and L_:(2)= ¿L_;(2)+1+ Lo(2) (27) Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one immediately concludes from (26) that the general solution of (26) is (1– (1 – 42)/2\o+1 (1+(1– 42)2\+1 + A2(2) | " (28) L,= A,(2) 2 2 The problem now is to determine the constants A(2) and A2(i). From the definition (25), it is clear that for any p the first term in the sum L,(2) is proportional to i?*'. This is compatible with the result (28) only if A2{2) =0 (29) The expression of A,(1) is then easy to find from the boundary condition (27) and one ends up with - (1 - 42)/2\P+3 1 L„(1)= (- (30) 2
and L_:(2)= ¿L_;(2)+1+ Lo(2) (27) Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one immediately concludes from (26) that the general solution of (26) is (1– (1 – 42)/2\o+1 (1+(1– 42)2\+1 + A2(2) | " (28) L,= A,(2) 2 2 The problem now is to determine the constants A(2) and A2(i). From the definition (25), it is clear that for any p the first term in the sum L,(2) is proportional to i?*'. This is compatible with the result (28) only if A2{2) =0 (29) The expression of A,(1) is then easy to find from the boundary condition (27) and one ends up with - (1 - 42)/2\P+3 1 L„(1)= (- (30) 2
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
Can you substitude equation 28 into equation 27 and try and achieve equation 30.
![and
L-1(2)= ¿L_,(2)+ 1+ Lo(2)
(27)
Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one
immediately concludes from (26) that the general solution of (26) is
1+(1– 42)/2\P+1
(28)
p+1
(1– 42)'/2)
L, = 4,(4)(-
A2(2) (T
+.
2
The problem now is to determine the constants A,(2) and A2(i).
From the definition (25), it is clear that for any p the first term in the sum
L,(2) is proportional to iP+!. This is compatible with the result (28)
only if
A2(2) =0
(29)
The expression of A,(2) is then easy to find from the boundary condition
(27) and one ends up with
´1-(1– 42)/2\P+3 1
L„(1) =(
|
wwww
(30)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F449fe0b2-a310-4385-aa4a-fe9b6697a61a%2Fe0590591-257c-40c2-9b9c-8c05f67050ab%2Fz7lybhg_processed.png&w=3840&q=75)
Transcribed Image Text:and
L-1(2)= ¿L_,(2)+ 1+ Lo(2)
(27)
Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one
immediately concludes from (26) that the general solution of (26) is
1+(1– 42)/2\P+1
(28)
p+1
(1– 42)'/2)
L, = 4,(4)(-
A2(2) (T
+.
2
The problem now is to determine the constants A,(2) and A2(i).
From the definition (25), it is clear that for any p the first term in the sum
L,(2) is proportional to iP+!. This is compatible with the result (28)
only if
A2(2) =0
(29)
The expression of A,(2) is then easy to find from the boundary condition
(27) and one ends up with
´1-(1– 42)/2\P+3 1
L„(1) =(
|
wwww
(30)
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 6 steps
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,
