By Stephen W. Goode and Scott A. Annin

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Extra resources for Differential Equations and Linear Algebra Stephen W. Goode and Scott A. Annin SOLN MANUAL

Example text

5, r1 = 6, and r2 = 4. Then ∆V = r1 ∆t − r2 ∆t =⇒ = 2 =⇒ dt dA 4A dA 2A d V (t) = 2(t + 50) since V (0) = 100. Then + = 3 =⇒ + = 3 =⇒ [(t + 50)2 A] = dt 2(t + 50) dt t + 50 dt 125000 3(t+50)2 =⇒ (t+50)2 A = (t+50)3 +c but A(0) = 100 so c = 125000 and therefore A(t) = t+50+ . (t + 50)2 The tank is full when V (t) = 200, that is when 2(t+50) = 200 so that t = 50 min. Therefore the concentration 9 A(50) just before the tank overflows is: = g/L. V (50) 16 5. 2. V (5) dV (a) ∆V = r1 ∆t − r2 ∆t =⇒ = 1 =⇒ V (t) = t + 10 since V (0) = 10.

10. TRUE. Since P (t) = kP , then P (t) = kP (t) = k 2 P > 0 for all t. Therefore, the concavity is always positive, and does not change, regardless of the initial population. Problems: dP 1. = kP =⇒ P (t) = P0 ekt . Since P (0) = 10, then P = 10ekt . Since P (3) = 20, then 2 = e3k =⇒ k = dt ln 2 . Thus P (t) = 10e(t/3) ln 3 . Therefore, P (24) = 10e(24/3) ln 3 = 10 · 28 = 2560 bacteria. 3 2. Using P (t) = P0 ekt we obtain P (10) = 5000 =⇒ 5000 = p0 e10k and P (12) = 6000 =⇒ 6000 = P0 e12k which implies that e2k = 56 =⇒ k = 12 ln 65 .

Integrating we obtain dx dx 2 2 u = sin x + c, so that y(x) = csc x(sin x + c). 31. The associated homogeneous equation is dy 1 − y = 0, with solution yH = cx. We determine the dx x dy du function u(x) such that y(x) = xu(x) is a solution of the given differential equation. We have = x + u. dx dx dy 1 du Substituting into − y = x ln x and simplifying yields = ln x, so that u = x ln x − x + c. Consequently, dx x dx 32. The associated homogeneous equation is 50 y(x) = x(x ln x − x + c). Problems 33 - 39 are easily solved using a differential equation solver such as the dsolve package in Maple.