Chapter 7: Problem 2
Express the general solution of the given system of equations in terms of real-valued functions. In each of Problems 1 through 6 also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-1} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x} $$
Short Answer
Expert verified
Solving the characteristic equation, we get:
$$
(-1-\lambda)((-1)-\lambda) - (-4)(1) = 0
$$
Expanding and simplifying,
$$
\lambda^2 + 2\lambda + 1 - 4 = 0 \Rightarrow \lambda^2 + 2\lambda - 3 = 0
$$
Factoring the quadratic equation gives:
$$
(\lambda - 1)(\lambda + 3) = 0
$$
So, the eigenvalues are:
$$
\lambda_1 = 1 \quad \text{and} \quad \lambda_2 = -3
$$
#tag_title#Step 2: Calculate the eigenvectors of the matrix#tag_content#Now we'll find the eigenvectors for each eigenvalue. For each eigenvalue, we'll solve the following system of linear equations:
$$
(A - \lambda I)\vec{v} = 0
$$
For $\lambda_1 = 1,$ we have:
$$
\left(\begin{array}{rr}{-1-1} & {-4} \\\ {1} & {-1-1}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
$$
\left(\begin{array}{rr}{-2} & {-4} \\\ {1} & {-2}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
Solving this system of equations, we get the eigenvector:
$$
\vec{v}_{1} = \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
For $\lambda_2 = -3,$ we have:
$$
\left(\begin{array}{rr}{-1-(-3)} & {-4} \\\ {1} & {-1-(-3)}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
$$
\left(\begin{array}{rr}{2} & {-4} \\\ {1} & {2}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
Solving this system of equations, we get the eigenvector:
$$
\vec{v}_{2} = \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
#tag_title#Step 3: Construct the general solution#tag_content#Now that we have the eigenvalues and eigenvectors, we can construct the general solution of the given system of linear differential equations using the following formula:
$$
\vec{x}(t) = c_{1} e^{\lambda_1 t} \vec{v}_{1} + c_{2} e^{\lambda_2 t} \vec{v}_{2}
$$
Substituting the eigenvalues and eigenvectors:
$$
\vec{x}(t) = c_{1} e^{\left(1\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right) + c_{2} e^{\left(-3\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
#tag_title#Step 4: Analyze the behavior as \(t \rightarrow \infty\) #tag_content#To analyze the behavior of the general solution as \(t \rightarrow \infty\), we consider the exponential terms:
- For the first term, $e^{(1)t}$, as \(t \rightarrow \infty\), the term will grow exponentially, making the solution unbounded.
- For the second term, $e^{(-3)t}$, as \(t \rightarrow \infty\), the term will tend to zero, making its contribution in the general solution nearly negligible.
Thus, as \(t \rightarrow \infty\), the behavior of the solution is noted to grow exponentially, primarily driven by the first term. The general solution becomes unbounded.
In conclusion, the general solution of the given system of linear differential equations is:
$$
\vec{x}(t) = c_{1} e^{\left(1\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right) + c_{2} e^{\left(-3\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
And as \(t \rightarrow \infty\), the solution is unbounded, growing exponentially due to the first term.
Step by step solution
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