Question

Berechnen Sie die Lösung des Anfangswertproblems $$ \left(\begin{array}{l} y_{1}^{\prime} \\ y_{2}^{\prime} \end{array}\right)=\left(\begin{array}{ll} 1 & 2 \\ 4 & 3 \end{array}\right)\left(\begin{array}{l} y_{1} \\ y_{2} \end{array}\right)+\left(\begin{array}{l} 2 \\ 0 \end{array}\right), \quad\left(\begin{array}{l} y_{1}(0) \\ y_{2}(0) \end{array}\right)=\left(\begin{array}{l} 1 \\ 0 \end{array}\right) $$

Short answer

The solution is \( \begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix} = \frac{11}{9} \begin{pmatrix} 1 \\ 2 \end{pmatrix} e^{5t} + \frac{7}{9} \begin{pmatrix} -1 \\ 2 \end{pmatrix} e^{-t} + \begin{pmatrix} -2 \\ \frac{8}{3} \end{pmatrix} \).

Step-by-step solution

Write the System of Differential Equations

The given system can be interpreted as two simultaneous differential equations:1. \( y_1' = 1\cdot y_1 + 2\cdot y_2 + 2 \)2. \( y_2' = 4\cdot y_1 + 3\cdot y_2 \).

Solve the Homogeneous System

First, solve the homogeneous system without the constant vector \( \begin{pmatrix} 2 \ 0 \end{pmatrix} \):\[ \begin{pmatrix} y_1' \ y_2' \end{pmatrix} = \begin{pmatrix} 1 & 2 \ 4 & 3 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix} \].To solve this, find the eigenvalues of the matrix:\[ \begin{vmatrix} 1-\lambda & 2 \ 4 & 3-\lambda \end{vmatrix} = 0 \].This simplifies to \( (1-\lambda)(3-\lambda) - 8 = \lambda^2 - 4\lambda - 5 = 0 \), giving eigenvalues \( \lambda_1 = 5, \lambda_2 = -1 \).

Find Eigenvectors

For \( \lambda_1 = 5 \), solve:\[ \begin{pmatrix} 1-5 & 2 \ 4 & 3-5 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]. This simplifies to finding the null space of the matrix \( \begin{pmatrix} -4 & 2 \ 4 & -2 \end{pmatrix} \), giving eigenvector \( \begin{pmatrix} 1 \ 2 \end{pmatrix} \).For \( \lambda_2 = -1 \), solve:\[ \begin{pmatrix} 1+1 & 2 \ 4 & 3+1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]. This results in eigenvector \( \begin{pmatrix} -1 \ 2 \end{pmatrix} \).

Construct General Solution for Homogeneous System

The general solution of the homogeneous system is:\[ \begin{pmatrix} y_1(t) \ y_2(t) \end{pmatrix} = C_1 \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{5t} + C_2 \begin{pmatrix} -1 \ 2 \end{pmatrix} e^{-t} \],where \(C_1\) and \(C_2\) are constants to be determined from initial conditions.

Determine Particular Solution

Assume a particular solution of the form:\( \begin{pmatrix} y_1 \ y_2 \end{pmatrix} = \begin{pmatrix} A \ B \end{pmatrix} \).Substitute into the equation:\[ \begin{pmatrix} 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 & 2 \ 4 & 3 \end{pmatrix} \begin{pmatrix} A \ B \end{pmatrix} + \begin{pmatrix} 2 \ 0 \end{pmatrix} \].Solve \( A + 2B + 2 = 0 \) and \( 4A + 3B = 0 \) to find \(A = -2, B = \frac{8}{3}\).

