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18.03 Problem Set 2: Part II Solutions
Part I points: 4. 4, 5. 6 , 6. 8, 7. 6.
- (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 + (^2) 1 �x(t)�t. Thus y˙ = −μy + (^2) 1 �x. Endium receives
half the decayed Startium and all the decayed Midium: z˙ = 1 2 �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 = 1 2 �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 � 2 μ−� 2 μ−� 2 μ−�(e−�t^ −e−μt).
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 μ−� μ−�e−μt
(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 2 2 μ−� 2 �e−�t^.^ Some^ algebra
leads to �e−^ �t^ = μe−μt^ , so e( μ−�)t^ = μ/�, so t ln^ μ−ln^ � max =^ 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 = − 2 dt/t, so ln |x| = − 2 ln |t| + c =
ln(t−^2 ) + c and x = C/t^2. So the general solution of the original equation is et^ + C/t^2.
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- (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)/
3 i)/ 2 (−1 + i)/
�i/ 4
2 e −
2 e�i/^6
e^5 �i/^4 �i/ 3 e e^3 �i/^4
k�i/ 4 (b) [8] (i) ± 1 ± i; or
2 e where k = 1, 3 , 5 , 7. (ii) − 1 ± i.
- (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) = Ree 4 it = Re((e it ) 4 ) = Re((cos t + i sin t) 4 ). By the binomial theorem,
(a+bi)^4 = a^4 +4a^3 bi− 6 a^2 b^2 − 4 ab^3 i+b^4 , so we find cos(4t) = cos^4 t− 6 cos^2 t sin 2 t+sin 4 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.
3 it
- (a) [8] �
e
3 + i
− i) (cos(3t) + i sin(3t)) has real part
� 4
(^3) cos(3t) + 1 4 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
e^3 it
2 6
3 + i 2
ei(3t−�/6)^ : B = 1 , � = �
and Re(Bei(3t−�)) = B cos(3t − �), so you get the same answer.
(b) [5] Substituting z = we^2 it^ , e^2 it^ = w 2 ie^2 it^ +3we^2 it, so 1 = w(2i+3) or w = (^2) i^1 +3^ = 3 − 132 i^.
Thus a solution of the desired form is z^3 −^2 i^2 it^3 t p =^13 e^.^ The^ general^ solution^ is^ zp +^ ce
3 − 2 i 2 it (c) [5] If x = Rez, the real part of z˙ + 3z = e 2 it is x˙ + 3x = cos(2t). Now zp = 13 e = 3 − 2 i (cos(2t) + i sin(2t)) has real part xp = 1 13 13 (3^ cos(2t)^ +^2 sin(2t)).^ The^ general^ solution is then x = xp + ce − 3 t .
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