This is the fourth part of a four-part problem. If the given solutions 7₁ (t) = [ 26² +6e-1]. 2e³t 3e3t +15e-t' 9 ÿ' = [15 form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem ÿ(t) = (0)[, -4 -7 - 3e³t ₂ (t) = = ÿ, ÿ(0) 2e³t+6e7 impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks. +15e-t [4e³t +2e- 6e³t +5e-t + = [28] " [4e³t +2e-t 6est + 5e7
This is the fourth part of a four-part problem. If the given solutions 7₁ (t) = [ 26² +6e-1]. 2e³t 3e3t +15e-t' 9 ÿ' = [15 form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem ÿ(t) = (0)[, -4 -7 - 3e³t ₂ (t) = = ÿ, ÿ(0) 2e³t+6e7 impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks. +15e-t [4e³t +2e- 6e³t +5e-t + = [28] " [4e³t +2e-t 6est + 5e7
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![This is the fourth part of a four-part problem.
If the given solutions
7₁ (1) = [ 2² +6e-1],
2e³t
[3e3t +15e-t'
9
ÿ' = [15
form a fundamental set (i.e., linearly independent set) of solutions for the initial value
problem
-4
-7
7(0) = 0[
-
3e³t
₂ (t) =
=
2e³t+6e7
+15e-t
ÿ, ÿ(0)
[4e³t +2e-
6e³t
+5e-t
impose the given initial condition and find the unique solution to the initial value problem.
If the given solutions do not form a fundamental set, enter NONE in all of the answer
blanks.
+
=
[28]
"
[4e³t +2e-t
6est + 5e7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6890296c-2fc0-4a83-b66a-ee1a85d807a1%2F5a6a609e-d872-43d9-8df2-0d78766bd7fb%2F5v2cjs8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:This is the fourth part of a four-part problem.
If the given solutions
7₁ (1) = [ 2² +6e-1],
2e³t
[3e3t +15e-t'
9
ÿ' = [15
form a fundamental set (i.e., linearly independent set) of solutions for the initial value
problem
-4
-7
7(0) = 0[
-
3e³t
₂ (t) =
=
2e³t+6e7
+15e-t
ÿ, ÿ(0)
[4e³t +2e-
6e³t
+5e-t
impose the given initial condition and find the unique solution to the initial value problem.
If the given solutions do not form a fundamental set, enter NONE in all of the answer
blanks.
+
=
[28]
"
[4e³t +2e-t
6est + 5e7
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