Review Problems
C1 C2 C1 + C2 cos t + sin t, 2 2 C1 + C2 C1 C2 sin t + cos t. z (t) = C3 et 2 2 y (t) = C3 et + To nd constants C1 , C2 , and C3 , we use the initial conditions. So we get 0 = x(0) = C1 cos 0 + C2 sin 0 = C1 , C1 C2 C1 C2 C1 + C2 0 = y (0)
Exercises 2.2
separates if we divide by P and multiply by dt. 1 r dP = P 100 dt ln P = r t+C 100 P (t) = Kert/100 ,
where K is the initial amount of money in the savings account, K = $1000, and r % is the interest rate, r = 5. This results in P (t) = 1000
Exercises 3.5
(e) Using the equation ve = we have ve = 2gR = 2 9.81 m/sec2 (1 km/1000 m)(6370 km) 11.18 km/sec. 2gR for the escape velocity and converting meters to kilometers
(f) Similarly to (e), we nd ve = EXERCISES 3.5: 2(g/6)R = 2(9.81/6)(1/1000)(173
Exercises 4.8
and so y (t) = 3 5 cos 8t sin 8t e8t = e8t sin(8t + ), 4 4
where tan = (3/4)/(1) = 3/4 and cos = 1 < 0. Thus, = + arctan(3/4) 3.785 . The damping factor is (5/4)e8t , the quasiperiod is P = 2/8 = /4, and the quasifrequency is 1/P = 4/ . 9. S
Chapter 5
A general solution to the corresponding homogeneous equation is given in (3) on page 243 of the text: yh (t) = c1 et/2 + c2 et/6 . A particular solution has the form yp (t) C , which results 3(C ) + 2(C ) + 1 C = 1.2 4 C = 4.8 .
Therefore, yp (t
Chapter 2
where vB is driver Bs constant velocity, and k > 0 is a positive constant. Separating variables we get dvA = k dt 2 vA dvA = 2 vA k dt 1 = kt + C . vA (t)
From the initial condition we nd 1 1 =k0+C =C = vB vA (0) Thus vA (t) = The function s(t)
Chapter 3
3. In this RC circuit, R = 100 , C = 1012 F, the initial charge of the capacitor is Q = q (0) = 0 coulombs, and the applied constant voltage is V = 5 volts. Thus we can use a general equation for the charge q (t) of the capacitor derived in Exam
Chapter 4
The maximum displacement of the mass is found by determining the rst time the velocity of the mass becomes zero. Therefore, we have
y (t) = 0 = 2 + 0.2 5 e2 5t 2 5 0.1 + 2 + 0.2 5 t e2 5t ,
which gives 2 1 = . t= 2 5(2 + 0.2 5) 5(2 + 0.2 5) Thu
Review Problems
that is, x1 = x2 , x2 = x3 , 1 5 + et x1 2x2 . x3 = 3 11. This system is equivalent to x =ty y , y =x x . Next, we introduce, as additional unknowns, derivatives of x(t) and y (t): x1 (t) := x(t), x4 (t) := y (t), With new variables, the s
Exercises 3.6
(c) Capacitor. Here, with q (t) denoting the electrical charge on the capacitor, V (t) = EC (t) = and so PC = d [CEC (t)2 /2] d [CEC (t)] C dEC (t) C d [EC (t)2 ] EC (t) = = . 2EC (t) = dt 2 dt 2 dt dt 1 q (t) C q (t) = CEC (t) I (t) = d [CE
Exercises 2.3
Since driver B was 3 miles behind driver A at time t = 0, and his speed remained constant, he nished the race at time tB = (3 + 2)/vB = 5/vB . At this moment, driver A had already gone s(tB ) = 1 1 5 vB ln 2 + 1 ln (vB tB ln 2 + 1) = ln ln 2
Chapter 4
k = 30 In this case, r = 5 25 30 = 5 5i. A general solution has the form 30. y = (C1 cos 5t + C2 sin 5t)e5t . For constants C1 and C2 , we have the system y (0) = C1 cos 5t + C2 sin 5t e5t t=0 = C1 = 1 , y (0) = ( 5C2 5C1 ) cos 5t ( 5C1 + 5C2 )
Chapter 3
25. (a) From Newtons second law we have m GMm dv = . dt r2 dv gR2 , = dt r2 where g is the gravitational force of Earth, R is the radius of Earth and r is the distance between Earth and the projectile. (b) Using the equation found in part (a), l
Exercises 4.8
It follows that 2 2 = =, 4 2 2 natural frequency = =. 2 period = A general solution to (4.10), given in (4) on page 211 in the text, becomes y (t) = C1 cos t + C2 sin t = C1 cos 4t + C2 sin 4t. We nd C1 and C2 from the initial conditions. y
Chapter 2
1 a= 2 a=1 a=2
4
2
2
0
2
2
4
1 a= 2
a=1
a=2
Figure 2A: Solutions to the initial value problem y = xy 3 , y (0) = a, a 0.5, 1, and 2.
