Higher Order Differential Equation

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Melvin Williams
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2 How to find the solution of the following DE: (i). d 2 y/d t 2 = k y (ii). d n y/d t n + a 1 d n-1 y/d t n a n-1 dy/d t + a n y = b

3 Second Order DE with constant coefficient y'' (t) + a 1 y' (t) + a 2 y = b ; y' (t) = d y/d t ; y'' (t) = d 2 y/d t 2 a 1, a 2, b constant General Solution: y(t) = y c + y p y c : complementary function general solution of its homogeneous DE y p : particular solution

4 Finding Particular Solution Particular solution is obtained by assuming y(t) = a (the simplest form). So, y' (t) = y'' (t) = 0. Therefore, the DE becomes a 2 y = b or y = b/a 2 for a 2 0 Example: Find the particular solution of: y'' (t) + y' (t) 2y = -10 y p = -10/-2 = 5 since b = -10 and a 2 = -2

5 Now, what happened if a 2 = 0 as in y''(t) + y'(t) = -10 In this case, particular solution is obtained by assuming y(t) = kt ; so y' = k; y'' = 0. The DE becomes k = -10 or y p = -10 t. In general, y p = b /a 1 t What happened if a 2 = a 1 = 0 So, the DE: y''(t) = b In this case, the particular solution y(t) = k t 2 therefore y' (t) = 2k t and y''(t) = 2k. Then, the DE: 2k = b or k= b/2 and y p = b/2 t 2

6 Finding Complementary Function Complementary function is a general solution of the homogeneous DE: y'' (t) + a 1 y' (t) + a 2 y = 0 The solution: y = A e rt ; r and A will be obtained later Thus: y' (t) = A r e rt ; y'' (t) = A r 2 e rt So A r 2 e rt + a 1 A r e rt + a 2 A e rt = 0 A e rt (r 2 + a 1 r + a 2 ) = 0 or: r 2 + a 1 r + a 2 = 0 r 1, r 2 = {- a 1 ± (a 12 4a 2 ) ½ }/2 check: r 1 + r 2 = - a 1 ; r 1 r 2 = a 2

7 If r 1 r 2, the complementary function, y c = A 1 e r1t + A 2 e r2t If r 1 = r 2, the complementary function, y c = A 3 e rt + A 4 t e r2t A 1, A 2, A 3, A 4 constants that can be obtained by the initial values of its DE. If the roots are complex numbers (that is when a 12 < 4a 2 ) the complementary function has a special form that out of our coverage.

8 Example: (1). y'' (t) + y' (t) 2y = -10 In this case: a 1 = 1, a 2 = -2 ; b = 10 ; r 1 = 1; r 2 = -2 The complementery function, y c = A 1 e t + A 2 e -2t The particuar function, y p = b/a 2 = -10/-2 = 5 The general solution, y = y c + y p y(t) = A 1 e t + A 2 e -2t + 5; y'(t) = A 1 e t -2 A 2 e -2t How to find A 1 and A 2?

9 For example: initial condition y(0) = 12 and y'(0) = -2 So, y(0) = A 1 + A = 12 A 1 + A 2 = 7 y'(0) = A 1 2A 2 = -2 A 1 2 A 2 = -2 A 1 = 4 and A 2 = 3 The general solution: y (t) = 4 e t + 3 e -2t + 5 Verification: y' (t) = 4 e t 6 e -2t y'' (t) = 4 e t + 12e -2t y'' (t) + y'(t) 2y(t) = 4e t 6e -2t + (4e t + 12e -2t ) 2(4e t + 3e -2t + 5) = -10.

