� � � � � � 18.03 Hour Exam I Solutions: February 24, 2010 1. (a) x(t) = number of rats at time t; t measured in years. x˙ = kx. So x(t) = x(0)ekt . ex(0) = x(1) = x(0)ekt implies k = 1. xx (b) x˙ = k 1 −Rx = 1 −1000 x. x (c) x˙ = 1 −Rx − a. The pest control peo ple hope for an equilibrium at x = 43 R. x˙ = 0 at equilib-rium, so a = (1 −3 )3 R = 3 R = 375. 4 4 16 2. (a) The phase line shows unsta ble critical points at x = −2 and x = 1 and a stable critical point at x = 0. The arrows of time are directed up above 1 and between −2 and 0, and down between 0 and 1 and below −2. (b) There are seven basic solution types: three equilibria; a solution rising above x = 1, a solution falling from 1 towards 0, a solution rising from −2 towards 0, and a solution falling away fr om −2. (f) True. After the solution crosses the nullcline, it is “inside” the parabola and its derivative is p ositive. If it were to c ross the nullcline again it would have to c ross the upper bra nch, from below. But the slope of the nullcline is positive, while at the moment of crossing the slope of the s olution would have to be zero. So it does not cross again; it stays below the upper branch of the nullcline, which has equation y = √x. k 0 3. (a) 1 2 3 xk 0 1/2 1 3/2 yk 1 3/2 5/2 17/4 mk = xk + yk 1 4 7/2 hmk 1/2 1 Ans: 17/4. 7/4 c + sin t c(b) The equa tion is d (tx) = cos t , so tx = sin t + c and x = . 1 = x(π) = so c = π and dt πt π + sin t x =. t 1 3−2i 3 24. (a) 3+2i = 32 +22 : a = 13 , b = −13 . π(b) r = |1 − i| = √2. θ = Arg(1 − i) = −4 . π 8π(c) |1 − i| = √2 and Arg(1 − i) = 4 , so |(1 − i)8| = (√2)8 = 16 and Arg((1 − i)8) = 4 = 2π, so (1 − i)8 = 16: a = 16, b = 0. (d) If (a+bi)3 = −1 then a+bi3 = (a+bi)3= = 1 so a+bi = 1, and 3Arg(a+bi) = Arg(−1) = π| | π | |5π |−1| | |(or 3π or 5π) so Arg (a + bi) = 3 or π or 3 . The first is the one with positive imaginar y part, so √ a = cos π 3 = 12 , b = sin π 3 = 23 . (e) eln 2+iπ = eln 2eiπ = 2(−1) = −2: a = −2, b = 0. π(f) A, φ a re the polar coordinates of (a, b) = (2, −2): A = 2√2, φ = −4 5. (a) Try x = Ae2t, so that ˙x = A2e2t and e2t = A2e2t + 3Ae2t = 5Ae2t so A = 15 : xp = 15 e2t . Thetransient is ce−3t, so x = 15 e2t + ce−3t is a valid solution for any c as well. (b) 1 = x(0) = 51 + c implies c = 54 : this pa rticular solution is x = 51 e2t + 54 e−3t (c) z˙ + 3z = e2it . Ae2itA2ie2it 2it 2it 1(d) Try z = : z˙ = , so e = z˙ + 3z = A(3 + 2i)e . This gives A = 3+2i and solution 12it 3−2i 3 2 zp = 3+2ie = 13 (cos(2t) + i sin(2t)), which has real par t xp = 13 cos(2t) + 13 sin(2t).MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations���� Spring 2010 For information about citing these materials or our Terms of Use, visit:
View Full Document
Unlocking...