = 1, with f (t) =  1 for 0 t < 5 2 for t 5

y 4y+ 4y = f (t); y(0) = 2, y (0) = 1, with f (t) =  t for 0 t < 3 t + 2 for t 3

Determine the output voltage in the circuit of Figure 3.18, assuming that at time zero the capacitor is charged to a potential of 5 volts and the switch is opened at time zero and closed 5 seconds later. Graph this output.

Determine the output voltage in the RL circuit of Figure 3.20 if the current is initially zero and E(t) =  0 for 0 t < 5 2 for t 5. Graph this output function

Solve for the current in the RL circuit of Problem 34 if the current is initially zero and E(t) =  k for 0 t < 5 0 for t 5.

Show that Heavisides formula can be written L1 [F](t) = n j=1 p(aj) q (aj) ea j t . Hint: Write (s aj) p(s) q(s) = p(s) (q(s) q(aj))/(s aj) .

In each of Problems 1 through 8, use the convolution theorem to help compute the inverse Laplace transform of the function. Wherever they occur, a and b are positive constants. 1 (s2 + 4)(s2 4)

s2 + 16 e2s

s (s2 + a2)(s2 + b2)

s2 (s 3)(s2 + 5)

1 s(s2 + a2)2

1 s4(s 5)

1 s(s + 2) e4s

s2(s2 + 5)

y 5y+ 6y = f (t); y(0) = y (0) = 0

y+ 10y+ 24y = f (t); y(0) = 1, y (0) = 0

y 8y+ 12y = f (t); y(0) = 3, y (0) = 2

y 4y 5y = f (t); y(0) = 2, y (0) = 1

y+ 9y = f (t); y(0) = 1, y (0) = 1

y k2 y = f (t); y(0) = 2, y (0) = 4

y(3) y 4y+ 4y = f (t); y(0) = y (0) = 1, y(0) = 0

y(4) 11y+ 18y = f (t); y(0) = y (0) = y(0) = y(3) (0) = 0

In each of Problems 17 through 23, solve the integral equation. y(4) 11y+ 18y = f (t); y(0) = y (0) = y(0) = y(3) (0) = 0

f (t) = t + t 0 f (t )sin( )d

f (t) = et + t 0 f (t )d

f (t) = 1 + t 2 t 0 f (t )sin( )d

f (t) = 3 + t 0 f ( ) cos(2(t ))d

f (t) = cos(t) + e2t t 0 f ( )e2 d

Solve for the replacement function r(t) if f (t) = A, constant, and m(t) = ekt with k a positive constant. Graph r(t).

Solve for the replacement function r(t) if f (t) = A + Bt and m(t) = ekt . Graph r(t).

Solve for the replacement function r(t) if f (t) = A + Bt + Ct 2 and m(t) = ekt . Graph r(t).

Prove the convolution theorem. Hint: First write F(s)G(s) = 0 F(s)es g( ) d.Show that F(s)G(s) = 0 L[H(t ) f (t )](s)g( ) d. Use the definitions of the Heaviside function and of the transform to obtain F(s)G(s) = 0 est g( ) f (t ) d. Reverse the order of integration to obtain F(s)G(s) = 0 t 0 est g( ) f (t ) d dt = 0 est( f g)(t) dt. From this, show that L[ f g](s) = F(s)G(s). 3.5

In each of Problems 1 through 5, solve the initial value problem and graph the solution. y+5y+6y =3(t 2)4(t 5); y(0)= y (0)=0

y 4y+ 13y = 4(t 3); y(0) = y (0) = 0

y+ 4y+ 5y+ 2y = 6(t); y(0) = y (0) = y(0) = 0

y+ 16y= 12(t 5/8); y(0) = 3, y (0) = 0

y+ 5y+ 6y = B(t); y(0) = 3, y (0) = 0

An object of mass m is attached to the lower end of a spring of modulus k. Assume that there is no damping. Derive and solve an equation of motion for the object, assuming that at time zero it is pushed down from the equilibrium position with an initial velocity v0. With what momentum does the object leave the equilibrium position?

Suppose, in the setting of Problem 6, the object is struck a downward blow of magnitude mv0 at time 0. How does the position of this object compare with that of the object in Problem 6 at any positive time t?

A 2 pound weight is attached to the lower end of a spring, stretching it 8/3 inches. The weight is allowed to come to rest in the equilibrium position. At some later time, which we call time 0, the weight is struck a downward blow of magnitude 1/4 pound (an impulse). Assume no damping in the system. Determine the velocity with which the weight leaves the equilibrium position as well as the frequency and magnitude of the oscillations.

Prove the filtering property of the delta function (Theorem 3.6). Hint: Replace (t a) with lim 0 1  (H(t a ) H(t a)) in the integral and interchange the limit and the integral.

