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.\(\mathbf{y}^{\prime}=\left[\begin{array}{rr}0 & -1 \\ -1 & 0\end{array}\right] \mathbf{y}+\left[\begin{array}{c}t \\ e^{2 t}\end{array}\right], \quad \mathbf{y}_{0}=\left[\begin{array}{l}0 \\\ 1\end{array}\right] . \quad \operatorname{Try} \mathbf{y}_{p}(t)=e^{2 t} \mathbf{a}+t \mathbf{b}+\mathbf{c}\)

Short answer

Question: Find the solution of the given initial value problem for the system of first-order linear differential equations: \(\mathbf{y}^{\prime} = A\mathbf{y} + \mathbf{g}(t)\), where \(A = \left[\begin{array}{rr}0 & -1\\ -1 & 0 \end{array}\right]\) \(\mathbf{g}(t) = \left[\begin{array}{c}t\\ e^{2t}\end{array}\right]\) with the initial condition \(\mathbf{y}(0) = \mathbf{y}_{0} = \left[\begin{array}{c}0 \\1\end{array}\right]\). Answer: The solution to the initial value problem is: \(\mathbf{y}(t) = \operatorname{Re}\left[-\frac{1}{2} e^{i t}\left[\begin{array}{c}1 \\ i\end{array}\right] + \frac{1}{2} e^{-i t}\left[\begin{array}{c}1 \\ -i\end{array}\right]\right] + e^{2t}\left[\begin{array}{c} 0 \\ -\frac{1}{2}\end{array}\right] + t\left[\begin{array}{c} -1 \\ -1\end{array}\right] + \left[\begin{array}{c} 0 \\ 1\end{array}\right]\)

Step-by-step solution

Find the complementary solution

To find the complementary solution, set the function \(\mathbf{g}(t)\) equal to zero and solve the homogeneous equation \(\mathbf{y}^{\prime}=A\mathbf{y}\). We can diagonalize the matrix \(A\) and use the eigenvectors and eigenvalues to find the complementary solution. The matrix A is given by: \(A = \left[\begin{array}{rr}0 & -1\\ -1 & 0 \end{array}\right]\) The characteristic polynomial of A is: \(p(\lambda) = \det(\mathbf{A}-\lambda I)=\lambda^2 + 1\) Solving the characteristic polynomial for eigenvalues: \(\lambda^2 + 1=0\) \(\lambda_1 = i\), \(\lambda_2 = -i\) Next, find the eigenvectors of the matrix A for these eigenvalues. For \(\lambda_1 = i\), \((A - iI)v = 0\) gives, \(\left(\begin{array}{rr} -i & -1 \\ -1 & -i \end{array}\right) \left(\begin{array}{c} x \\ y \end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \end{array}\right)\) Solving this system, we find the eigenvector \(\mathbf{v}_{1} = \left[\begin{array}{c}1 \\ i\end{array}\right]\) For \(\lambda_2 = -i\), \((A - (-i)I)v = 0\) gives, \(\left(\begin{array}{rr} i & -1 \\ -1 & i \end{array}\right) \left(\begin{array}{c} x \\ y \end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \end{array}\right)\) Solving this system, we find the eigenvector \(\mathbf{v}_{2} = \left[\begin{array}{c}1 \\ -i\end{array}\right]\) The complementary solution is given by: \(\mathbf{y}_{c}(t) = \operatorname{Re}[C_1 e^{i t}\mathbf{v}_1 + C_2 e^{-i t}\mathbf{v}_2]\)

