Question

Consider the initial value problem \(\mathbf{y}^{\prime}=A \mathbf{y}+\mathbf{g}(t)\),\(\mathbf{y}(0)=\mathbf{y}_{0}\)(a) Form the complementary solution. (b) Construct a particular solution by assuming the form suggested and solving for the undetermined constant vectors \(\mathbf{a}, \mathbf{b}\), and \(\mathbf{c}\). (c) Form the general solution. (d) Impose the initial condition to obtain the solution of the initial value problem.\(\mathbf{y}^{\prime}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \mathbf{y}+\left[\begin{array}{c}t \\\ -1\end{array}\right], \quad \mathbf{y}_{0}=\left[\begin{array}{r}2 \\\ -1\end{array}\right] . \quad \operatorname{Try} \mathbf{y}_{P}(t)=t \mathbf{a}+\mathbf{b}\)

Short answer

In this exercise, we solved an initial value problem involving a linear system of first-order differential equations. We followed the steps, which included finding the complementary solution, constructing a particular solution, forming the general solution, and imposing the initial condition. After solving for the constants \(c_1\) and \(c_2\), the specific solution to the initial value problem is: $$ \mathbf{y}(t) = \frac{3}{2}e^{t}\left[\begin{array}{c}1 \\\ 1\end{array}\right]+\frac{1}{2}e^{-t}\left[\begin{array}{c}1 \\\ -1\end{array}\right]+t\left[\begin{array}{r}-1 \\\ 1\end{array}\right]+\left[\begin{array}{c}0 \\\ 1\end{array}\right] $$

Step-by-step solution

Find the complementary solution

To find the complementary solution, we will find the eigenvalues and eigenvectors for the matrix \(A\). The matrix \(A\) is given by: $$ A = \left[\begin{array}{ll}0 & 1 \\\ 1 & 0\end{array}\right] $$ To find the eigenvalues, we can solve the characteristic equation: $$ \text{det}(A-\lambda I) = 0 $$ Where \(I\) is the identity matrix. Substituting \(A\), $$ \text{det}\left(\left[\begin{array}{cc}0-\lambda & 1 \\\ 1 & 0-\lambda\end{array}\right]\right) = (\lambda)(-\lambda) - (1)(1) = \lambda^2 -1 = 0 $$ Solving this equation, we get the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = -1\). Now, for each eigenvalue, we want to find the eigenvector associated with it. For \(\lambda_1 = 1\): $$ (A - \lambda_1 I)\mathbf{x} = 0 $$ Substituting, we get: $$ \left[\begin{array}{cc}-1 & 1 \\\ 1 & -1\end{array}\right]\left[\begin{array}{c}x_1 \\\ x_2\end{array}\right] = \left[\begin{array}{c}0 \\\ 0\end{array}\right] $$ By solving this system, we get eigenvector \(\mathbf{x}_1 = \left[\begin{array}{c}1 \\\ 1\end{array}\right]\). Next, for \(\lambda_2 = -1\): $$ (A - \lambda_2 I)\mathbf{x} = 0 $$ Substituting, we get: $$ \left[\begin{array}{cc}1 & 1 \\\ 1 & 1\end{array}\right]\left[\begin{array}{c}x_1 \\\ x_2\end{array}\right] = \left[\begin{array}{c}0 \\\ 0\end{array}\right] $$ By solving this system, we get eigenvector \(\mathbf{x}_2 = \left[\begin{array}{c}1 \\\ -1\end{array}\right]\). Now we can write down our complementary solution: $$ \mathbf{y}_C(t) = c_1e^{t}\left[\begin{array}{c}1 \\\ 1\end{array}\right]+c_2e^{-t}\left[\begin{array}{c}1 \\\ -1\end{array}\right] $$

Construct a particular solution

To find a particular solution, we need to try the given form, \(\mathbf{y}_P(t) = t\mathbf{a}+\mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors that need to be determined. We start by finding the derivative \(\mathbf{y}_P'(t)\): $$ \mathbf{y}_P'(t) = \mathbf{a} $$ Next, we substitute \(\mathbf{y_P}\) and \(\mathbf{y_P}'\) back into the original equation: $$ \mathbf{a} = A(t\mathbf{a}+\mathbf{b})+\mathbf{g}(t) $$ Substituting \(A\) and \(\mathbf{g}(t)\), we get: $$ \mathbf{a} = \left[\begin{array}{cc}0 & 1 \\\ 1 & 0\end{array}\right](t\mathbf{a}+\mathbf{b})+\left[\begin{array}{c}t \\\ -1\end{array}\right] $$ This results in the two equations: $$ a_2 = tb_1+a_1, \quad a_1 = tb_2+a_2-1 $$ To eliminate \(t\) and simplify the system, we set the coefficients of \(t\) equal to each other: $$ b_1 = 0, \quad b_2 = 1 $$ Substituting these values back into our previous system of equations, we obtain the constant vectors \(\mathbf{a}\) and \(\mathbf{b}\): $$ \mathbf{a} = \left[\begin{array}{r}-1 \\\ 1\end{array}\right], \quad \mathbf{b} = \left[\begin{array}{c}0 \\\ 1\end{array}\right] $$ And the particular solution is: $$ \mathbf{y}_P(t) = t\left[\begin{array}{r}-1 \\\ 1\end{array}\right]+\left[\begin{array}{c}0 \\\ 1\end{array}\right] $$