Apply Initial Conditions

The full solution is \( \begin{pmatrix} y_1(t) \ y_2(t) \end{pmatrix} = C_1 \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{5t} + C_2 \begin{pmatrix} -1 \ 2 \end{pmatrix} e^{-t} + \begin{pmatrix} -2 \ \frac{8}{3} \end{pmatrix} \).Using initial condition \( y_1(0)=1 \) and \( y_2(0)=0 \), solve:\[ C_1 \begin{pmatrix} 1 \ 2 \end{pmatrix} + C_2 \begin{pmatrix} -1 \ 2 \end{pmatrix} + \begin{pmatrix} -2 \ \frac{8}{3} \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \].This results in \( C_1 = \frac{11}{9}, C_2 = \frac{7}{9} \).

Write Final Solution

The final solution to the system is:\( \begin{pmatrix} y_1(t) \ y_2(t) \end{pmatrix} = \frac{11}{9} \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{5t} + \frac{7}{9} \begin{pmatrix} -1 \ 2 \end{pmatrix} e^{-t} + \begin{pmatrix} -2 \ \frac{8}{3} \end{pmatrix} \).

Key concepts

Homogeneous System

A homogeneous system of differential equations is a system in which all the terms are linear and dependent only on the system variables and their derivatives, with no added constants. Let's break it down:

Consider a system of differential equations given by \( \mathbf{Y}' = A \mathbf{Y} \), where \( \mathbf{Y} \) represents a vector of functions, \( A \) is a matrix of coefficients, and \( \mathbf{Y}' \) is the derivative with respect to time \( t \). The system is called homogeneous because there are no constant inputs; all terms stem from the linear interactions among the system variables.

When solving such a system, we aim to find the eigenvalues and eigenvectors of the matrix \( A \). These values help us derive the general solution that describes the system's behavior over time.

In our exercise, the homogeneous part of the system was given by the differential equations excluding the constant vector:\[ \begin{pmatrix} y_1' \ y_2' \end{pmatrix} = \begin{pmatrix} 1 & 2 \ 4 & 3 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix} \].

This form is crucial for finding the characteristic equation, which provides the eigenvalues and allows the construction of the system's complementary solution.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are key concepts in solving systems of differential equations, especially homogeneous systems. They help express complex systems in manageable ways.

Let's break it down:

• **Eigenvalues**: These are solutions to the characteristic equation, given by \( |A - \lambda I| = 0 \), where \( A \) is the matrix from the system, \( \lambda \) are the eigenvalues, and \( I \) is the identity matrix.


• **Eigenvectors**: For each eigenvalue, an eigenvector is a non-zero vector \( v \) that satisfies \( (A - \lambda I)v = 0 \). These vectors provide the direction associated with each eigenvalue.

In our exercise, the matrix \( A \) was \( \begin{pmatrix} 1 & 2 \ 4 & 3 \end{pmatrix} \). Solving \( (A - \lambda I) = 0 \) produced the eigenvalues \( \lambda_1 = 5 \) and \( \lambda_2 = -1 \). For these eigenvalues, the respective eigenvectors \( \begin{pmatrix} 1 \ 2 \end{pmatrix} \) and \( \begin{pmatrix} -1 \ 2 \end{pmatrix} \) were found.

These results enable us to formulate the general solution of the homogeneous system as a sum of terms involving these eigenvectors and \( e^{\lambda t} \), each term weighted by constants determined by initial conditions.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation with given conditions at a specific starting point. It seeks a particular solution that fits both the differential equation and the initial conditions.

Let's examine our problem:

We started with a system of differential equations with initial conditions \( y_1(0) = 1 \) and \( y_2(0) = 0 \). The goal was to find functions \( y_1(t) \) and \( y_2(t) \) that satisfy these conditions.

The process is as follows:

• **Solve The Homogeneous System**: Discover the general solution using eigenvalues and eigenvectors.


• **Find A Particular Solution**: Assume a form that includes any constant vector terms in the right side of the equation.


• **Apply Initial Conditions**: Use the initial conditions to determine unknown constants in the general solution.

In our exercise, this involved combining the homogeneous and particular solutions then solving for constants \( C_1 \) and \( C_2 \) so the initial conditions were satisfied. This resulted in our specific solution fitting all given requirements, ensuring it accurately reflects the system's behavior from the start.