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314 CHAPTER 10. TARGET-ZONE MODELSlemmads(t) = dG[f(t)]= G 0 [f(t)]df (t)+ σ22 G00 [f(t)]dt= G 0 [f(t)][ηdt + σdz(t)] + σ22 G00 [f(t)]dt. (10.28)Taking expectations conditioned on time-t information you getE t [ds(t)] = G 0 [f(t)]ηdt + σ22 G00 [f(t)]dt. Dividing this result throughby dt you getE t [ ús(t)] = ηG 0 [f(t)] + σ22 G00 [f(t)]. (10.29)Now substitute (10.27) and (10.29) into the monetary model (10.21)and re-arrange to get the second-order differential equation in GG 00 [f(t)] + 2ησ 2 G0 [f(t)] − 2ασ G[f(t)] = − 2 f(t). 2 ασ2 (10.30)Digression on second-order differential equations. Consider the secondorderdifferential equation,y 00 + a 1 y 0 + a 2 y = bt (10.31)A trial solution to the homogeneous part (y 00 + a 1 y 0 + a 2 y = 0) isy = Ae λt , which implies y 0 = λAe λt and y 00 = λ 2 Ae λt , andAe λt (λ 2 +√a 1 λ + a 2 ) = 0, for which√there are obviously two solutions,λ 1 = −a 1+ a 2 1 −4a 2and λ2 2 = −a 1− a 2 1 −4a 2. If you let y2 1 = Ae λ1t andy 2 = Be λ2t , then clearly, y ∗ = y 1 + y 2 also is a solution because(y ∗ ) 00 + a 1 (y ∗ ) 0 + a 2 (y ∗ )=0.Next, you need to Þnd the particular integral, y p ,whichcanbeobtained by undetermined coefficients. Let y p = β 0 + β 1 t. Thenyp 00 =0,yp 0 = β 1 and yp 00 + a 1 yp 0 + a 2 y p = a 1 β 1 + a 2 β 0 + a 2 β 1 t = bt.It follows that β 1 = ba 2,andβ 0 = − a 1b.a 2 2Since each of these pieces are solutions to (10.31), the sum of thesolutionsisalsobeasolution.Thusthegeneralsolutionis,y(t) =Ae λ1t + Be λ2t − a 1b+ b t. (10.32)a 2 2 a 2