Question

Find the general solution of each of the linear systems in Exercises 1-26. $$ \begin{aligned} &x^{\prime}=x-3 y \\ &y^{\prime}=3 x+y \end{aligned} $$

Short answer

The general solution of the linear system \( \begin{aligned} &x^{\prime}=x-3 y \\ &y^{\prime}=3 x+y \end{aligned} \) is given by: \[ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = C_1 e^{4t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \] where C_1 and C_2 are constants determined by initial conditions.

Step-by-step solution

Write the system in matrix form

To write the given system of linear differential equations in matrix form, we need to express them as a single matrix equation: \[ \begin{pmatrix} x'\\ y' \end{pmatrix} = \begin{pmatrix} 1 & -3 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \] The resulting matrix equation is: \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix} \] where \[ A = \begin{pmatrix} 1 & -3 \\ 3 & 1 \end{pmatrix} \]

Find the eigenvalues and eigenvectors of A

To find the eigenvalues (λ) of A, we need to solve the characteristic polynomial equation: \[ \text{det}(A-\lambda I)=0 \] Where I is the identity matrix. For the matrix A, this equation becomes: \[ \text{det}\begin{pmatrix} 1-\lambda & -3 \\ 3 & 1-\lambda \end{pmatrix} = (1-\lambda)^{2} - (-3)(3) = 0 \] Solving for λ, we obtain the eigenvalues: \[ \lambda_1 = 4, \quad \lambda_2 = -2 \] Now, for each eigenvalue, we need to find the corresponding eigenvector. For λ_1 = 4, we solve the equation (A - 4I)x_1 = 0: \[ \begin{pmatrix} -3 & -3 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] We can choose the eigenvector x_1 = (1,1). For λ_2 = -2, we solve the equation (A + 2I)x_2 = 0: \[ \begin{pmatrix} 3 & -3 \\ 3 & 3 \end{pmatrix} \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] We can choose the eigenvector x_2 = (1, -1).

Write the general solution

Using the eigenvalues and their corresponding eigenvectors, we can now determine the general solution of the given system of linear differential equations. The general solution is given by: \[ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = C_1 e^{\lambda_1 t} x_1 + C_2 e^{\lambda_2 t} x_2 \] Plugging in our eigenvalues and eigenvectors, we get: \[ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = C_1 e^{4t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \] Where C_1 and C_2 are constants that can be determined by initial conditions. This is the general solution for the given linear system of differential equations.

Key concepts

Eigenvalues and Eigenvectors

When solving linear differential equations using matrices, eigenvalues and eigenvectors play a central role. They help us understand the behavior of the system. The matrix \(A\) from our exercise, \[\begin{pmatrix}1 & -3 \3 & 1\end{pmatrix}\] is key to finding these values.
The eigenvalues \(\lambda\) are found by solving the characteristic equation \(\text{det}(A - \lambda I) = 0\), which simplifies to a quadratic equation. In our case: \[(1-\lambda)^{2} + 9 = 0\]
Solved, this gives two solutions, \(\lambda_1 = 4\) and \(\lambda_2 = -2\).
Next, we find eigenvectors by substituting each eigenvalue back into the equation \((A - \lambda I)x = 0\). This yields directions that the system follows as it evolves over time. For example, for \(\lambda_1 = 4\), we have the eigenvector \(\begin{pmatrix} 1 \ 1 \end{pmatrix}\), while for \(\lambda_2 = -2\), we find \(\begin{pmatrix} 1 \ -1 \end{pmatrix}\).
These results provide insight into the solution's growth or decay patterns, helping us understand system dynamics more clearly.

General Solution

The general solution is an expression that includes all possible solutions of the differential equation system. In our context, \[\begin{pmatrix} x(t) \ y(t) \end{pmatrix} = C_1 e^{\lambda_1 t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + C_2 e^{\lambda_2 t} \begin{pmatrix} 1 \ -1 \end{pmatrix}\]
reflects this concept.
This expression is constructed using the eigenvalues and eigenvectors found earlier. Here, \(C_1\) and \(C_2\) are constants. They are arbitrary without specific initial conditions.
The terms \(e^{\lambda_1 t}\) and \(e^{\lambda_2 t}\) show exponential behavior. Positive eigenvalues, like \(\lambda_1 = 4\), indicate growth. Negative eigenvalues, like \(\lambda_2 = -2\), imply decay. Using these properties, the general solution captures all possible cases of the system's evolution.

Matrix Form of Differential Equations

Expressing a system of linear differential equations in matrix form simplifies analysis. It condenses the system into a compact expression. Originally given as:\[\begin{aligned}x' &= x - 3y \y' &= 3x + y\end{aligned}\]
This was rewritten in matrix form as \[\begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} 1 & -3 \ 3 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}\].
Here, the left side of the equation represents the derivatives \(x'\) and \(y'\), indicating change over time. The right side involves matrix multiplication, with matrix \(A\) acting on vector \(\begin{pmatrix} x \ y \end{pmatrix}\).
Using matrix form makes it more straightforward to apply strategies like eigenvalue analysis. It's a powerful tool for visualizing and solving systems of differential equations.

Initial Conditions

Initial conditions provide specific values at \(t=0\) to determine unique solutions. Until specific initial conditions are given, the solution involves arbitrary constants \(C_1\) and \(C_2\).
To find these constants, set initial values for \(x(0)\) and \(y(0)\). For example, if given \(x(0) = 2\) and \(y(0) = 3\), substitute these into the general solution:
1. Use \(\begin{pmatrix} x(0) \ y(0) \end{pmatrix} = C_1 \begin{pmatrix} 1 \ 1 \end{pmatrix} + C_2 \begin{pmatrix} 1 \ -1 \end{pmatrix}\).
2. Solve for \(C_1\) and \(C_2\) using known values.
Initial conditions help frame specific scenarios, tailoring the general solution to particular instances or experiments.