newell_lec11-15_100713

7 and the general solution is p p y t ae 2 t bte 2 t

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Unformatted text preview: ￿2 rt L[y (r, t)] = (r2 + pr + q )aert = r + ae . 2 p Clearly L[y (r = −p/2, t)] = 0. And therefore y1 (t) = ae− 2 t is a solution. But, note also that ￿ ￿ ∂ ∂y L[y (r, t)] = L (r, t) ∂r ∂r = (r + p/2)2 atert + 2(r + p/2)aert = 0 also when r = −￿ 2. Therefore, at a double root, both y1 (t) = y (r = −p/2, t) and p/ ￿ y2 (t) = ∂ y (r, t) are solutions. ∂r r =−p/2 ￿ p ∂y ￿ ￿ ￿ = atert ￿ = ate− 2 t . ￿ ∂ r r=−p/2 r =−p/2 p p Thus y1 (t) = e− 2 t and y2 (t) = te− 2 t are two essentially different solutions of (6.7) and the general solution is p p y (t) = Ae− 2 t + Bte− 2 t when p2 = 4q . We will come to the precise meaning of essentially different shortly. Example: y (0) = 2, y ￿ (0) = −1. 9y ￿￿ − 12y ￿ + 4y = 0, y ￿￿ − 4 y ￿ + 4 y = 0. 3 9 y = ert :⇒ r2 − 4 r + 3 2 4 9 =0⇒ r= 2 twice. 3 2 Solutions: y1 (t) = e 3 t , y2 (t) = te 3 t . 2 2 General solution: y (t) = Ae 3 t + Bte 3 t . Initial conditions: y (0) = 2 = A y ￿ (0) = −1 = 2 A + B 3 ⇒ A = 2, B = −7/3. Thus y ( t) = 2 e 2 3t 72 − te 3 t = 3 8 ￿ ￿ 2 7 2 − t e 3 t. 3 Reduction of order. We interrupt the discussion at this point to introduce a very useful new idea, reduction of order, discussed in NSS §4.7 pg. 198. The idea is this: If we know one solution y = y1 (t) of (6.2) d2 y dy + p( t ) + q (t)y = 0, dt2 dt (6.2) we can find another one y2 (t) by setting y 2 ( t ) = y 1 ( t) v ( t ) . (6.11) Differentiate (6.11) twice dy2 dy1 dv = v + y1 dt dt dt (6.12a) d 2 y2 d 2 y1 dy1 dv d2 v = v+2 + y1 2 2 2 dt dt dt dt dt (6.12b) and substitute (6.12) into (6.2). We find, on collecting terms ￿ ￿ ￿2 ￿ d2 v dy1 dv d y1 dy1 y1 2 + 2 + p( t) y 1 + + p( t ) + q (t)y1 v = 0. dt dt dt dt2 dt But the last bracket is zero because y1 (t) solves (6.2). Therefore we are left with a first order ode for u(t) = dv dt ￿ ￿ du dy1 y1 +2 + p( t ) y 1 u = 0 (6.13) dt dt (6.13) is a linear, first order ode for u(t) which we solve using what we learned in Lectures 5-6. Divide across by y1 . Then the integrating factor for (6.13) is ￿￿ ￿ 2 dy1 IF = exp + p(t) dt y1 dt ￿ ￿ ￿t = exp 2 ln |y1 | + p(t￿ )dt￿ 2 y1 exp = Therefore d dt or ￿ 2 uy1 ￿ exp t ￿ p(t￿ )dt￿ . t ￿ p(t )dt ￿ ￿ =0 ￿ ￿t ￿ dv 1 = 2 exp − p(t￿ )dt￿ . dt y1 Therefore a second solution y2 (t) of (6.2) is ￿ ￿t ￿ ￿ dt ￿ ￿ y 2 ( t ) = y 1 ( t) exp − p(t )dt . 2 y 1 ( t) u= 9 Furthermore, we will find when we get to define what linearly independent means, y1 (t) and y2 (t) are linearly independent (essentially different) on any interval on which p(t), q (t) are continuous. Application to double root situation. Let q = p2 /4. We know one solution to d2 y dy p2 +p + y=0 dt2 dt 4 is y1 (t) = e−p/2 t. Let y2 (t) = ve−p/2 t . Then ￿ ￿￿ y2 (t) = v ￿ e−p/2 t − p/2 ve−p/2 t , y2 = v ￿￿ e−p/2 t − p...
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