Mathematics for Economists by Simon Blume 习题解答 answers4

2
1 ϭ y(0) ϭ k1 ϩ k2 , 0 ϭ y ˙ (0) ϭ Ϫ3k1 Ϫ 2k2 =⇒ k1 ϭ Ϫ2, k2 ϭ 3. r 2 Ϫ 6r ϩ 9 ϭ (r Ϫ 3)2 ϭ 0. y ˙ ϭ 3k1 e3t ϩ k2 e3t ϩ 3k2 te3t . 0ϭy ˙ (0) ϭ 3k1 ϩ k2 =⇒ k1 ϭ 1, k2 ϭ Ϫ3. 1 3 r 2 ϩ r ϩ 1 ϭ 0, r ϭ Ϫ Ϯ i . 2 2 k1 cos Ϫ 3 3 t ϩ k2 sin t 2 2
d) y ¨ ϩ 5˙ y ϩ 6 y ϭ 0, y ϭ k1 eϪ3t ϩ k2 eϪ2t , y ϭ Ϫ2eϪ3t ϩ 3eϪ2t . e) y ¨ Ϫ 6˙ y ϩ 9 y ϭ 0, y ϭ k1 e3t ϩ k2 te3t , 1 ϭ y(0) ϭ k1 , y ϭ e3t Ϫ 3te3t . f) y ¨ϩy ˙ ϩ y ϭ 0, y ϭ eϪt y ˙ ϭ eϪt Ϫ
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MATHEMATICS FOR ECONOMISTS
Ϫ4A Ϫ 4B ϭ 1,
Ϫ4B ϩ 4A ϭ 0 =⇒ A ϭ B ϭ Ϫ 1 8.
1 y ϭ k1 ϩ k2 eϪ2t Ϫ 1 8 sin 2t Ϫ 8 cos 2t .
e) y ¨ ϩ 4 y ϭ sin 2t . The general solution of the homogeneous equation is: y ϭ k1 cos 2t ϩ k2 sin 2t . A particular solution of the nonhomogeneous equation is: yp ϭ t (A sin 2t ϩ B cos 2t ). y ˙ p ϭ A sin 2t ϩ B cos 2t ϩ 2tA cos 2t Ϫ 2tB sin 2t, y ¨ p ϭ 4A cos 2t Ϫ 4B sin 2t Ϫ 4tA sin 2t Ϫ 4tB cos 2t, y ¨ p ϩ 4 yp ϭ 4A cos 2t Ϫ 4B sin 2t ϭ sin 2t. A ϭ 0, B ϭ Ϫ1 4. So y ϭ k1 cos 2t ϩ k2 sin 2t Ϫ (t 4) cos 2t . f) y ¨ Ϫ y ϭ et . General solution of the homogeneous equation: r 2 Ϫ 1 ϭ 0 =⇒ y ϭ k1 et ϩ k2 eϪt . Particular solution of the nonhomogeneous equation: yp ϭ Atet . y ˙ p ϭ Aet ϩ Atet , y ¨ p ϭ 2Aet ϩ Atet . y ¨ p Ϫ yp ϭ 2Aet ϭ et =⇒ A ϭ 1 2.
yϪy ˙ ϩ 2 y ϭ 0. Look for the solution y ϭ ert . 24.17 y[3] Ϫ 2¨ ert (r 3 Ϫ 2r 2 Ϫ r ϩ 2) ϭ 0. (r Ϫ 1)(r ϩ 1)(r Ϫ 2) ϭ 0 =⇒ r ϭ 1, Ϫ1, 2. y ϭ k1 et ϩ k2 eϪt ϩ k3 e2t . 24.18 y ϭ AeϪt , y ˙ ϭ ϪAeϪt , y ¨ ϭ AeϪt .
0ϭy ˙ (0) ϭ k2 Ϫ k1 .
y ϭ e (cos t ϩ sin t ).
ห้องสมุดไป่ตู้
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MATHEMATICS FOR ECONOMISTS
c) 4¨ y Ϫ 4˙ y ϩ y ϭ 0, y ϭ k1 et y(t ) ϭ et
2
4r 2 Ϫ 4r ϩ 1 ϭ (2r Ϫ 1)2 ϭ 0.
t y ˙ϭ1 2 k1 e 2
3
6r 2 Ϫ r Ϫ 1 ϭ (3r ϩ 1)(2r Ϫ 1) ϭ 0.
Ϫt y ˙ ϭ Ϫ1 3 k1 e 3 t 2 ϩ1 2 k2 e . 1 0ϭy ˙ (0) ϭ Ϫ 1 3 k1 ϩ 2 k2 . 3 t 2 ϩ2 5e .
