lan-2 for n > 2 with initial conditions ao = 3, a1 = 5. The first step as usual is to find the characteristic equation an = -2an-1 by trying a solution of the "geometric" format an = r". (We assume also r + 0). In this case we get: p" = -2r"-1 – 1r"-2. Since we are assuming r O we can divide by the smallest power of r, i.e., r"-2 to get the characteristic equation: p2 = -2r – 1. (Notice since our Ihcc recurrence was degree 2, the characteristic equation is degree 2.) This characteristic equation has a single root r. (We say the root has multiplicity 2). Find r. Since the root is repeated, the general theory (Theorem 4 in section 7.2 of Rosen) tells us that the general solution to our Ihcc recurrence looks like: an = a1(r)" + azn(r)" for suitable constants a1, a2. To find the values of these constants we have to use the initial conditions an = 3, aj = 5. These yield by using n=0 and n=1 in the formula above: 3 = a1 (r)° + a20(r)° and 5 = a1 (r)' + az1(r)' By plugging in your previously found numerical value for r and doing some algebra, find a1, a2: aj = a2
lan-2 for n > 2 with initial conditions ao = 3, a1 = 5. The first step as usual is to find the characteristic equation an = -2an-1 by trying a solution of the "geometric" format an = r". (We assume also r + 0). In this case we get: p" = -2r"-1 – 1r"-2. Since we are assuming r O we can divide by the smallest power of r, i.e., r"-2 to get the characteristic equation: p2 = -2r – 1. (Notice since our Ihcc recurrence was degree 2, the characteristic equation is degree 2.) This characteristic equation has a single root r. (We say the root has multiplicity 2). Find r. Since the root is repeated, the general theory (Theorem 4 in section 7.2 of Rosen) tells us that the general solution to our Ihcc recurrence looks like: an = a1(r)" + azn(r)" for suitable constants a1, a2. To find the values of these constants we have to use the initial conditions an = 3, aj = 5. These yield by using n=0 and n=1 in the formula above: 3 = a1 (r)° + a20(r)° and 5 = a1 (r)' + az1(r)' By plugging in your previously found numerical value for r and doing some algebra, find a1, a2: aj = a2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:(1 point) We will find the solution to the following Ihcc recurrence:
-2an-1 - la,n-2 for n > 2 with initial conditions ao = 3, a1 = 5. The first step as usual is to find the characteristic equation
by trying a solution of the "geometric" format an = r". (We assume also r + 0). In this case we get:
An =
pn
= -2r"-1 – 1p"–2.
Since we are assuming r + 0 we can divide by the smallest power of r, i.e., r"-2 to get the characteristic equation:
— — 2г — 1.
(Notice since our Ihcc recurrence was degree 2, the characteristic equation is degree 2.)
This characteristic equation has a single root r. (We say the root has multiplicity 2). Find r.
r =
Since the root is repeated, the general theory (Theorem 4 in section 7.2 of Rosen) tells us that the general solution to our Ihcc
recurrence looks like:
an = a1(r)" + a,n(r)"
for suitable constants a1, a2·
To find the values of these constants we have to use the initial conditions ao =
3, a1 = 5. These yield by using n=0 and n=1 in the
formula above:
3 = a1(r)º + a20(r)°
and
5 = a1 (r)' + az1(r)'
By plugging in your previously found numerical value for r and doing some algebra, find a1, a2:
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