18.03 Problem Set 2: Part II Solutions Part I points: 4. 4, 5. 6 , 6. 8, 7. 6. 4. (a) [2] x falls; y rises and then falls; and z rises. With � = .1, µ = .2: (b) [4] Startium obeys the natural decay equation, ˙x = −�x, with solution x = x(0)e−�t . To relate � to its half-life, solve for it in x(0)/2 = x(0)e−�tS to find � = (ln 2)/tS . Similarly, µ = (ln 2)/tM . Midium decays as well, but in each small time interval gets half the decayed Startium added: so y(t + �t) � −µy(t)�t + 21 �x(t)�t. Thus y˙ = −µy + 21 �x. Endium receives half the decayed Startium and all the decayed Midium: z˙ = 12 �x + µy. Adding these three equations gives x˙ + ˙y + ˙z = 0. (c) [4] Using x(0) = 1, we know that x = e−�t . Thus y˙ + µy = 12 �e−�t . An integrating factor is given by eµt: d (eµty) = 1 �e(µ−�)t . Integrating, eµty = 1 � e(µ−�)t + c or y = dt 2 2 µ−� 1 � e−�t +ce−µt. The initial condition is y(0) = 0, so c = −1 � : y = 1 � (e−�t −e−µt).2 µ−� 2 µ−� 2 µ−� We could solve for z in the same way, but it’s easier to calculate z = 1 − x − y = 1 + �/2−µ e−�t + �/2 e−µt µ−� µ−� (d) [4] From the differential equation for y, we know that a critical point occurs when µy = 1 �e−�t . Substitute the value for y: µ 1 � (e−�t − e−µt) = 1 �e−�t . Some algebra 2 2 µ−� 2 �t = µe−µt (µ−�)t ln µ−ln � leads to �e−, so e = µ/�, so tmax = � . µ−(e) [2] Everything gets doubled. (f) [4] If x = et then q(t) = tx˙ + 2x = tet + 2et = (t + 2)et . The associated homogeneous equation is tx˙ + 2x = 0, which is separable: dx/x = −2dt/t, so ln |x| = −2 ln |t| + c = ln(t−2) + c and x = C/t2 . So the general solution of the original equation is et + C/t2 .�5. (a) [10] and 6. (a) The rectangular expression gives the coordinates for the little pictures. Any angle may be altered by adding a multiple of 2�. 1 − i �3 + i (−1 − i)/�2 (1 + �3i)/2 (−1 + i)/�2 �2, −�/4 2, �/6 1, 5�/4 1, �/3 1, 3�/4 �i/4�2e−2e�i/6 5�i/4e�i/3e3�i/4ek�i/4(b) [8] (i) ±1 ± i; or �2e where k = 1, 3, 5, 7. (ii) −1 ± i. 6. (a) [2] above. (b) [3] ea+bi = eaebi so |ea+bi| = |ea||ebi| = ea . Since | − 2| = 2, a = ln 2. Arg(ea+bi) = b up to adding multiples of 2�. Arg(−1) = �, so b is any odd multiple of �. Answer: ln 2 + b�i, b = ±1, ±3, . . .. (c) [3] cos(4t) = Ree4it = Re((eit)4) = Re((cos t + i sin t)4). By the binomial theorem, (a+bi)4 = a4 +4a3bi−6a2b2 −4ab3i+b4, so we find cos(4t) = cos4 t−6 cos2 t sin2 t+sin4 t. (d) [8] (i) w = 2�i. The trajectory is the unit circle. (ii) w = −1. The trajectory is the positive real axis. (iii) w = −1 + 2�i. The trajectory is a spiral, spiralling in towards the origin in a counterclockwise direction and passing though 1. (iv) w = 0. The trajectory is the single point 1. 3it 7. (a) [8] �e3 + i =(�34 − i)(cos(3t) + i sin(3t)) has real part �43 cos(3t) + 14 sin(3t). Form the right triangle with sides a = 43 and b = 41 . The hypotenuse is A = 1/2 and the angle is � = �/6. �3+i = 2e�i/6 (by essentially the same triangle), so e3it =1 2 6 ,�3 + i 2 e i(3t−�/6): B = 1 , � = � and Re(Bei(3t−�)) = B cos(3t − �), so you get the same answer. (b) [5] Substituting z = we2it , e2it = w2ie2it +3we2it, so 1 = w(2i+3) or w = 1 = 3−2i .2i+3 13 3−2i 2it 3tThus a solution of the desired form is zp = 13 e . The general solution is zp + ce−. 3−2i 2it(c) [5] If x = Rez, the real part of ˙z + 3z = e2it is ˙x + 3x = cos(2t). Now zp = 13 e = 3−2i (cos(2t) + i sin(2t)) has real part xp = 1 (3 cos(2t) + 2 sin(2t)). The general solution 13 13 is then x = xp + ce−3t .MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations���� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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