کتاب حل المسائل ریاضی جیمز استوارت دانشگاهی

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404 O CHAPTER 9 DIFFERENTIAL EQUATIONS

15. (a) Using (4) and (7) in Section 9.4, we see that for a = 0.1P

with P(0) = 100, we have k = 0.1,

2000

K = 2000, P. = 100, and A = 2000 - 100 = 19. Thus, the solution of the initial-value problem is

100 2000

P(t)=1+19e-0.1

E and P(20) = + 19e-2

2000

1o | 09

-1 =

2000 (b) P = 1200 - 1200 =- 1200 F 1200 = 100 + 1 + 19e1 ==19e-0.lt 1+19e-0.1t

1200 e 0.1* = (3)/19 = -0.lt = Int = t=-10 In 33.5. 17. (a * * L-= = k(L. --L) = = | dt = -In|L. - L| = k + C =

In |L- L| = -kt –C = |L. - L| =e**-C = L. – L = Aekt = L= L- Ae-**

At t = 0, L= L(0) = L. - A = A = L. - L(0) + L(t) = L. - [L. - L(0)]e-et(b) Loo = 53 cm, L(0) = 10 cm, and k = 0.2 = L(t) = 53 - (53 - 10)e=0.2* - 53 - 43e0.2.

19. Let P represent the population and I the number of infected people. The rate of spread dl/dt is jointly proportional to I and

to P-I, so for some constant k,

= kl(P – I) = (kP

From Equation 9.4.7 with K = P and k replaced by

IP kP, we have I(t) = A-kPEI, + (P-I,)e-kPa

Now, measuring t in days, we substitute t = 7, P = 5000, IG = 160 and I(7) = 1200 to find k: 160 - 5000

2000 160 + (5000 - 160)e-5000-7-& F &= 1En

160 + 4840-35,000 = 480+14,520e-35,000 – 2000

1200 ==

حل المسائل ریاضی جیمز استوارت

25. *? =y+1 2° = 1(y+1). This is an equation of a parabola with 4p = 1, so p=. The vertex is (0, -1) and the

focus is (0,-4).

27. *? = 4y – 24? - ? + 2y - 4y = 0 + x + 2(y? – 2y+1) = 2 + 2° + 2(y – 1) = 2 +

**+ (Y-1)-1. This is an equation of an ellipse with vertices at (vā, 1). The foci are at (+V2 -1, 1) = (+1, 1).

29. y’ + 2y = 4x’ +3 + y + 2y+1= 4x + 4 + (y+1)? – 41° = 4 + (y-r =1. This is an equation

of a hyperbola with vertices (0, -1+ 2) = (0, 1) and (0, -3). The foci are at (0, -1+V4+1) = (0, -12V5).

31. The parabola with vertex (0,0) and focus (0, -2) opens downward and has p= -2, so its equation is zo = 4py = -8y.
32. The distance from the focus (-4,0) to the directrix x = 2 is 2- (-4) = 6, so the distance from the focus to the vertex is

(6) = 3 and the vertex is (-1,0). Since the focus is to the left of the vertex, p = -3. An equation is y = 4p(x+1) =

y = -12(x+1).

35. A parabola with vertical axis and vertex (2, 3) has equation y–3 = a(x - 2)o. Since it passes through (1,5), we have

5-3 = a(1-2)2 + a = 2, so an equation is y-3 = 2(x - 2)?. 37. The ellipse with foci (+2,0) and vertices (+5,0) has center (0,0) and a horizontal major axis, with a = 5 and c = 2,

so b? -a? -c = 25 – 4 = 21. An equation is

39. Since the vertices are (0,0) and (0,8), the ellipse has center (0,4) with a vertical axis and a = 4. The foci at (0,2) and (0,6)

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