Assume that Y₁ = k₁e-2 [1] and Y₂ = k₂e6t [1] are linearly inde- pendent solutions to the system = AY. Find the values of k₁ and k₂ for the solution with initial condition Y (0) = [ }]. k₁ + = k₂ = + ▾
Assume that Y₁ = k₁e-2 [1] and Y₂ = k₂e6t [1] are linearly inde- pendent solutions to the system = AY. Find the values of k₁ and k₂ for the solution with initial condition Y (0) = [ }]. k₁ + = k₂ = + ▾
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
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![Assume that Y₁ = k₁e-2 [1]
and Y₂ = k₂e6t
[²]
pendent solutions to the system = AY. Find the values of k₁ and
Y(O) = [²].
k₂
for the solution with initial condition Y (0)
k₁ +
=
k₂
=
+
▾
are linearly inde-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e6acd23-120e-474d-b2a7-11ec3e3e21ae%2F3f6a4771-a1cf-4d67-bb84-7bc8d5ce9de5%2Ffsyfzff_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Assume that Y₁ = k₁e-2 [1]
and Y₂ = k₂e6t
[²]
pendent solutions to the system = AY. Find the values of k₁ and
Y(O) = [²].
k₂
for the solution with initial condition Y (0)
k₁ +
=
k₂
=
+
▾
are linearly inde-
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