Dynamic Problem Set 5 Solutions
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1 Dynamic Problem Set 5 Solutions Jonathan Kreamer 211 Question 1 Solve the following two differential equations. (i) y 1 (t) + y(t) = 2 + 2t, y() = 1 Method 1: We can work from the general solution to non-autonomous ODEs. y(t) = y()e (t) + b(s)e (t) (s) ds (t) = a(s)ds For our setup (t) = t ( 1)ds = t. Then the solutions is y(t) = y()e (t) + b(s)e (t) (s) ds = 1 e t + (2 + 2s)e t+s ds = e t + 2e t [se s ] t = e t + 2e t [ te t] = e t + 2t Method 2: Start with the guess y(t) = A + Bt + Ce dt. Then we have y (t) = B + Cde dt y (t) = B + d [ Ce dt + Bt + A ] Ad Bdt where the term in brackets is y(t). Then this can be written y (t) dy(t) = B Ad Bdt So we have d = 1. Plugging this in yields y (t) + y(t) = B + A + Bt 1
2 Which implies B = 2, which in turn implies A =. So the general solution is y(t) = 2t + Ce t It s good practice at this point to check that this satisfies the initial condition. To find the particulat solution, we use the boundary condition y() = 1, which implies y() = C = 1 So the particular solution is y p (t) = 2t + e t (ii) 3y 1 (t) + y(t) = e 2t, y() = 1 This is equivalent to y 1 (t) = 2y(t) (e 2t) Method 1: Use the general method. In this setup (t) = t ( 2)ds = 2t, then the general solution becomes y(t) = y()e (t) + = 1 e 2t + = e 2t e 2t b(s)e (t) (s) ds 1 3 (e 2t)e 2t+2s ds = e 2t e 2t ( e + 2t)e 2s ds ( e 1)e 2s ds + (1 + 2t)e 2s ds = e 2t + 1 [ 3 e 2t e + 1 ] 2 (e2s ) t + (se 2s ) t = e 2t + 1 [ 3 e 2t e + 1 ] 2 (e2t 1) + te 2t = e 2t + e + 1 (1 e 2t ) t = 5 e e 2t + e t Method 2: To start with, let s divide by 3 to write the initial expression as y 1 (t) + 2y(t) = e t We again start with the guess y(t) = A + Bt + Ce dt. Then we have y (t) = B + Cde dt y (t) = B + d [ Ce dt + Bt + A ] Ad Bdt where the term in brackets is y(t). Then this can be written y (t) dy(t) = B Ad Bdt 2
3 This gives us d = 2. Plugging this in yields Which implies B = 1 3. Plugging this in yields which implies 2A 1 3 = e 3 y (t) + 2y(t) = B + 2A + 2Bt y (t) + 2y(t) = A 2 3 t e+1, or A =. So we have the general solution y(t) = e + 1 t 3 + Ce 2t which we should again check to be sure it satisfies the initial condition. To find the particular solution, we use the boundary condition y() = e C = 1 Solving for C, we obtain C = 5 e. So the particular solution is Question 2 y p (t) = e + 1 Reduce the nth-order linear differential equation t ( ) 5 e 3 + e 2t a n x n (t) + a n 1 x n 1 (t) a x(t) = to a system of first-order differential equations. Consider n 1 new functions x 1 (t)...x n 1 (t) in addition to x(t) where x i (t) = x i (t). Then our equations can be re-written as system of n first-order differential equations: Question 3 a n x 1 n 1(t) + a n 1 x n 1 (t) + a n 2 x n a 1 x 1 (t) + a x(t) = x 1 i (t) x i+1 (t) = i = 1,..., n 2 x 1 (t) x 1 (t) = Reconsider the Solow-Swan growth model that we have discussed in the lectures, and relax two of the assumption we made, namely that the capital stock does not depreciate, and that there is no technical progress. To allow for depreciation, specify the law of motion for the capital stock as K 1 (t) = I(t) δ for δ (, 1). To allow for technical progress, suppose that the production function is given by Y (t) = F [, A(t)L(t)] = [] [ e µt L(t) ] 1 for µ >, (, 1). (i) Derive the implied law of motion for the effective capital-labor ratio, k(t) = A(t)L(t). 3
