dX -(t) = CX(t) 2 C = c=(; ;). 3 0 -1 (). 1 vi = and v2 =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![Problem 9
§4.5, Exercise 3. Consider
\[
\frac{dX}{dt} = CX(t) \quad (4.5.10)
\]
where
\[
C = \begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}.
\]
Let
\[
v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix},
\]
and let
\[
Y(t) = e^{2t}v_1 \quad \text{and} \quad Z(t) = e^{-t}v_2.
\]
(a) Show that \(Y(t)\) and \(Z(t)\) are solutions to (4.5.10). Use the principle of superposition to observe that \(X(t) = \alpha Y(t) + \beta Z(t)\) is a solution to (4.5.10).
(b) Using the solution found in (a), find a solution \(X(t)\) to (4.5.10) such that
\[
X(0) = \begin{pmatrix} 3 \\ -1 \end{pmatrix}.
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9fd5ba49-7d01-4178-9fd2-555b265ff7cd%2F4bba6972-6c10-46ce-9b90-824828402883%2Fw5f09tf_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 9
§4.5, Exercise 3. Consider
\[
\frac{dX}{dt} = CX(t) \quad (4.5.10)
\]
where
\[
C = \begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}.
\]
Let
\[
v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix},
\]
and let
\[
Y(t) = e^{2t}v_1 \quad \text{and} \quad Z(t) = e^{-t}v_2.
\]
(a) Show that \(Y(t)\) and \(Z(t)\) are solutions to (4.5.10). Use the principle of superposition to observe that \(X(t) = \alpha Y(t) + \beta Z(t)\) is a solution to (4.5.10).
(b) Using the solution found in (a), find a solution \(X(t)\) to (4.5.10) such that
\[
X(0) = \begin{pmatrix} 3 \\ -1 \end{pmatrix}.
\]
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