ϩ k2 et 2 ,
1 ϭ y(0) ϭ k1 ϩ k2 , b) y ¨ ϩ 2˙ y ϩ 2 y ϭ 0,
24.21 General solution of the homogeneous equation: r 2 Ϫ r Ϫ 2 ϭ (r Ϫ 2)(r ϩ 1) ϭ 0 y ϭ k1 e2t ϩ k2 eϪt . Particular solution #1: yp1 ϭ At ϩ B, y ˙ p1 ϭ A, y ¨ p1 ϭ 0, y ¨ p1 Ϫ y ˙ p1 Ϫ 2 yp1 ϭ ϪA Ϫ 2(At ϩ B ) ϭ 6t. Ϫ 2A ϭ 6 and Ϫ A Ϫ 2B ϭ 0 =⇒ A ϭ Ϫ3, B ϭ ϩ3 2.
c1 ϭ 2, c2 ϭ 1, so y ϭ eϪt (2 cos 3t ϩ sin 3t ). y ϭ k1 cos 3t ϩ k2 sin 3t ,
y ϭ e␣t (c1 cos ␤t ϩ c2 sin ␤t ), y ˙ ϭ e␣t [(␣c1 ϩ ␤c2 ) cos ␤t ϩ (␣c2 Ϫ ␤c1 ) sin ␤t ], y ¨ ϭ e␣t [(␣ 2 Ϫ ␤ 2 )c1 ϩ 2␣␤c2 ] cos ␤t ϩ e␣t [(␣ 2 Ϫ ␤ 2 )c2 Ϫ 2␣␤c1 ] sin ␤t.
Ϫt
2 3 Ϫt So k1 ϭ 3 5 , k2 ϭ 5 ; and y ϭ 5 e
r 2 ϩ 2r ϩ 2 ϭ 0,
Ϫt
r ϭ Ϫ1 Ϯ i. So k1 ϭ k2 ϭ 1.
y ϭ e (k1 cos t ϩ k2 sin t ), 1 ϭ y(0) ϭ k1 ,
Ϫt
y ˙ ϭ e [(k2 Ϫ k1 ) cos t Ϫ (k1 ϩ k2 ) sin t ].
ANSWERS PAMPHLET
151
2 ϭ y(0) ϭ c1 ; b) y ¨ ϩ 9 y ϭ 0; 2 ϭ y(0) ϭ k1 , k1 ϭ 2, k2 ϭ 24.15
1 3.
1ϭy ˙ (0) ϭ Ϫc1 ϩ 3c2 . r 2 ϩ 9 ϭ 0, r ϭ Ϯ3i. y ˙ ϭ Ϫ3k1 sin 3t ϩ 3k2 cos 3t . 1ϭy ˙ (0) ϭ 3k2 . y ϭ 2 cos 3t ϩ 1 3 sin 3t .
y ¨ Ϫ 2˙ y Ϫ 3 y ϭ AeϪt ϩ 2AeϪt Ϫ 3AeϪt ϵ 0, but y ¨ Ϫ 2˙ y Ϫ 3 y must equal Ϫt 8e .
ANSWERS PAMPHLET
153
24.19 a) y ¨ Ϫ 2˙ y Ϫ y ϭ 7. For the general solution of the homogeneous equation: r 2 Ϫ 2r Ϫ 1 ϭ 0 =⇒ r ϭ 1 Ϯ 2. A particular solution of the nonhomogeneous equation is yp ϭ Ϫ7. So, y ϭ c1 e(1ϩ
ay ¨ ϩ by ˙ ϩ cy ϭ e␣t cos ␤t [a(␣ 2 Ϫ ␤ 2 )c1 ϩ 2a␣␤c2 ϩ b␣c1 ϩ b␤c2 ϩ cc1 ] ϩ e␣t sin ␤t [a(␣ 2 Ϫ ␤ 2 )c2 Ϫ 2a␣␤c1 ϩ b␣c2 Ϫ b␤c1 ϩ cc2 ] ϭ e␣t cos ␤t [c1 (a(␣ 2 Ϫ ␤ 2 ) ϩ b␣ ϩ c) ϩ c2 (2a␣␤ ϩ b␤ )] ϩ e␣t sin ␤t [Ϫc1 (2a␣␤ ϩ b␤ ) ϩ c2 (a(␣ 2 Ϫ ␤ 2 ) ϩ b␣ ϩ c)]. On the other hand, since r ϭ ␣ ϩ i␤ is a root of ar 2 ϩ br ϩ c ϭ 0, 0 ϭ a(␣ ϩ i␤ )2 ϩ b(␣ ϩ i␤ ) ϩ c ϭ [a(␣ 2 Ϫ ␤ 2 ) ϩ b␣ ϩ c] ϩ (2a␣␤ ϩ b␤ ). So a(␣ 2 Ϫ ␤ 2 ) ϩ b␣ ϩ c ϭ 0 and 2a␣␤ ϩ b␤ ϭ 0. But these are precisely the coefficients of c1 and c2 in the previous expression, and so that expression equals zero. 24.16 a) 6¨ yϪy ˙ Ϫ y ϭ 0, y ϭ k1 eϪt
2
k1 k2 3 k2 3 3 3 cos tϩ cos t Ϫ sin t 2 2 2 2 2 2
3 k1 3 sin t . 2 2 0ϭy ˙ (0) ϭ Ϫ k1 k2 3 ϩ =⇒ k1 ϭ 1, k2 ϭ 1 2 2 3.
