| (ri)º (r2)° (r3)0* (ri)! (r2)' (r3)! (ri)² (r2)? (r3)². a2 -2 a2 = Az =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:a, = a1(ri)" + a2(r2)" + a3(r3)"
for suitable constants a1, a2, ɑz .
To find the values of these constants we have to use
the initial conditions ao = 0, a = -2, a2 = 0.
These yield by using n=0,n=1 and n=2 in the formula
above:
0 = a,(r¡)° + az(r2)º + a3(r3)º
and
-2 = a,(r)' + az(r2)' + az(r3)'
and
0 = a,(r1)? + az(r2)² + az(r3)²
By plugging in your previously found numerical values
for r1, r2 and r3 and doing some algebra, find
A1, a2, A3:
Note: Ad hoc substitution should work to find the a;
but for those who know linear algebra, note the
system of equations above can be written in matrix
form as:
(ri)º (r2)º (r3)°
(ri)' (r2)' (r3)'
[(r)? (r2)? (r3)?
-2
az
aj =
a2 =
az =
Note the final solution of the recurrence is:
a, = a1(r1)" + a2(r2)" + a3(r3)"
where the numbers r;, a; have been found by your
work. This gives an explicit numerical formula in terms
of n for the an .
Transcribed Image Text:За,-1 + 1а,-2 — За,-з for n 23 with initial
-2, az = 0. The first step
An
conditions ao
:0, a1
as usual is to find the characteristic equation by trying
a solution of the "geometric" format a, = r". (We
assume alsor+ 0). In this case we get:
p" = 3r"-1 + 1r"-2 – 3r"-3.
Since we are assuming r # 0 we can divide by the
smallest power of r, i.e.,
pl-3
to get the characteristic
equation:
p3 = 3r² + 1r – 3.
(Notice since our Ihcc recurrence was degree 3, the
characteristic equation is degree 3.)
Find the three roots of the characteristic equation
r1, r2 and r3. When entering your answers use
ri =
,r3 =
Since the roots are distinct, the general theory
(Theorem 3 in section 5.2 of Rosen) tells us that the
general solution to our Ihcc recurrence looks like:
An = a¡(r¡)" + a2(r2)" + az(r3)"
for suitable constants a1, a2, az.
To find the values of these constants we have to use
the initial conditions ao = 0, a1 = -2, az = 0.
These yield by using n=0,n=1 and n=2 in the formula
above:
0 = a,(r,)° + az(r2)° + a3(r3)º
%3D
and
-2 = a,(r))' + az2(r2)' + a3(r3)'
and
0 = a;(ri)² + az(r2)² + az(r3)²
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