4 Note that µ is the growth rate of the economy. Let the population growth rate be n, meaning that L(t) = e nt L. Finally, suppose that the savings rate is s, so that we have I(t) = sy (t). Let k(t) =. Then using the quotient rule we have A(t)L(t) Rearranging, this is which we can write as k (t) = K (t)a(t)l(t) [A (t)l(t) + A(t)L (t)] [A(t)L(t)] 2 k (t) = K (t) [ ] [ ] [ ] A (t) L (t) A(t)L(t) A(t) A(t)L(t) L(t) A(t)L(t) k (t) k(t) = K (t) A (t) A(t) L (t) L(t) Which is a standard expression saying that percentage growth of k is the sum of percentage growth in its multiplicative components. Note that we have A (t) A(t) = µeµt e µt = µ L (t) L(t) = nent e nt = n Finally, we have K (t) = I(t) δ. Using I(t) = sy (t) = s [] [A(t)L(t)] 1, we obtain K (t) = s [] [A(t)L(t)] 1 [ ] 1 δ A(t)L(t) = s δ which we can write as Substituting into the expression above, we have K (t) = s k(t) 1 δ k (t) k(t) = s [ k(t) ] 1 δ µ n k (t) = s [ k(t) ] (δ + µ + n) k(t) (ii) Compute the (nontrivial) steady state for the effective capital-labor ratio and analyze its stability properties. The steady state defined implicitly as k ss that satisfy k (t) =. Plugging into the differential equation above, we obtain s [k ss ] = (δ + µ + n) k ss Which we can express as ( k ss = s δ + µ + n ) 1 1 Which is the steady state. We can draw a phase diagram with k on the x-axis, and the curve s k and the line (n + δ + µ) k. Since s k satisfies the Inada conditions, there is guaranteed to be a nontrivial intersection of these lines. Moreover, since the curve s k lies above the line when k < k ss, and then cuts it from above and lies below the line for k > k ss, we have a convergent steady state. 4
5 Question 4 Consider the expectation augmented Phillips curve π(t) = f (u(t)) + π e (t) with f ( ) > and (, 1], where π is the observed inflation rate, u is the unemployment rate, and π e is the expected inflation rate. Assume that expectations are gradually corrected acording to with β >. π e (t) = β [π(t) π e (t)] (i) Obtain the (reduced form) differential equation as a function of observed quantities and characterize it. Observed quantities are u and π, not π e. We can rearrange the top equation to read Differentiating yields π e (t) = π(t) f (u(t)) π e (t) = π (t) Substituting into the second equation yields π (t) Simplifying this expression, we obtain f (u(t)) u (t) f (u(t)) u (t) [ = β π(t) π(t) + f (u(t)) ] π (t) = f (u(t)) u (t) + βf (u(t)) β (1 ) π(t) This gives us the path of the inflation rate in response to the path of the unemployment rate. (ii) Now assume that the government wishes to maintain the unemployment rate at a certain target u. In the short run it is known that < 1, but in the long run = 1. Comment on the feasibility of this policy taking into account the stability properties of the differential equation. If we fix the unemployment rate at a level u, then we have u (t) =, so the differential equation governing the inflation rate becomes π (t) = βf (u) β (1 ) π(t) In the long run, we have = 1 meaning that this equation becomes still simpler π (t) = βf (u) This equation suggests that if we choose u such that f (u) =, this is stable. If we choose a lower value of u, we get a constantly increasing inflation rate, while if we choose a higher value we get accelerating deflation. In the short-run we do get a stable inflation rate. Setting π (t) = in the equation above, we obtain π = f (u) 1 as the fixed inflation rate. Since we have that f is a decreasing function, it follows that a lower level of unemployment requires a higher level of inflation to achieve. We can achieve u < u N by enduring a constant level of inflation, and u > u N by enduring a constant level of deflation. 5
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