1 ϭ y(0) ϭ k1 , y ϭ eϪt
2
cos
1 3 3 sin ϩ t . 2 2 3
ϩ k2 tet 2 ,
t 2 Ϫ1 2 te .
ϩ k2 et
2
t 2 ϩ1 2 k2 te .
1 ϭ y(0) ϭ k1 ,
2
0ϭy ˙ (0) ϭ 1 ⇒ k1 ϭ 1, k2 ϭ Ϫ 1 2 k1 ϩ k2 = 2. r 2 ϩ 5r ϩ 6 ϭ (r ϩ 3)(r ϩ 2) ϭ 0. y ˙ ϭ Ϫ3k1 eϪ3t Ϫ 2k2 eϪ2t .
t y ϭ k1 et ϩ k2 eϪt ϩ 1 2 te .
24.20
d d2 a( yp1 ϩ yp2 ) ϩ b( yp1 ϩ yp2 ) ϩ c( yp1 ϩ yp2 ) dt 2 dt ϭ ay ¨ p1 ϩ b y ˙ p1 ϩ cyp1 ϩ a y ¨ p2 ϩ b y ˙ p2 ϩ cyp2 ϭ g1 (t ) ϩ g2 (t ).
2)t
ϩ c2 e(1Ϫ
2)t
Ϫ 7.
b) y ¨ϩy ˙ Ϫ 2 y ϭ 6t . General solution of the homogeneous equation: r 2 ϩ r Ϫ 2 ϭ (r ϩ 2)(r Ϫ 1) ϭ 0, y ϭ k1 eϪ2t ϩ k2 et . Particular solution of the nonhomogeneous equation: y ϭ At ϩ B, y ˙ ϭA =⇒ A Ϫ 2At Ϫ 2B ϭ 6t , Ϫ2A ϭ 6, A Ϫ 2B ϭ 0 =⇒ A ϭ Ϫ3, B ϭ Ϫ3 2. So, y ϭ k1 eϪ2t ϩ k2 et Ϫ 3t Ϫ 3 2. c) y ¨Ϫy ˙ Ϫ 2 y ϭ 4eϪt . General solution of the homogeneous equation: r 2 Ϫ r Ϫ 2 ϭ (r Ϫ 2)(r ϩ 1) ϭ 0, y ϭ k1 e2t ϩ k2 eϪt . Particular solution of the nonhomogeneous equation: not AeϪt , but yp ϭ AteϪt . y ˙ p ϭ AeϪt Ϫ AteϪt , Ϫ3A ϭ 4, y ¨ p ϭ (Ϫ2A)eϪt ϩ AteϪt . yp ϭ (Ϫ4 3)teϪt . y ¨p Ϫ y ˙ p Ϫ 2 yp ϭ eϪt [Ϫ2A ϩ At Ϫ (A Ϫ At ) Ϫ 2At ] ϭ 4eϪt . A ϭ Ϫ4 3, So y ϭ k1 e2t ϩ k2 eϪt Ϫ (4 3)teϪt . d) y ¨ ϩ 2˙ y ϭ sin 2t . General solution of the homogeneous equation: r 2 ϩ 2r ϭ 0 =⇒ y ϭ k1 e0t ϩ k2 eϪ2t Particular solution of the nonhomogeneous equation: y ϭ A sin 2t ϩ B cos 2t . y ˙ ϭ ϩ2A cos 2t Ϫ 2B sin 2t, y ¨ ϭ Ϫ4A sin 2t Ϫ 4B cos 2t. y ¨ ϩ 2˙ y ϭ (Ϫ4A Ϫ 4B ) sin 2t ϩ (Ϫ4B ϩ 4A) cos 2t ϭ sin 2t.