Chapter 4: Problem 5
In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(-1)=\left[\begin{array}{l} 2 \\ 2 \end{array}\right] $$
Short Answer
Expert verified
Based on the given homogeneous linear system and initial value problem, the solution is as follows:
$$\mathbf{y}(t)=\frac{2(1+e^4)}{2e^3}e^{-t}\begin{bmatrix} 1 \\ -1 \end{bmatrix} + \frac{2(1+e^4)}{2e^3}e^{3t}\begin{bmatrix} 1 \\ 1\end{bmatrix}.$$
Step by step solution
01
Find Eigenvalues and Eigenvectors
In order to find the eigenvalues, we need to solve the characteristic equation given by $$\text{det}(A - \lambda I)=0$$ where $$A=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$$ and $$I$$ denotes the identity matrix. The characteristic equation can be written as
$$(1-\lambda)^2 - 2^2 = (\lambda - 1)^2 - 4 = \lambda^2 - 2\lambda - 3 = 0.$$
Solving this equation, we find two eigenvalues, which are $$\lambda_1 = -1$$ and $$\lambda_2 = 3$$.
Next, we need to find the eigenvectors associated with these eigenvalues. For $$\lambda_1 = -1$$, we solve the linear system $$ (A - (-1)I) \mathbf{x} = 0$$ and for $$\lambda_2 = 3$$, we solve the linear system $$(A - 3I) \mathbf{x} = 0$$.
For $$\lambda_1 = -1$$, the system becomes
$$\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}\mathbf{x}=\mathbf{0},$$
which gives us the eigenvector $$\mathbf{x}^{(1)}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$$.
For $$\lambda_2 = 3$$, the system becomes
$$\begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}\mathbf{x}=\mathbf{0},$$
which gives us the eigenvector $$\mathbf{x}^{(2)}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$.
02
Find the General Solution
Now that we have found the eigenvalues and eigenvectors, we can find the general solution of the homogeneous linear system. The general solution is given by
$$\mathbf{y}(t)=c_1 e^{\lambda_1 t}\mathbf{x}^{(1)} + c_2 e^{\lambda_2 t}\mathbf{x}^{(2)},$$
where $$c_1$$ and $$c_2$$ are constants to be determined. Substituting the eigenvalues and eigenvectors we found in Step 1, we obtain
$$\mathbf{y}(t)=c_1 e^{-t}\begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{3t}\begin{bmatrix} 1 \\ 1\end{bmatrix}.$$
03
Determine the Constants
Finally, we need to impose the initial condition $$\mathbf{y}(-1)=\begin{bmatrix} 2 \\ 2 \end{bmatrix}$$ to find the values of $$c_1$$ and $$c_2$$. We have
$$\mathbf{y}(-1)=c_1 e^{-(-1)}\begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{3(-1)}\begin{bmatrix} 1 \\ 1\end{bmatrix} = c_1 e\begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{-3}\begin{bmatrix} 1 \\ 1\end{bmatrix}=\begin{bmatrix} 2 \\ 2 \end{bmatrix}.$$
From this equation, we can find the linear system $$\begin{bmatrix} e & e^{-3} \\ -e & e^{-3} \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}.$$ Solving this system, we find that $$c_1 = \frac{2(1+e^4)}{2e^3}$$ and $$c_2 = \frac{2(1+e^4)}{2e^3}$$.
Therefore, the solution to the initial value problem is
$$\mathbf{y}(t)=\frac{2(1+e^4)}{2e^3} e^{-t}\begin{bmatrix} 1 \\ -1 \end{bmatrix} + \frac{2(1+e^4)}{2e^3} e^{3t}\begin{bmatrix} 1 \\ 1\end{bmatrix}.$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are crucial concepts in linear algebra, especially when dealing with linear systems of differential equations. Let's break it down:
- Eigenvalues: These are scalar values (\(\lambda\)) that give us information about the scaling factor applied to an eigenvector during a linear transformation. In simpler terms, they tell us how much the direction represented by an eigenvector will stretch or shrink when a transformation is applied.
- Eigenvectors: These are non-zero vectors that do not change direction in a transformation, although they may change in magnitude. They align with the directions in which the linear transformation associated with the matrix is just a scaling transformation.
General Solution
The general solution to a homogeneous linear system involves an expression that encompasses all possible solutions to the system. Once eigenvalues and eigenvectors have been identified, they can be used to construct the solution. In our scenario, having determined the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 3\) and their corresponding eigenvectors \(\mathbf{x}^{(1)} = \begin{bmatrix} 1 \ -1 \end{bmatrix}\) and \(\mathbf{x}^{(2)} = \begin{bmatrix} 1 \ 1 \end{bmatrix}\), the general solution is given in the form: \[\mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{x}^{(1)} + c_2 e^{\lambda_2 t}\mathbf{x}^{(2)}\] This equation comprises arbitrary constants \(c_1\) and \(c_2\) which adjust the contribution of each eigenvector to the solution, allowing the equation to capture the full range of solutions. For this case: \[\mathbf{y}(t) = c_1 e^{-t}\begin{bmatrix} 1 \ -1 \end{bmatrix} + c_2 e^{3t}\begin{bmatrix} 1 \ 1 \end{bmatrix}\] The role of this general solution is to describe every possible behavior of the system for any constants \(c_1\) and \(c_2\).
Initial Value Problem
An initial value problem (IVP) is a type of differential equation accompanied by specific initial conditions that allow us to find a unique solution. In our problem, we need to determine the particular solution that is consistent with the initial condition given in the exercise. For the given problem, we have the initial condition \(\mathbf{y}(-1) = \begin{bmatrix} 2 \ 2 \end{bmatrix}\). Our general solution \(\mathbf{y}(t) = c_1 e^{-t}\begin{bmatrix} 1 \ -1 \end{bmatrix} + c_2 e^{3t}\begin{bmatrix} 1 \ 1 \end{bmatrix}\) needs to be evaluated at \(t = -1\) and equated to this initial condition. By plugging \(t = -1\) into the general solution, we form a system of linear equations that will allow us to solve for \(c_1\) and \(c_2\). \[\begin{bmatrix} 2 \ 2 \end{bmatrix} = c_1 e\begin{bmatrix} 1 \ -1 \end{bmatrix} + c_2 e^{-3}\begin{bmatrix} 1 \ 1 \end{bmatrix}\] Solving this system provides specific values for \(c_1\) and \(c_2\) that satisfy both the differential equation and the initial condition, offering the precise solution that fits the problem constraints.
Characteristic Equation
The characteristic equation is the foundation for finding eigenvalues of a matrix and plays a pivotal role in understanding linear systems. It comes from attempting to find solutions to the equation \(\det(A - \lambda I) = 0\), where \(A\) is the system's matrix and \(I\) is the identity matrix of the same size. The characteristic equation transforms the problem of matrix analysis into finding the roots of a polynomial. The polynomial is derived from \(\det(A - \lambda I)\), which for our example is: \[\det(\begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}) = (1-\lambda)^2 - (2 \cdot 2)\] This simplifies to \(\lambda^2 - 2\lambda - 3 = 0\), a quadratic equation. Solving it provides the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 3\). These eigenvalues help in determining how the system changes over time and lays the groundwork for further analysis, including determining eigenvectors and constructing the general solution.