We begin by finding the EL equation for the auxiliary functional S[y] = = - S₁ 12 dx (xy2+ Ay - λy) = = · L dx F ( dxF(x,y,y', 2). d 0 = (Fy') - Fy (2xy') – (A – 1). dx dx Integrating gives A -λ B y' + 2 x and integrating again gives A - λ y = -x+Blogx+C, 2 77 come where C is another constant. The initial condition y(1) = 0 gives C = -(A - λ)/2. We now apply the natural boundary condition (HB pg 24) given by Fy = 0 at x = v. That is 0 = Fy | x=v = 2xy' | x=v' that is y'(v) = 0. Now A - X 0 = y' (v) = إن B v(A - λ) + hence B = 2 v Hence A - λ y(x) = K(x − 1 - vlogx) where K = 2
We begin by finding the EL equation for the auxiliary functional S[y] = = - S₁ 12 dx (xy2+ Ay - λy) = = · L dx F ( dxF(x,y,y', 2). d 0 = (Fy') - Fy (2xy') – (A – 1). dx dx Integrating gives A -λ B y' + 2 x and integrating again gives A - λ y = -x+Blogx+C, 2 77 come where C is another constant. The initial condition y(1) = 0 gives C = -(A - λ)/2. We now apply the natural boundary condition (HB pg 24) given by Fy = 0 at x = v. That is 0 = Fy | x=v = 2xy' | x=v' that is y'(v) = 0. Now A - X 0 = y' (v) = إن B v(A - λ) + hence B = 2 v Hence A - λ y(x) = K(x − 1 - vlogx) where K = 2
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![We begin by finding the EL equation for the auxiliary functional
S[y] =
=
- S₁
12
dx (xy2+ Ay - λy) =
=
· L dx F (
dxF(x,y,y', 2).
d
0 =
(Fy') - Fy
(2xy') – (A – 1).
dx
dx
Integrating gives
A -λ
B
y'
+
2
x
and integrating again gives
A - λ
y =
-x+Blogx+C,
2
77
come
where C is another constant.
The initial condition y(1) = 0 gives C = -(A - λ)/2.
We now apply the natural boundary condition (HB pg 24) given by Fy = 0 at x = v. That is
0 = Fy | x=v
=
2xy' | x=v'
that is y'(v) = 0.
Now
A - X
0 = y' (v) =
إن
B
v(A - λ)
+
hence B =
2
v
Hence
A - λ
y(x) = K(x − 1 - vlogx)
where K
=
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Fe22f741b-d78c-4f4c-9e70-429967591605%2Fstu51l8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:We begin by finding the EL equation for the auxiliary functional
S[y] =
=
- S₁
12
dx (xy2+ Ay - λy) =
=
· L dx F (
dxF(x,y,y', 2).
d
0 =
(Fy') - Fy
(2xy') – (A – 1).
dx
dx
Integrating gives
A -λ
B
y'
+
2
x
and integrating again gives
A - λ
y =
-x+Blogx+C,
2
77
come
where C is another constant.
The initial condition y(1) = 0 gives C = -(A - λ)/2.
We now apply the natural boundary condition (HB pg 24) given by Fy = 0 at x = v. That is
0 = Fy | x=v
=
2xy' | x=v'
that is y'(v) = 0.
Now
A - X
0 = y' (v) =
إن
B
v(A - λ)
+
hence B =
2
v
Hence
A - λ
y(x) = K(x − 1 - vlogx)
where K
=
2
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