In each of Problems 1 through 11, use the Laplace transform to solve the initial value problem x 2y= 1, x+ y x = 0; x(0) = y(0) = 0

2x 3y + y= 0, x+ y= t; x(0) = y(0) = 0

x+ 2y y = 1, 2x+ y = 0; x(0) = y(0) = 0

x+ y x = cos(t), x+ 2y= 0; x(0) = y(0) = 0

3x y = 2t, x+ y y = 0; x(0) = y(0) = 0

x+ 4y y = 0, x+ 2y = et ; x(0) = y(0) = 0

x+ 2x y= 0, x+ y + x = t 2 ; x(0) = y(0) = 0

x+ 4x y = 0, x+ y= t; x(0) = y(0) = 0

x+ y+ x y = 0, x+ 2y+ x = 1; x(0) = y(0) = 0

x+ 2y x = 0, 4x+ 3y+ y = 6; x(0) = y(0) = 0

y 1 2y 2 + 3y1 = 0 y1 4y 2 + 3y 3 = t y1 2y 2 + 3y 3 = 1 y1(0) = y2(0) = y3(0)= 0

Solve for the currents in the circuit of Figure 3.28 assuming that the currents and charges are initially zero and that E(t) = 2H(t 4) H(t 5). E(t) 2 3 1 4

Solve for the currents in the circuit of Figure 3.28 if the currents and charges are initially zero and E(t) = 1 H(t 4)sin(2(t 4))

Solve for the displacement functions of the masses in the system of Figure 3.29. Neglect damping and assume zero initial displacements and velocities and external forces f1(t) = f2(t) = 0. k1 = m1 = 1 k2 = 2 m2 = 1 k3 = 3

Solve for the displacement functions in the system of Figure 3.29 if f1(t) = 1 H(t 2), f2(t) = 0 and the initial displacements and velocities are zero.

Consider the system of Figure 3.30. Let M be subjected to a periodic driving force f (t) = A sin(t). The masses are initially at rest in the equilibrium position. M k m 2 k1 y1 y2 FIGURE 3.30 Mass/spring system in Problem 16, Section 3.6. (a) Derive and solve the initial value problem for the displacement functions for the masses. (b) Show that, if m and k2 are chosen so that = k2/m, then the mass m cancels the forced vibrations of M. In this case, we call m a vibration absorber.

Two objects of masses m1 and m2 are attached to opposite ends of a spring having spring constant k (Figure 3.31). The entire apparatus is placed on a highly varnished table. Show that, if the spring is stretched and released from rest, the masses oscillate with respect to each other with period 2 m1m2 k(m1 + m2) . m1 m2

Solve for the currents in the circuit of Figure 3.32 if E(t) = 5H(t 2) and the initial currents are zero. E(t) 20 H 30 H 10 10 i2 i1

Solve for the currents in the circuit of Figure 3.32 if E(t) = 5(t 1).

Two tanks are connected by a series of pipes as shown in Figure 3.33. Tank 1 initially contains 60 gallons of brine in which 11 pounds of salt are dissolved. Tank 2 initially contains 7 pounds of salt dissolved in 18 gallons of brine. Beginning at time zero, a mixture containing 1/6 pound of salt for each gallon of water is pumped into tank 1 at the rate of 2 gallons per minute, while salt water solutions are interchanged between the two tanks and also flow out of tank 2 at the rates shown in the diagram. Four minutes after time zero, salt is poured into tank 2 at the rate of 11 pounds per minute for a period of 2 minutes Determine the amount of salt in each tank for any time t 0

Two tanks are connected by a series of pipes as shown in Figure 3.34. Tank 1 initially contains 200 gallons of brine in which 10 pounds of salt are dissolved. Tank 2 initially contains 5 pounds of salt dissolved in 100 gallons of water. Beginning at time zero, pure water is pumped into tank 1 at the rate of 3 gallons per minute, while brine solutions are interchanged between the tanks at the rates shown in the diagram. Three minutes after time zero, 5 pounds of salt are dumped into tank 2. Determine the amount of salt in each tank for any time t 0

t 2 y 2y = 2 Hint: First set u = 1/t

y+ 4t y 4y = 0; y(0) = 0, y (0) = 7

y 16t y+ 32y = 0; y(0) = y (0) = 0

y+ 8t y 8y = 0; y(0) = 0, y (0) = 4

t y+ (t 1)y+ y = 0; y(0) = 0

y+ 2t y 4y = 6; y(0) = y (0) = 0

y+ 8t y= 0; y(0) = 4, y (0) = 0

y 4t y+ 4y = 0; y(0) = 0, y (0) = 10

y 8t y+ 16y = 3; y(0) = y (0) = 0

(1 t)y+ t y y = 0; y(0) = 3, y (0) = 1

Review the derivation of the solution of Bessels equation of order n for n a positive integer. Are any steps taken that would prevent n being an arbitrary positive number, not necessarily an integer? Could n be negative?