S[y1, y2] = =S Hence From the first part we have y' + (y₁ + 3y2) = 0, Using the linear transformation Z1 = y1 + 3y2, we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix Rearranging leads to y1 = dx ((v₁)² + (v₂)² − (y₁ + 3y2)²). S[Z₁, 2] = | Z1 - - 3z2 1 - За = / ₁x (1 dx (( y + 3(y1 + 3y2) = 0. " Z2 = ayı + y2 y2 = 2 2 - 3z2 Sta.al - / + (( =)² + (-³) ²-4) S[Z₁, Z2] = dx 1 3a 3a 1 az1 - Z2 За - 1 - The only terms involving both dependant variables and their derivatives are the zz½ terms: -6z/z2 - 2aziz. Taking a = -3 (note that then 1 - 3a 0) kills these 'cross terms', and we have z² 3 (1³) a how come (z)² + 9(z)² +9(z₁)² + (z₂)² 100 is invertible. ??
S[y1, y2] = =S Hence From the first part we have y' + (y₁ + 3y2) = 0, Using the linear transformation Z1 = y1 + 3y2, we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix Rearranging leads to y1 = dx ((v₁)² + (v₂)² − (y₁ + 3y2)²). S[Z₁, 2] = | Z1 - - 3z2 1 - За = / ₁x (1 dx (( y + 3(y1 + 3y2) = 0. " Z2 = ayı + y2 y2 = 2 2 - 3z2 Sta.al - / + (( =)² + (-³) ²-4) S[Z₁, Z2] = dx 1 3a 3a 1 az1 - Z2 За - 1 - The only terms involving both dependant variables and their derivatives are the zz½ terms: -6z/z2 - 2aziz. Taking a = -3 (note that then 1 - 3a 0) kills these 'cross terms', and we have z² 3 (1³) a how come (z)² + 9(z)² +9(z₁)² + (z₂)² 100 is invertible. ??
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
100%
![S[y1, y2] =
=/
Hence
From the first part we have y₁ + (y₁ + 3y2) = 0, y + 3(y₁ + 3y2) = 0.
Using the linear transformation
Z1 = y1 + 3y2,
we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix
Rearranging leads to
dx ((v₁)² + (v₂)² − (y1 + 3y2)²) .
y1 =
S[21, 2] = |
Z1 - - 3z2
1 - 3a
= [ ₁x (1
dx
((
Z2 = ayı + y2
"
y2 =
2
2
3z2
az
Sta.al - / + (( =)² + (K = ³ ) ² --)
S[Z1, Z2] = dx
-
1 3a
За - 1
az1 - Z2
3a - 1
The only terms involving both dependant variables and their derivatives are the zz½ terms:
-6z/z2 - 2azz2₂.
Taking a = -3 (note that then 1 – 3a 0) kills these 'cross terms', and we have
z₁
3
(1³)
a
how come
(z₁)² +9(z)² +9(z₁)² + (z₂)²
100
is invertible.
??](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59d44c96-efb1-4f3c-83b3-5a6a84cf94cb%2F4b8ed948-b0fc-4c93-8318-85a8d875c7ea%2Ftroazd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:S[y1, y2] =
=/
Hence
From the first part we have y₁ + (y₁ + 3y2) = 0, y + 3(y₁ + 3y2) = 0.
Using the linear transformation
Z1 = y1 + 3y2,
we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix
Rearranging leads to
dx ((v₁)² + (v₂)² − (y1 + 3y2)²) .
y1 =
S[21, 2] = |
Z1 - - 3z2
1 - 3a
= [ ₁x (1
dx
((
Z2 = ayı + y2
"
y2 =
2
2
3z2
az
Sta.al - / + (( =)² + (K = ³ ) ² --)
S[Z1, Z2] = dx
-
1 3a
За - 1
az1 - Z2
3a - 1
The only terms involving both dependant variables and their derivatives are the zz½ terms:
-6z/z2 - 2azz2₂.
Taking a = -3 (note that then 1 – 3a 0) kills these 'cross terms', and we have
z₁
3
(1³)
a
how come
(z₁)² +9(z)² +9(z₁)² + (z₂)²
100
is invertible.
??
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 with 1 images
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,
