Problem 2 - Arbeitsbuch Mathematik für Ingenieure

Question

Bestimmen Sie die allgemeine reelle Lösung des folgenden Systems linearer Differentialgleichungen $$ \begin{aligned} &y_{1}^{\prime}=2 \cdot y_{1}+y_{2} \\ &y_{2}^{\prime}=-y_{1}+2 \cdot y_{2} \end{aligned} $$

Step-by-Step Solution

Answer

The general solution is \[ Y(t) = e^{2t} \begin{pmatrix} c_1 \cos(t) + c_2 \sin(t) \ c_2 \cos(t) - c_1 \sin(t) \end{pmatrix} \].

Step 1: Identify Eigenvalues

For a system of linear differential equations, we first write the system in matrix form. Here we have: \[ Y' = \begin{pmatrix} y_1' \ y_2' \end{pmatrix} = \begin{pmatrix} 2 & 1 \ -1 & 2 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix} \]To find the general solution, we start by finding the eigenvalues of the coefficient matrix \(A = \begin{pmatrix} 2 & 1 \ -1 & 2 \end{pmatrix} \). This involves solving the characteristic equation \( \text{det}(A - \lambda I) = 0 \). So we need: \[ \text{det} \begin{pmatrix} 2 - \lambda & 1 \ -1 & 2 - \lambda \end{pmatrix} = 0 \]}\[ (2 - \lambda)^2 + 1 = 0 \]\[ \lambda^2 - 4\lambda + 5 = 0 \]

Step 2: Solve for Eigenvalues

Solve the characteristic polynomial \( \lambda^2 - 4\lambda + 5 = 0 \) using the quadratic formula: \[ \lambda = \frac{4 \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{4 \pm \sqrt{-4}}{2} = 2 \pm i \].Thus, the eigenvalues are \( \lambda_1 = 2 + i \) and \( \lambda_2 = 2 - i \).

Step 3: Find Eigenvectors

Now, find the corresponding eigenvectors. For \( \lambda = 2 + i \), solve the equation \( (A - (2+i)I)v = 0 \). This yields: \[ \begin{pmatrix} -i & 1 \ -1 & -i \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \]From the first equation: \( -iv_1 + v_2 = 0 \) gives \( v_2 = iv_1 \). Hence, an eigenvector corresponding to \( 2 + i \) is \( v = \begin{pmatrix} 1 \ i \end{pmatrix} \).For \( \lambda = 2 - i \), following a similar process, we get an eigenvector \( v = \begin{pmatrix} 1 \ -i \end{pmatrix} \).

Step 4: Construct General Solution

The general solution is given by: \[ Y(t) = c_1 e^{(2+i)t} \begin{pmatrix} 1 \ i \end{pmatrix} + c_2 e^{(2-i)t} \begin{pmatrix} 1 \ -i \end{pmatrix} \].Combining real and imaginary parts, the solution simplifies to: \[ Y(t) = e^{2t} (c_1 \begin{pmatrix} \cos(t) \ -\sin(t) \end{pmatrix} - i\sin(t) \begin{pmatrix} 1 \ 0 \end{pmatrix}) + c_2 e^{2t} (\begin{pmatrix} \cos(t) \ \sin(t) \end{pmatrix} + i\sin(t) \begin{pmatrix} 0 \ 1 \end{pmatrix}) \].Simplify further to get real solutions.

Key Concepts

Eigenvalues

Eigenvalues are key numbers associated with a square matrix. They are essential in solving linear differential equations. To find eigenvalues, you need to form the characteristic equation from the coefficient matrix. In our problem, the coefficient matrix is: $$ A = \begin{pmatrix} 2 & 1 \ -1 & 2 \end{pmatrix} $$.
To find its eigenvalues, solve the characteristic equation$$ \text{det}(A - \lambda I) = 0 $$.
This results in solving$$ \text{det} \begin{pmatrix} 2 - \lambda & 1 \ -1 & 2 - \lambda \end{pmatrix} = 0 $$which simplifies to $$ (2 - \lambda)^2 + 1=0 \Rightarrow \lambda^2 - 4\lambda + 5 = 0 . $$
Using the quadratic formula, we get:$$ \lambda = \frac{4 \pm \sqrt{16 - 20}}{2} = 2 \pm i.$$
Thus, the eigenvalues are complex numbers \( 2 + i \) and \( 2 - i \).

Eigenvectors

Eigenvectors accompany the eigenvalues and are non-zero vectors that, when multiplied by the matrix, scale by the eigenvalue rather than change direction. Once the eigenvalues are determined, we can find their corresponding eigenvectors.
For \( \lambda = 2 + i \), solve$$ (A - (2 + i)I)v = 0 $$,
leading to the system:$$ \begin{pmatrix} -i & 1 \ -1 & -i \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 .$$
From the equation, we get:$$ -i \, v_1 + v_2 = 0 \Rightarrow v_2 = i \, v_1.$$
An eigenvector corresponding to \( 2 + i \) is thus \( v = \begin{pmatrix} 1 & i \end{pmatrix}^T \).
Repeating for \( \lambda = 2 - i \), we find the eigenvector:$$ v = \begin{pmatrix} 1 & -i \end{pmatrix}^T .$$

Characteristic Equation

The characteristic equation is essential for finding eigenvalues. It is formed by setting the determinant of \( A - \lambda I \) to zero, where \( A \) is the coefficient matrix, \( \lambda \) is a scalar (eigenvalue), and \( I \) is the identity matrix of the same size as \( A \).
For our problem: $$ A = \begin{pmatrix} 2 & 1 \ -1 & 2 \end{pmatrix} , \ $$ the identity matrix is $$ I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} ,$$ so we compute:$$ \text{det}(A - \lambda I) = \text{det} \begin{pmatrix} 2 - \lambda & 1 \ -1 & 2 - \lambda \end{pmatrix} .$$
Solving the determinant results in the quadratic equation: $$ \lambda^2 - 4\lambda + 5 = 0,$$ whose roots yield the eigenvalues \( 2 + i \) and \( 2 - i \).

General Solution

The general solution to a system of linear differential equations is constructed using the eigenvalues and eigenvectors. It is a combination of exponential functions multiplied by the eigenvectors.
For the matrix in our problem, the eigenvalues are \( \lambda_1 = 2 + i \) and \( \lambda_2 = 2 - i \), and their corresponding eigenvectors are \( \begin{pmatrix} 1 \ i \end{pmatrix} \) and \( \begin{pmatrix} 1 \ -i \end{pmatrix} \), respectively.
Therefore, the general solution is:$$ Y(t) = c_1 e^{(2+i)t} \begin{pmatrix} 1 \ i \end{pmatrix} + c_2 e^{(2-i)t} \begin{pmatrix} 1 \ -i \end{pmatrix}. $$
By expressing these in terms of real functions (sine and cosine), and splitting into real and imaginary parts, we get the complete form of the general solution in real terms:

$$ Y(t) = e^{2t} \begin{pmatrix} c_1 (\text{cos}(t) + i\text{sin}(t)) \ c_2 (\text{cos}(t) - i\text{sin}(t)) \end{pmatrix} \Rightarrow e^{2t} \begin{pmatrix} A\text{cos}(t) - B\text{sin}(t) \ B\text{cos}(t) + A\text{sin}(t) \end{pmatrix}. $$