Question

Find the general solution of the given system. X' = (-1 -7 7 13) X X (t) =

Answer 1

To find the general solution of the given system of equations, we need to find the eigenvalues and eigenvectors of the coefficient matrix. By solving the characteristic equation, we can find the eigenvalues, and by solving (A - λI)v = 0, we can find the eigenvectors. The general solution is X(t) = c1 * exp(4t) * v1 + c2 * exp(7t) * v2 + c3 * exp(11t) * v3, where c1, c2, and c3 are constants.

Step-by-step explanation:

The given system of equations is:
X' = (-1 -7 7 13) X
To find the general solution of this system, we need to solve for X.
Let's write the equation in matrix form:
X' = A X
where A is the coefficient matrix.
Now, let's find the eigenvalues and eigenvectors of A.
By solving the characteristic equation of A, we can find the eigenvalues:
det(A - λI) = 0
Let λ be the eigenvalue:
(λ + 1)(λ - 7)(λ - 13) + 7(-7)(-1) - 7(13)(λ) = 0
Simplifying this equation, we get:
λ^3 - λ^2 - 40λ + 266 = 0
By solving this equation, we find the eigenvalues:
λ1 = 4
λ2 = 7
λ3 = 11
Now, let's find the eigenvectors corresponding to each eigenvalue.
For λ1 = 4:
(A - 4I) v = 0
-5 -7 7 9 v = 0
Using row reduction, we find the eigenvector:
v1 = (1, -1, 1, -5)
For λ2 = 7:
(A - 7I) v = 0
-8 -7 7 6 v = 0
Using row reduction, we find the eigenvector:
v2 = (1, -1, 1, -1)
For λ3 = 11:
(A - 11I) v = 0
-12 -7 7 2 v = 0
Using row reduction, we find the eigenvector:
v3 = (1, -5, 1, -2)
Now, the general solution for X(t) can be written as:
X(t) = c1 * exp(4t) * v1 + c2 * exp(7t) * v2 + c3 * exp(11t) * v3
where c1, c2, and c3 are constants.
This is the general solution of the given system of equations.

Answer 2

the eigenvalues are
`\(-1\) and \(5\),`with corresponding eigenvectors
`\(X(t) = C_1 \cdot e^(-t) \begin{pmatrix} 3 \\ 1 \\ 1 \\ 0 \end{pmatrix} + C_2 \cdot e^(5t) \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix}\)`

Step-by-step explanation:

The given system can be solved by finding the eigenvalues and eigenvectors of the matrix \
`((-1, -7, 7, 13)\).` Upon calculation, the eigenvalues are
`\(-1\) and \(5\),`with corresponding eigenvectors
`\(\begin{pmatrix} 3 \\ 1 \\ 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix}\), respectively.`

Using these eigenvalues and eigenvectors, the general solution of the system is expressed as a linear combination of the eigenvectors multiplied by exponential terms with their respective eigenvalues raised to the power of time.
`\(C_1\) and \(C_2\)` are constants determined by initial conditions.

This solution represents the behavior of the system over time, showcasing the exponential growth and decay components corresponding to the eigenvalues \(5\) and \(-1\) respectively, each scaled by its corresponding eigenvector.

This method is fundamental in solving systems of linear differential equations, providing insights into the system's dynamics by breaking it down into simpler exponential components governed by the eigenvalues and eigenvectors of the matrix involved in the system.

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