Question

Find the solution of the given initial value problem. Draw the trajectory of the solution in the \(x_{1} x_{2}-\) plane and also the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{3} & {9} \\ {-1} & {-3}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{2} \\ {4}\end{array}\right) $$

Short answer

Based on the given initial value problem, we found a solution of: $$ \mathbf{x}(t)=-e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+2e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$ When analyzing the behavior of the solution as \(t \rightarrow \infty\), we observe that \(\mathbf{x}(t)\) approaches the zero vector: $$ \lim_{t \to \infty} \mathbf{x}(t) = \mathbf{0} $$

Step-by-step solution

Finding Eigenvalues

We start by finding the eigenvalues \(\lambda\) of the matrix: $$ A = \left(\begin{array}{rr}{-3} & {2} \\\ {-1} & {-1}\end{array}\right) $$ We find the eigenvalues by solving the characteristic equation \(\text{det}(A-\lambda I) =0\). In our case, for the given matrix: $$ \text{det}\left(\begin{array}{cc}{(-3-\lambda)} & {2} \\\ {-1} & {(-1-\lambda)}\end{array}\right) = (-3-\lambda)((-1)-\lambda) - 2(-1) = 0 $$

Solving the characteristic equation

Now we solve the characteristic equation: $$ (-3-\lambda)((-1)-\lambda) +2= 0 \Rightarrow \lambda^2 +4\lambda +3 = 0 $$ Applying the quadratic equation formula to find the eigenvalues, we get: $$ \lambda_1 = \frac{-4+\sqrt{(-4)^2-4(3)}}{2} = -1 \\ \\ \lambda_2 = \frac{-4-\sqrt{(-4)^2-4(3)}}{2} = -3 $$

Finding eigenvectors

Next, we find the eigenvectors corresponding to each eigenvalue by solving the system \((A-\lambda_iI)\mathbf{v}_i = \mathbf{0}\). For \(\lambda_1 = -1\): $$ \left(\begin{array}{cc}{-2} & {2} \\\ {-1} & {0}\end{array}\right) \left(\begin{array}{c}{v_{11}} \\\ {v_{12}}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ The corresponding eigenvector is: $$ \mathbf{v}_{1}=\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) $$ For \(\lambda_2 = -3\): $$ \left(\begin{array}{cc}{0} & {2} \\\ {-1} & {2}\end{array}\right) \left(\begin{array}{c}{v_{21}} \\\ {v_{22}}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ The corresponding eigenvector is: $$ \mathbf{v}_{2}=\left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$

General solution of the system

With the eigenvalues and eigenvectors, we write down the general solution of the system: $$ \mathbf{x}(t)=c_{1} e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+c_{2} e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$

Apply the initial condition

Next, we use the initial condition \(\mathbf{x}(0)=\left(\begin{array}{r}{1} \\\ {-2}\end{array}\right)\) to find the coefficients \(c_1\) and \(c_2\). Plugging in \(t=0\): $$ \left(\begin{array}{c}{1} \\\ {-2}\end{array}\right) = c_{1} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + c_{2} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$ Solving the system of linear equations, we get \(c_1=-1\) and \(c_2=2\). Thus, the solution to the initial value problem is: $$ \mathbf{x}(t)=-e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+2e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$

Behavior as \(t\rightarrow \infty\)

As \(t \rightarrow \infty\), both exponentials, \(e^{-t}\) and \(e^{-3t}\), tend to zero. Therefore, the solution \(\mathbf{x}(t)\) approaches the zero vector: $$ \lim_{t \to \infty} \mathbf{x}(t) = \lim_{t \to \infty} \left( -e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+2e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) \right) = \mathbf{0} $$