Question

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

Short answer

The general solution to the linear system of differential equations is given by: \[ \begin{pmatrix} x(t)\\y(t)\end{pmatrix} = c_1 e^t \begin{pmatrix} 1\\ 2 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} 1\\ 1\end{pmatrix} \]

Step-by-step solution

Write the system in matrix form

Write the given system of linear differential equations in matrix form, such that $$\begin{pmatrix} x' \\ y' \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}$$, where A is the matrix of coefficients. This gives us: \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \]

Find eigenvalues and eigenvectors

Compute the eigenvalues and eigenvectors for the matrix A. To find the eigenvalues, solve the characteristic equation, which is given by $$\text{det}(A-\lambda I) = 0$$. \[ \begin{vmatrix} 3-\lambda & -1 \\ 1 &1-\lambda \end{vmatrix} = (3-\lambda)(1-\lambda) - (-1)(1) = \lambda^2 - 4\lambda + 4 -1 = \lambda^2 -4\lambda +3 \] The characteristic equation becomes $$\lambda^2 -4\lambda +3 = 0$$. To find its roots, we factor it: $$(\lambda-1)(\lambda-3)=0$$. Hence, the eigenvalues are 1 and 3. Now, we will find the eigenvectors associated with each eigenvalue. For $$\lambda_1=1$$, we have: \[ (A-\lambda_1 I) \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 1 & 0\end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} 0\\ 0\end{pmatrix} \] The eigenvector corresponding to eigenvalue 1 is \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\). For $$\lambda_2=3$$, we have: \[ (A-\lambda_2 I) \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & -2\end{pmatrix} \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0\\ 0\end{pmatrix} \] The eigenvector corresponding to eigenvalue 3 is \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

Express the general solution

Express the general solution using the eigenvalues and eigenvectors found in the previous step. The general solution to a system of linear differential equations is given by: $$\begin{pmatrix} x(t)\\y(t)\end{pmatrix} = c_1 e^{\lambda_1 t} \begin{pmatrix} x_1\\ y_1 \end{pmatrix} + c_2 e^{\lambda_2 t} \begin{pmatrix} x_2\\ y_2 \end{pmatrix}$$ Substituting the eigenvalues and eigenvectors, we obtain: $$\begin{pmatrix} x(t)\\y(t)\end{pmatrix} = c_1 e^{1\cdot t} \begin{pmatrix} 1\\ 2 \end{pmatrix} + c_2 e^{3\cdot t} \begin{pmatrix} 1\\ 1\end{pmatrix}$$ Thus, the general solution to the linear system of differential equations is given by: $$\begin{pmatrix} x(t)\\y(t)\end{pmatrix} = c_1 e^t \begin{pmatrix} 1\\ 2 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} 1\\ 1\end{pmatrix}$$

Key concepts

Eigenvalues

Eigenvalues are key to solving linear differential equations involving matrices. To find eigenvalues, solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \). Here, \( A \) is a matrix and \( I \) is the identity matrix of the same size. Solving the determinant gives a polynomial equation in terms of \( \lambda \), the eigenvalue. For the system \[ \begin{pmatrix} 3 & -1 \ 1 & 1 \end{pmatrix} \], the characteristic equation \( \lambda^2 - 4\lambda + 3 = 0 \) yields eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \). These values are crucial in finding specific solutions that simplify the differential system.

Eigenvectors

Once eigenvalues are determined, eigenvectors are needed to fully describe the solution to the differential equation. Eigenvectors work as direction indicators associated with each eigenvalue. For each eigenvalue \( \lambda \), solve \( (A - \lambda I) \mathbf{v} = \mathbf{0} \), where \( \mathbf{v} \) is an eigenvector. For \( \lambda_1 = 1 \), solve \( \begin{pmatrix} 2 & -1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \ y_1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). The corresponding eigenvector is \( \begin{pmatrix} 1 \ 2 \end{pmatrix} \). For \( \lambda_2 = 3 \), using \( \begin{pmatrix} 0 & -1 \ 1 & -2 \end{pmatrix} \), the eigenvector is \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \). This process connects each eigenvalue to its respective solution path.

Matrix Form

The matrix form of a linear system streamlines the process of solving differential equations. Representing the system \( x' = 3x - y \) and \( y' = x + y \) in matrix form, you have \[ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} 3 & -1 \ 1 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} \]. This compact representation allows us to utilize linear algebra techniques, simplifying the task of finding solutions using eigenvalues and eigenvectors. Each column of the matrix indicates the coefficient of the variables (\( x \) and \( y \)), making the transition from equations to a matrix straightforward and efficient for computations.

General Solution

The general solution to the differential system is built from the eigenvalues and eigenvectors we previously found. Using \( \begin{pmatrix} x(t) \ y(t) \end{pmatrix} = c_1 e^{\lambda_1 t} \begin{pmatrix} x_1 \ y_1 \end{pmatrix} + c_2 e^{\lambda_2 t} \begin{pmatrix} x_2 \ y_2 \end{pmatrix} \), substitute \( \lambda_1 = 1 \), \( \lambda_2 = 3 \), and their corresponding eigenvectors. The resulting general solution is \[ \begin{pmatrix} x(t) \ y(t) \end{pmatrix} = c_1 e^t \begin{pmatrix} 1 \ 2 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \]. This solution incorporates constants \( c_1 \) and \( c_2 \), representing initial conditions and the influence of each eigencomponent over time.