This is the fourth part of a four-part problem. If the given solutions [t² = 2t], in(t) = T₂(t) = [7¹1]. y' = in₁ (t) = [t² = 2t] - form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem 21-2 -2t-2 12t¹+2t -27 21-¹2-2 ÿ, ÿ(5) = -53] -32] - 2t ( ) [²³² 2 ² ² ] + (( 2t " t> 0, impose the given initial condition and find the unique solution to the initial value problem for t > 0. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks. y(t) = ( [7]
This is the fourth part of a four-part problem. If the given solutions [t² = 2t], in(t) = T₂(t) = [7¹1]. y' = in₁ (t) = [t² = 2t] - form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem 21-2 -2t-2 12t¹+2t -27 21-¹2-2 ÿ, ÿ(5) = -53] -32] - 2t ( ) [²³² 2 ² ² ] + (( 2t " t> 0, impose the given initial condition and find the unique solution to the initial value problem for t > 0. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks. y(t) = ( [7]
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 15EQ
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please solve it on paper
![This is the fourth part of a four-part problem.
If the given solutions
ÿ₁(t)
ÿ'
=
-2
2t
-2t-2
t²
2t, ÿ₂(t) =
T2(t) = [† 7²¹].
1
form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem
1 – 2t-¹
2t-¹-2-2
2t
+2t
2
ÿ(5) = [-32]
"
t> 0,
impose the given initial condition and find the unique solution to the initial value problem for t > 0. If the given solutions do not form a fundamental
set, enter NONE in all of the answer blanks.
y(t) = (
2t
-
) [²³² 2 ² ² ] + (
2t
-
[t
[1¹].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05295cc1-71f0-4e3b-b6f8-2c1731c6d617%2Fae00a22c-bfe8-4d5d-8dc9-0654994e1fc5%2Fhj4zs7m_processed.png&w=3840&q=75)
Transcribed Image Text:This is the fourth part of a four-part problem.
If the given solutions
ÿ₁(t)
ÿ'
=
-2
2t
-2t-2
t²
2t, ÿ₂(t) =
T2(t) = [† 7²¹].
1
form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem
1 – 2t-¹
2t-¹-2-2
2t
+2t
2
ÿ(5) = [-32]
"
t> 0,
impose the given initial condition and find the unique solution to the initial value problem for t > 0. If the given solutions do not form a fundamental
set, enter NONE in all of the answer blanks.
y(t) = (
2t
-
) [²³² 2 ² ² ] + (
2t
-
[t
[1¹].
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