Solutions to Differential Equations: Examples and Applications, Exams of Differential Equations

Download Solutions to Differential Equations: Examples and Applications and more Exams Differential Equations in PDF only on Docsity! TEST 1, Introduction to Diff. Equations - Spring 2012 - CRN 14344 Name:............................................................................................................ Date: Thursday, February 16 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 1. [15 pts] . If y(t) = 3et 2 is known to be the solution of the initial value problem y′ + p(t)y = 0, y(0) = y0 what must the function p(t) and the constant y0 be? Sol. Since y(t) = 3et 2 satisfies the initial value problem we have (3et 2 )′ + p(t) · 3et 2 = 0, and y0 = 3e 02 then, 6tet 2 + p(t) · 3et 2 = 0, and y0 = 3 thus, ( 6t+ 3p(t) ) et 2 = 0 Since et 2 is never zero, p(t) must be −2t  2. [15 pts] Solve the following I.V.P. by using an integration factor. y′ + 2ty = 3t, y(0) = 1. Sol. Integration Factor: e ∫ 2t dt = et 2 . Then et 2 y′ + 2tet 2 y = 3tet 2 is precisely Dt(e t2y) = 3 2 (2tet 2 ) Integrating, we have et 2 y = 3 2 et 2 + C whence y = 3 2 + Ce−t 2 Since 1 = y(0) = 3 2 + C, then C = −1/2. Therefore, y(t) = 3 2 − 1 2 e−t 2  1 3. A tank is initially filled with 100 gallons of fresh water. At time t = 0, we begin filling the tank with salt water which has a concentration of 1lb of salt per gallon of water at a rate of 1 gallon per minute. Also at time t = 0, a leak in the tank develops. Water leaks out at the rate of ro = 2 gal per minute. a)[4 pts] Deduce that the total volume V (t) of the solution in the tank at time t is given by V (t) = 100− t. V (t) = V (0) + ∫ t 0 ri − ro dt = 100 + ∫ t 0 1− 2 dt = 100− t  b)[12 pts] Find Q(t), the amount of salt in the tank after t minutes. It is clear that ri = 1, ro = 2, ci = 1, and Q(0) = 0. Recall that co = Q V . By concentration law, we have Q′(t) = (1)(1)− (2)Q(t) V (t) That is, Q′ + 2 100− t Q = 1, with Q(0) = 0. This is in standard form and p(t) = 2(100− t)−1, then ∫ p(t) dt = −2 ln(100− t) = ln((100− t)−2) Thus, the integration factor is eln((100−t) −2) = (100− t)−2. Multiplying we have (100− t)−2Q′ + 2(100− t)−3Q = (100− t)−2 Integrating, (100− t)−2Q = (100− t)−1 +C. Since 0 = Q(0), then 0 = (100− 0)−1 +C and so C = −1/100. Therefore, Q(t) = (100− t)− 1 100 (100− t)2  c)[4 pts] At what time does Q(t) reach its maximum? Note that Q′(t) + 2 100− t Q(t) = 1. If Q′(t) = 0, then 2 100− t Q(t) = 1. Thus Q(t) (100− t) = 1 2 and so 1− 1 100 (100− t) = 1 2 whence t/100 = 1/2 and therefore t = 50 min.  2