10 (2). y'' (t) + 6 y'(t) + 9y =27 a 1 = 6 ; a 2 = 9 ; b = 27; r 1 = r 2 = -3 Complementary function, y c = A 3 e -3t + A 4 t e -3t Particular solution, y p = b/a 2 = 27/9 = 3 General Solution, y(t) = y c +y p A 3 and A 4? = A 3 e -3t + A 4 t e -3t + 3

11 If the initial conditions: y(0) = 5 and y'(0) = -5, so y(0) = A = 5 A 3 = 2 y'(t) = -3 A 3 e -3t + A 4 (e -3t 3t e -3t ) y'(0) = -6 + A 4 = -5 A 4 = 1 General Solution: y(t) = 2 e -3t + t e -3t + 3 Verivication: y'(t) = -6 e -3t + (e -3t 3 t e -3t ) y'' (t) = +18 e -3t 3 e -3t 3 (e -3t -3 t e -3t ) = 12 e -3t + 9 t e -3t y'' (t) + 6 y'(t) + 9 y(t) = 12 e -3t + 9 t e -3t + 6(-5 e -3t 3 t e -3t ) + 9(2 e -3t +t e -3t + 3) = 27

12 Back to our market model: Market Model with Price Expectation Models that we have studied assume that Q d and Qs only depend on the current price, P. But, buyers and sellers often concern about future price trends. How to incorporate this into the model? Recall: P'(t) = dp/dt P''(t) = d 2 P/dt 2 price change. represents price change represents an acceleration of

13 So, if buyers and sellers concerns about the trend of future price, the model is: Q d = D [P(t), P'(t), P''(t)] Q s = S [P(t), P'(t), P''(t)] In explicit forms, we can represent the above relationships in the linear form (for simplicity), as: Q d = α - βp + m P' + n P'' (α, β > 0) Q s = -γ + δ P + up' + wp'' ( γ, δ > 0)

14 Comments: 1. If m > 0, if price increases (P' > 0), Q d will increase also. In this case, buyers have expectation that price will keep increasing in the future, so the buyers should buy now before the price will increase again. 2. However, if m < 0, if price increase, Q d will decrease. In this case, buyers have expectation that price will keep decreasing in the future, so the buyers should postpone or reduce their buying

15 Simplified Model To simplify the analysis, it is assumed that only demand function depends on price expectation. In other words, it is assumed that m 0; n 0; u = w = 0, so the simplified model is: Q d = α - βp + m P' + n P'' (α, β > 0) Q s = -γ + δ P (γ, δ > 0)

16 In a market equilibrium, Q d = Q s becomes: the model P'' + m/n P' (β + δ )/n P = - (α + γ) / n Or P'' + a 1 P' + a 2 P = b; with: a 1 = m/n; a 2 = (β+δ)/n and b = -(α+γ)/n This is a Differential Equation Order 2 in P (price) and the general solution is: P(t) = Pp + Pc

17 Finding Particular Solution P p From earlier discussion, P p (t) = b/a 2 = (a + γ) / (β + δ ) > 0 This solution represents stationary eqauilibrium. Finding complementary function, P c (i). If the roots are different, or if (m/n) 2 > -4 (β + δ )/n P c = A 1 e r1t + A 2 e r2t r 1, r 2 = [-m/n ± {(m/n) (β + δ )/n } 1/2 ]/2 Then, P(t) = P c + P p = A 1 e r1t + A 2 e r2t +(α + γ)/(β + δ)

18 (ii). If the roots are the same, or if (m/n) 2 = -4(β + δ )/n, then r = -m/2n the complementary function, P c = A 3 e -mt/2n + A 4 t e -mt/2n The general solution: P(t) = A 3 e -mt/2n + A 4 t e -mt/2n +(α+ γ)/(β + δ)

19 Example with numbers: Q d = 42 4P 4P' + P'' Q s = P Initial condition: P(0) = 6 and P' (0) = 4; Find P(t)? In this example, α = 42, β = 4, γ = 6, δ = 8, m =-4, n=1 r 1, r 2 = [4/1 ± { (4+8)} 1/2 ]/2 = (4±8)/2 r 1 = 6 ; r 2 = -2

20 The general solution: P(t) = A 1 e 6t + A 2 e -2t + (42 + 6)/(4+8) = A 1 e 6t + A 2 e -2t + 4 Initial condition P(0) = A 1 + A = 6 P'(t) = 6A 1 e 6t 2A 2 e -2t P'(0) = 6 A 1-2A 2 = 4 Then, from above equations, A 1 = A 2 = 1 and therefore, P(t) = e 6t + e -2t + 4.

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22 The End

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