First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. -27 -31 -1 x' = 1 x x₁ = e X2 X3 24 28 -24 -24 3 =0=×₁' 4e-3t -27 -31 -1 24 28 1 3 -24 -24 -27-31 -1 24 28 1 3 H -24 -24
First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. -27 -31 -1 x' = 1 x x₁ = e X2 X3 24 28 -24 -24 3 =0=×₁' 4e-3t -27 -31 -1 24 28 1 3 -24 -24 -27-31 -1 24 28 1 3 H -24 -24
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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![### Verifying Solutions and Linear Independence in a System of Differential Equations
#### System of Differential Equations
Given the system of differential equations:
\[ x' = \begin{pmatrix}
-27 & -31 & -1 \\
24 & 28 & 1 \\
-24 & -24 & 3
\end{pmatrix} x \]
We need to verify that the given vectors are solutions for this system.
#### Given Vectors
The given vectors are:
\[ x_1 = e^{-3t} \begin{pmatrix}
5 \\
4 \\
4
\end{pmatrix}, \quad x_2 = e^{3t} \begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}, \quad x_3 = e^{4t} \begin{pmatrix}
1 \\
-1 \\
0
\end{pmatrix} \]
#### Verification of Solutions
To verify, we substitute these vectors in the differential equation and check the equality as follows:
For \( x_1 \):
\[
\begin{pmatrix}
-27 & -31 & -1 \\
24 & 28 & 1 \\
-24 & -24 & 3
\end{pmatrix}
\begin{pmatrix}
5 e^{-3t} \\
4 e^{-3t} \\
4 e^{-3t}
\end{pmatrix}
\]
Results in:
\[
\boxed{
= x_1'
}
\]
...
(The rest of the verification process continues similarly for \( x_2 \) and \( x_3 \)).
#### Wronskian Matrix
To demonstrate that the vectors \( x_1 \), \( x_2 \), and \( x_3 \) are linearly independent, we need to compute their Wronskian.
\[ W(x_1, x_2, x_3) = \begin{vmatrix}
5e^{-3t} & e^{3t} & e^{4t} \\
4e^{-3t} & e^{3t} & -e^{4t} \\
4e^{-3t} & e^{3t} & 0
\end{vmatrix} \]
By evaluating the determinant of the Wronskian matrix, if it is non-zero, it implies that the vectors \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06bbd155-c9d1-4bdc-8881-0cd893ac016a%2Ff6e6ae9d-5387-4fb0-9546-581d15d1255b%2F9r2ckfo_processed.png&w=3840&q=75)
Transcribed Image Text:### Verifying Solutions and Linear Independence in a System of Differential Equations
#### System of Differential Equations
Given the system of differential equations:
\[ x' = \begin{pmatrix}
-27 & -31 & -1 \\
24 & 28 & 1 \\
-24 & -24 & 3
\end{pmatrix} x \]
We need to verify that the given vectors are solutions for this system.
#### Given Vectors
The given vectors are:
\[ x_1 = e^{-3t} \begin{pmatrix}
5 \\
4 \\
4
\end{pmatrix}, \quad x_2 = e^{3t} \begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}, \quad x_3 = e^{4t} \begin{pmatrix}
1 \\
-1 \\
0
\end{pmatrix} \]
#### Verification of Solutions
To verify, we substitute these vectors in the differential equation and check the equality as follows:
For \( x_1 \):
\[
\begin{pmatrix}
-27 & -31 & -1 \\
24 & 28 & 1 \\
-24 & -24 & 3
\end{pmatrix}
\begin{pmatrix}
5 e^{-3t} \\
4 e^{-3t} \\
4 e^{-3t}
\end{pmatrix}
\]
Results in:
\[
\boxed{
= x_1'
}
\]
...
(The rest of the verification process continues similarly for \( x_2 \) and \( x_3 \)).
#### Wronskian Matrix
To demonstrate that the vectors \( x_1 \), \( x_2 \), and \( x_3 \) are linearly independent, we need to compute their Wronskian.
\[ W(x_1, x_2, x_3) = \begin{vmatrix}
5e^{-3t} & e^{3t} & e^{4t} \\
4e^{-3t} & e^{3t} & -e^{4t} \\
4e^{-3t} & e^{3t} & 0
\end{vmatrix} \]
By evaluating the determinant of the Wronskian matrix, if it is non-zero, it implies that the vectors \(
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