2 will have enough free constants namely two so that

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Unformatted text preview: d 2 y1 dy1 d y2 dy2 A + p( t ) + q ( t) y 1 + B + p( t) + q ( t) y 2 dt2 dt dt2 dt A.0 + B.0 = 0. Hence Ay1 (t) + By2 (t) solves (6.2). We use this idea to build the general solution of (6.2). By general solution, we mean that the solution of (6.2) will have enough free constants, namely two, so that we can satisfy two initial conditions. dy y ( t0 ) = γ 0 , ( t0 ) = γ 1 dt given at some initial time t0 . We look for two “special” solutions y1 (t) and y2 (t) and then take an arbitrary linear combination of them to build the general solution y (t). We can’t use any solutions y1 (t), y2 (t) because they must have the property that we can find A, B such that Ay1 (t0 ) + By2 (t0 ) = γ0 ￿ ￿ Ay1 (t0 ) + By2 (t0 ) = γ1 ￿ ￿ for arbitrary γ0 , γ1 . For this to hole, we must have y1 (t0 )y2 (t0 ) − y1 (t0 )y2 (t0 ) ￿= 0. More on this later. We now illustrate this idea for 2nd order linear, homogeneous, constant coefficient ode’s. In this case, the functions p(t) and q (t) are constants p and q . 5 Methods for solving 2nd order, linear, homogeneous, constant coefficient ode’s. We want to solve L[ y ] = d2 y dy +p + qy = 0 dt2 dt (6.7) Look for a solution in the form y = y (t, r) = aert . (6.8) Substitute (6.8) into (6.7) and find that L[y (r, t) = aert ] ≡ (r2 + pr + q )aert = 0. Therefore (6.8) satisfies (6.7) if r is a root of the characteristic equation r2 + pr + q = 0. (6.9) Example: L[y ] ≡ y ￿￿ + y ￿ − 2y = 0; y (0) = 1, y ￿ (0) = 1 L[ert ] = (r2 + r − 2)ert = 0 if r = −2 or r = +1. Check: e−2t and et solve y ￿￿ + y ￿ − 2y = 0. Check: Ae−2t + Bet solves y ￿￿ + y ￿ − 2y = 0. General solution: y = Ae−2t + Bet . Initial conditions: y (0) = 1 = A + B, y ￿ (0) = 1 = −2A + B ⇒ A = 0, B = 1. Solution of y ￿￿ + y ￿ − 2y = 0 with initial conditions y (0) = y ￿ (0) = 1 is y (t) = et . Check: Use JOde. y = x dx dt = u, x(0) = 1 du dt = −u + 2x, u(0) = 1. In general, the roots of the characteristic equation (6.9) are ￿ ￿ 1￿ r= −p ± p2 − 4q . 2 There are 3 cases we must study. Case 1. p2 > 4q , roots real and distinct. ￿ ￿ 1￿ r1 ,2 = −p ± p2 − 4q . 2 6 (6.10) General solution of (6.7); √2 √2 1 1 y (t) = Ae 2 (−p+ p −4q)t + Be 2 (−p− p −4q)t . Example: L[y ] = y ￿￿ + 3y ￿ + 2y = 0, y (0) = 1, y ￿ (0) = 1. L[ert ] = (r2 + 3r + 2)ert = 0 General solution if r = −1, −2 y (t) = Ae−t + Be−2t Initial Conditions ⇒ y (0) = A + B = 1 ⇒ A = 3, B = −2 y ￿ (0) = −A − 2B = 1 y (t) = 3e−t − 2e−2t y ￿ (t) = −3e−t + 4e−2t See JOde graph for discussion. Question: Note that the solution approaches zero (0, 0) along the line y ￿ = −y in the JOde simulation. Why? 7 Case 2. p2 = 4q , double root r = −p/2. Now, the method has only one given p solution y1 (t) = e 2 t . Go back to (6.7), (6.8) for the case p2 = 4q . Then ￿ p...
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This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at University of Arizona- Tucson.

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