Form the general solution

The general solution is the sum of the complementary solution \(\mathbf{y}_C(t)\) and the particular solution \(\mathbf{y}_P(t)\): $$ \mathbf{y}(t) = \mathbf{y}_C(t) + \mathbf{y}_P(t) = c_1e^{t}\left[\begin{array}{c}1 \\\ 1\end{array}\right]+c_2e^{-t}\left[\begin{array}{c}1 \\\ -1\end{array}\right]+t\left[\begin{array}{r}-1 \\\ 1\end{array}\right]+\left[\begin{array}{c}0 \\\ 1\end{array}\right] $$

Impose the initial condition

To find the specific solution that satisfies the initial condition \(\mathbf{y}(0) = \mathbf{y}_0 = \left[\begin{array}{r}2 \\\ -1\end{array}\right]\), we substitute \(t=0\): $$ \left[\begin{array}{r}2 \\\ -1\end{array}\right] = c_1e^{0}\left[\begin{array}{c}1 \\\ 1\end{array}\right]+c_2e^{-0}\left[\begin{array}{c}1 \\\ -1\end{array}\right]+\left[\begin{array}{c}0 \\\ 1\end{array}\right] $$ Simplifying the equation, we get: $$ \left[\begin{array}{r}2 \\\ -1\end{array}\right] = c_1\left[\begin{array}{c}1 \\\ 1\end{array}\right]+c_2\left[\begin{array}{c}1 \\\ -1\end{array}\right]+\left[\begin{array}{c}0 \\\ 1\end{array}\right] $$ Solving for \(c_1\) and \(c_2\), we find: $$ c_1 = \frac{3}{2}, \quad c_2 = \frac{1}{2} $$ So the specific solution to the initial value problem is: $$ \mathbf{y}(t) = \frac{3}{2}e^{t}\left[\begin{array}{c}1 \\\ 1\end{array}\right]+\frac{1}{2}e^{-t}\left[\begin{array}{c}1 \\\ -1\end{array}\right]+t\left[\begin{array}{r}-1 \\\ 1\end{array}\right]+\left[\begin{array}{c}0 \\\ 1\end{array}\right] $$

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in the study of linear algebra and play a critical role in the analysis of linear systems. When we encounter a square matrix, like the one in our initial value problem, we're often interested in finding its eigenvalues and corresponding eigenvectors. An eigenvalue is a scalar \( \lambda \) that satisfies the equation \( A\textbf{x} = \lambda\textbf{x} \) for a given matrix \( A \) and nonzero vector \( \textbf{x} \), which is known as the eigenvector.

To find eigenvalues, we solve the characteristic equation, \( \text{det}(A - \lambda I) = 0 \), where \( I \) represents the identity matrix. Once the eigenvalues \( \lambda \) are known, we can compute the corresponding eigenvectors by solving \( (A - \lambda I)\textbf{x} = 0 \) for each \( \lambda \) to gain the set of vectors that are scaled by \( A \).

Eigenvalues reveal important properties about the transformation represented by matrix \( A \) and are especially useful when finding solutions to systems of differential equations. They can indicate, for example, the stability of a system or the behavior of the system over time. Eigenvectors associated with distinct eigenvalues are linearly independent and provide a basis for the space.

Complementary Solution

The complementary solution, often denoted as \( \textbf{y}_C(t) \), in the context of solving differential equations, refers to the part of the general solution that contains all solutions to the homogeneous equation, which is the equation without any added external forces or inputs, symbolically given as \( \textbf{y}' = A\textbf{y} \). The method to find the complementary solution typically involves using the eigenvalues and eigenvectors of the matrix \( A \).

Once the eigenvalues \( \lambda_1, \lambda_2, \ldots \) and associated eigenvectors \( \textbf{x}_1, \textbf{x}_2, \ldots \) are found, we construct the complementary solution by combining these eigenvectors with exponential factors involving the eigenvalues and time \( t \). This part of the solution is crucial as it represents the system's behavior influenced solely by its internal properties, independent of any external driving forces.

Particular Solution

When a differential equation includes a non-homogeneous part, such as an external input or force, we must find a particular solution, denoted as \( \textbf{y}_P(t) \), that alone satisfies the non-homogeneous equation. The specific form of the particular solution is generally suggested by the form of this external input represented by \( \textbf{g}(t) \) in our provided exercise.

The strategy involves assuming a form for \( \textbf{y}_P(t) \) with undefined constants or functions, then plugging this assumed form into the non-homogeneous equation. The goal is to determine the constants or functions in such a way that \( \textbf{y}_P(t) \) satisfies the equation for all \( t \).

This task often involves solving a system of equations derived from substituting the assumed form into the original equation and equating coefficients. The particular solution is vital for representing the impact of external influences on the system's behavior over time.

General Solution

The general solution of a differential equation is the expression that represents every possible solution to the equation. It is typically composed of two parts: the complementary solution \( \textbf{y}_C(t) \) and the particular solution \( \textbf{y}_P(t) \).

To form the general solution, we simply add the complementary and particular solutions together, as in \( \textbf{y}(t) = \textbf{y}_C(t) + \textbf{y}_P(t) \). This summation represents the superposition principle, where the individual solutions are combined since the original differential equation is linear.

The general solution encompasses not just the system's intrinsic behavior (complementary solution) but also the response due to external forces (particular solution). It includes undetermined constants which are then defined by applying initial conditions, hence uniquely specifying the solution to an initial value problem as we have done in our exercise.