So that S[z1,z2] S1 [21] + S₂[z2]. Hence the resulting 'uncoupled' EL equations are given by z + 10Z1 The solutions are y1 = 1 10 = (Z1 - 3z2) = 0, z = 0. Z1 = A. cos 10x + B. sin 10x, Z2 = Cx + D. Translating these back to give solutions in terms of the original dependent variables gives why not use ± √√To ? ?
So that S[z1,z2] S1 [21] + S₂[z2]. Hence the resulting 'uncoupled' EL equations are given by z + 10Z1 The solutions are y1 = 1 10 = (Z1 - 3z2) = 0, z = 0. Z1 = A. cos 10x + B. sin 10x, Z2 = Cx + D. Translating these back to give solutions in terms of the original dependent variables gives why not use ± √√To ? ?
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
![Hence the two required functionals are
So that
Changing Dependent Variables
The solutions are
S[Z1, Z2] S₁ [21] + S₂[z2].
Hence the resulting 'uncoupled' EL equations are given by
z₁ + 10Z₁ = 0, z = 0.
у1 =
Z1 = A. cos 10x + B. sin √10x, z2 = Cx + D.
Translating these back to give solutions in terms of the original dependent variables gives
=
1
=
dx
51/21-/ dr (10 (26²-2²).
S2[²2] = / dx (11(²2)²2).
-(Z1 - 3z2)
10
1
((A.C cos √√10x + B. sin √10x) − 3(Cx + D))
3C
3D
B
10
10
10
Here I'd absorb the constant factors and write this as y₁ = A cos √10x + B sin √10x + Cx + D.
10
A
10
cos √ 10x +
why not use
sin √10x
± √TO ? ?
-X-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59d44c96-efb1-4f3c-83b3-5a6a84cf94cb%2F385c334d-cdde-4ee7-9324-177adc81a20c%2Fkbjphtq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Hence the two required functionals are
So that
Changing Dependent Variables
The solutions are
S[Z1, Z2] S₁ [21] + S₂[z2].
Hence the resulting 'uncoupled' EL equations are given by
z₁ + 10Z₁ = 0, z = 0.
у1 =
Z1 = A. cos 10x + B. sin √10x, z2 = Cx + D.
Translating these back to give solutions in terms of the original dependent variables gives
=
1
=
dx
51/21-/ dr (10 (26²-2²).
S2[²2] = / dx (11(²2)²2).
-(Z1 - 3z2)
10
1
((A.C cos √√10x + B. sin √10x) − 3(Cx + D))
3C
3D
B
10
10
10
Here I'd absorb the constant factors and write this as y₁ = A cos √10x + B sin √10x + Cx + D.
10
A
10
cos √ 10x +
why not use
sin √10x
± √TO ? ?
-X-
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