By Abell M.L., Braselton J.P.

ISBN-10: 0123846641

ISBN-13: 9780123846648

**Read Online or Download Answers, .: to selected exercises for Introductory DE with BVP 3ed. PDF**

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**Additional resources for Answers, .: to selected exercises for Introductory DE with BVP 3ed.**

**Sample text**

3◦ F. 9. T (t) = −150 2 3 t/30 + 300; T (−30) = 75◦ F 11. Using Newton’s Law of Cooling, T (t) = (T0 − Ts )e−kt + Ts = (200 − 68)e−kt + 68. T (2) = 132e−2k + 68 = 170 ⇒ 132e−2k = 102 ⇒ e−2k = 17/22 ⇒ e−k = (17/22)1/2 . Thus, T (t) = 132 (17/22)t/2 + 68. 7 min. 12. This problem is a great class experiment if time allows. Depending upon your assumptions, the problem can be complicated. Is your model confirmed by real data? 13. u(t) = −5(9 + π2 )−1 −8π2 − (2π2 + 27)e−t/4 + 3π sin(πt/12) + 9 cos(πt/12) − 72 15.

Then T (t) = 70 · (5/7)t/15 + 30. 85 min. t= ln(5/7) = 49 50 CH A P T E R 3: Applications of First-Order Differential Equations 3. With T (t) = −35ek + 75, we find ek : T (5) = −35e5k + 75 = 50 −35e5k = −25 e5k = 5 so ek = 7 T (t) = −35 Next, we solve −35 5 t/5 + 75 7 5 7 1/5 5 7 t/5 + 75. 59 min. ln(5/7) 5. m. With this convention, T0 = 79, T (3) = 68, and Ts = 60. This means that T (t) = (79 − 60)e−kt + 60 = 19(e−k )t = 19e−kt + 60. Because T (3) = 68, T (3) = 19e−3k + 60 = 68 19e−3k = 8 so T (t) = 19 8 19 e−k = 8 19 8 19 t/3 t/3 + 60.

Y(1000) = 100e1000k = 50 so ek = 2−1/1000 and y(t) = 100 · 2−t/1000 . 71. 7. y(t) = 500ekt and y(6) = 500e6k = 600 so ek = (6/5)1/6 and y(t) = 500 · (6/5)t/6 . 8. 81. 9. y(5) = e5k = 1/2 so ek = 2−1/5 and y(t) = 2−t/5 . 92 and y(15) = 2−3 = 1/8. 11. Let H denote the half-life of the radioactive substance. Then, y(t) = eHk = 1/2 gives us t1 y(t) = (1/2)t/H . Thus, y(t1 ) = (1/2)t1 /H so ln y(t1 ) = − ln 2 and y(t2 ) = (1/2)t2 /H so H t2 ln y(t2 ) = − ln 2. Subtracting these two equations and solving for H gives the result.