newell_lec11-15_100713

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at Arizona.

Ask a homework question - tutors are online