# 6 a from 1 w t 2 w t 1 p t 1 p t γ βw t 1 γ βw t

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Chapter 4 / Exercise 16
Elementary and Intermediate Algebra: Algebra Within Reach
Larson
Expert Verified
6. (a) From [1], W t + 2 W t + 1 = P t + 1 P t = γ + βW t + 1 γ + βW t , and [2] follows immediately.(b) According to [2], the fraction W t + 1 /(γ + βW t ) is the same for all t . For t = 0 it is equal to W 1 /P 0 = c , and thus [3] follows. From [20.4], the solution is W t = W 0 1 (cβ) t + 1 ( 1 = cβ) (c) The equation is stable iff | | < 1. In this case, W t 1 as t → ∞ . 20.2 2. According to [4] in Example 20.5, the yearly repayment is z = 0 . 07 · 100 , 000 1 ( 1 . 07 ) 30 8058 . 64 In the first year the interest payment is 0 . 07 B = 7000, and so the principal repayment is 8058 . 64 7000 = 1058 . 64. In the last year, the interest payment is 0 . 07 b 29 8058 . 64[1 ( 1 . 07 ) 1 ] 527 . 20, and so the principal repayment is 8058 . 64 527 . 20 = 7531 . 44. 20.4 2. x t = A + B t is a solution: x t + 2 2 x t + 1 + x t = A + B (t + 2 ) 2[ A + B (t + 1 ) ] + A + B t = A + Bt + 2 B 2 A 2 Bt 2 B + A + B t = 0. Substituting t = 0 and t = 1 in x t = A + B t yields A = x 0 and A + B = x 1 , with solution A = x 0 and B = x 1 x 0 . So x t = A + B t is the general solution of the given equation. 4. x t = (A + B t) 2 t + 1 is a solution: x t + 2 4 x t + 1 + 4 x t = [ A + B (t + 2 ) ]2 t + 2 + 1 4 { [ A + B (t + 1 ) ]2 t + 1 + 1 } + 4[ (A + B t) 2 t + 1] = 4 A 2 t + 4 Bt 2 t + 8 B 2 t + 1 8 A 2 t 8 Bt 2 t 8 B 2 t 4 + 4 A 2 t + 4 Bt 2 t + 4 = 1. Substituting t = 0 and t = 1 in x t = A 2 t + B t 2 t + 1 yields A + 1 = x 0 and 2 A + 2 B + 1 = x 1 , with solution A = x 0 1 and B = 1 2 x 1 x 0 + 1 2 . So x t = A 2 t + B t 2 t + 1 is the general solution of the given equation. © Knut Sydsæter and Peter Hammond 2010
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Chapter 4 / Exercise 16
Elementary and Intermediate Algebra: Algebra Within Reach
Larson
Expert Verified
C H A P T E R 2 0 D I F F E R E N C E E Q U A T I O N S 61 20.5 2. (a) The characteristic equation m 2 + 2 m + 1 = (m + 1 ) 2 = 0 has the double root m = − 1, so the general solution of the homogeneous equation is x t = (C 1 + C 2 t)( 1 ) t . We find a particular solution by inserting u t = A 2 t . This yields A = 1, and so the general solution of the inhomogeneous equation is x t = (C 1 + C 2 t)( 1 ) t + 2 t . (b) By using the method of undetermined coefficients to determine the constants A , B , and C in the particular solution u t = A 5 t + B cos π 2 t + C sin π 2 t , we obtain A = 1 4 , B = 3 10 , and C = 1 10 . So the general solution to the given equation is x t = C 1 + C 2 2 t + 1 4 5 t + 3 10 cos π 2 t + 1 10 sin π 2 t . 4. Thecharacteristicequationis m 2 4 (ab + 1 )m + 4 a 2 b 2 = 0, withsolutions m 1 , 2 = 2 (ab + 1 ± 1 + 2 ab ) . The general solution is therefore D n = C 1 m n 1 + C 2 m n 2 . 6. Inserting x t = u t ( a/ 2 ) t into [20.21], assuming that b = 1 4 a 2 , we obtain the equation x t + 2 + ax t + 1 + 1 4 a 2 x t = u t + 2 ( a/ 2 ) t + 2 + au t + 1 ( a/ 2 ) t + 1 + 1 4 a 2 u t ( a/ 2 ) t = 1 4 a 2 ( a/ 2 ) t (u t + 2 2 u t + 1 + u t ) , which is 0 if u t + 2 2 u t + 1 + u t = 0. The general solution of this equation is u t = A + B t , so x t = u t ( a/ 2 ) t = (A + B t)( a/ 2 ) t , which is the result claimed for case 2 in the frame on page 751. 8. (a) The first two equations state that consumption and capital are proportional to the net national product in the previous period. The third equation states that net national product, Y t , is divided between con- sumption, C t , and net investment, K t K t 1 .