(e) For the values a = 1/2, 1, and 2 the solutions are found in (b); for a = 1, we just have to choose the negat
Chapter 5
Eliminating x by applying D to the rst equation and subtracting the second equation from it yields D D 2 + 1 D [y ] = 0 Thus on integrating 3 times we get y (t) = C3 + C2 t + C1 t2 . We substitute this solution into the rst equation of given sys
Chapter 3
Therefore, the velocity is given by v (t) = 6 Ke2t . Setting v = 1 when t = 0, we nd that 1=6K K = 5.
Thus the equation for velocity v (t) is v (t) = 6 5e2t . The limiting velocity of the sailboat under these conditions is found by letting time
Chapter 4
b=0 0. 64 = 4i. r= 2
A general solution has the form y = C1 cos 4t + C2 sin 4t. Constants C1 and C2 can be found from the initial conditions. y (0) = (C1 cos 4t + C2 sin 4t)
t=0
= C1 = 1 ,
t=0
y (0) = (4C1 sin 4t + 4C2 cos 4t) and so y (t) = cos
Exercises 2.2
dA A 120 A =3 = . dt 40 40 Separating this dierential equation and integrating yield 40 dA = dt 120 A 40 ln |120 A| = t + C Therefore,
ln |120 A| =
C t + C, where is replaced by C 40 40 120 A = Cet/40 , where C can now be positive or negati
Review Problems
We look for a particular solution of the form xp (t) = Aet . Substituting this function into the equation, we obtain Aet + 9Aet = et and so A= 1 10 xp (t) = et , 10
et . 10 To nd y , we multiply the rst equation in (5.40) by 3 and add to t
Exercises 3.4
Solving this linear equation and using the initial condition, v1 (0) = 2, we get v1 (t) = and so x1 (t) =
0 (A) t (A) (A)
65 49 4t/3 e , 8 8
65 49 4s/3 65 147 4t/3 e 1 . ds = e t 8 8 8 32
The boat A will have the velocity 5 m/sec at t = t sa
Chapter 5
Substituting x(t) into the rst equation of the system (5.41) yields y y = C1 sin t + C2 cos t. The general solution to the corresponding homogeneous equation, y y = 0, is yh (t) = C3 et . We look for a particular solution to (5.42) of the form y
Chapter 2
(a) We are given that M = 100. To nd C we must solve the equation T (0) = 40 = C + 100. This gives C = 60. Thus the equation becomes T (t) = 60e0.02180t + 100. We want to solve for t when T (t) = 90. This gives us 90 = 60e0.02180t + 100 0.0218t
Exercises 4.8
Therefore, the equation of motion is given by y (t) = 1 e0.1t sin 99.99t . 99.99
The maximum displacement to the right occurs at the rst point of local maximum of y (t). The critical points of y (t) are solutions to e0.1t y (t) = 99.99 cos 9
Chapter 5
x4 = x5 , x5 = x6 , x6 = x2 x3 . 13. With the notation used in (1) on page 264 of the text, f (x, y ) = 4 4y, g (x, y ) = 4x, and the phase plane equation (see equation (2) on page 265 of the text) can be written as g (x, y ) 4x x dy = = = . dx
Chapter 2
7. In this equation, P (x) 1 and Q(x) = e3x . Hence the integrating factor (x) = exp P (x)dx = exp (1)dx = ex .
Multiplying both sides of the equation by (x) and integrating, we obtain dy d (ex y ) x x 3x 2x e e y =e e =e = e2x dx dx 1 2x x 2x e
Exercises 4.9
Substitution into the original equation yields m A 2 cos t B 2 sin t + k (A cos t + B sin t) = F0 cos t A m 2 + k cos t + B m 2 + k sin t = F0 cos t A = F0 / (k m 2 ) , B=0 yp (t) = F0 cos t. k m 2
Therefore, a general solution to the origin
Chapter 2
At the initial point, x = /4, cos(/4) > 0 and, therefore, we can take (x) = (cos x)1 . Multiplying the standard form of the given equation by (x) gives sin x 1 dy + y = 2x cos x dx cos2 x 1 y = 2x dx = x2 + C cos x From the initial condition, we