Construct a particular solution

To find a particular solution, we have a suggested form: \(\mathbf{y}_{p}(t) = e^{2t}\mathbf{a} + t\mathbf{b} + \mathbf{c}\) Differentiating \(\mathbf{y}_{p}(t)\) with respect to \(t\), we get: \(\mathbf{y}^{\prime}_p(t) = 2e^{2t}\mathbf{a} + \mathbf{b}\) Now, substitute \(\mathbf{y}_{p}(t)\) and \(\mathbf{y}^{\prime}_p(t)\) into the original equation: \(2e^{2t}\mathbf{a} + \mathbf{b} = A(e^{2t}\mathbf{a} + t\mathbf{b} + \mathbf{c}) + \left[\begin{array}{c}t\\ e^{2t}\end{array}\right]\) Multiplying the matrices and then equating the components, we get: \(2e^{2t}\mathbf{a} + \mathbf{b} = \left[\begin{array}{c} -te^{2t} - t^2 \\ -te^{2t} - 2t \end{array}\right] + \left[\begin{array}{c} t \\ e^{2t}\end{array}\right]\) Solve the system of equations for \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\). \(\mathbf{a} = \left[\begin{array}{c} 0 \\ -\frac{1}{2}\end{array}\right]\) \(\mathbf{b} = \left[\begin{array}{c} -1 \\ -1\end{array}\right]\) \(\mathbf{c} = \left[\begin{array}{c} 0 \\ 1\end{array}\right]\) The particular solution is given by: \(\mathbf{y}_{p}(t) = e^{2t}\left[\begin{array}{c} 0 \\ -\frac{1}{2}\end{array}\right] + t\left[\begin{array}{c} -1 \\ -1\end{array}\right] + \left[\begin{array}{c} 0 \\ 1\end{array}\right]\)

Form the general solution

The general solution is a sum of the complementary solution \(\mathbf{y}_{c}(t)\) and the particular solution \(\mathbf{y}_{p}(t)\): \(\mathbf{y}(t) = \mathbf{y}_{c}(t) + \mathbf{y}_{p}(t)\) \(\mathbf{y}(t) = \operatorname{Re}[C_1 e^{i t}\mathbf{v}_1 + C_2 e^{-i t}\mathbf{v}_2] + e^{2t}\left[\begin{array}{c} 0 \\ -\frac{1}{2}\end{array}\right] + t\left[\begin{array}{c} -1 \\ -1\end{array}\right] + \left[\begin{array}{c} 0 \\ 1\end{array}\right]\)

Impose the initial condition

We now need to apply the initial condition \(\mathbf{y}(0) = \mathbf{y}_{0} = \left[\begin{array}{l}0 \\ 1\end{array}\right]\) to find the solution of the initial value problem. By calculating \(\mathbf{y}(0)\) from the expression of \(\mathbf{y}(t)\) and solving the system of equations for \(C_1\) and \(C_2\), we find that \(C_1 = -\frac{1}{2}\) and \(C_2 = \frac{1}{2}\). The solution to the initial value problem is: \(\mathbf{y}(t) = \operatorname{Re}\left[-\frac{1}{2} e^{i t}\left[\begin{array}{c}1 \\ i\end{array}\right] + \frac{1}{2} e^{-i t}\left[\begin{array}{c}1 \\ -i\end{array}\right]\right] + e^{2t}\left[\begin{array}{c} 0 \\ -\frac{1}{2}\end{array}\right] + t\left[\begin{array}{c} -1 \\ -1\end{array}\right] + \left[\begin{array}{c} 0 \\ 1\end{array}\right]\)

Key concepts

Complementary Solution

The process of finding a complementary solution involves solving the homogeneous version of a differential equation. In this context, where we have a system given by \( \mathbf{y}' = A \mathbf{y} + \mathbf{g}(t) \), we define the complementary solution by setting \( \mathbf{g}(t) = 0 \). This turns our equation into \( \mathbf{y}' = A \mathbf{y} \).

To find the complementary solution, we focus on the matrix \( A \), here given as \( A = \left[\begin{array}{rr}0 & -1 \ -1 & 0 \end{array}\right] \). The key to tackling this step is finding the eigenvalues and eigenvectors of \( A \). The characteristic polynomial helps us find eigenvalues, which in this case are complex: \( \lambda_1 = i \) and \( \lambda_2 = -i \).

Using these eigenvalues, we calculate corresponding eigenvectors. For \( \lambda_1 = i \), the eigenvector is \( \mathbf{v}_{1} = \left[\begin{array}{c}1 \ i\end{array}\right] \), and for \( \lambda_2 = -i \), it is \( \mathbf{v}_{2} = \left[\begin{array}{c}1 \ -i\end{array}\right] \). The complementary solution is constructed using these eigenvectors and involves terms like \( e^{i t} \) and \( e^{-i t} \), forming complex exponentials.

In general, the complementary solution is expressed as a linear combination of these exponentials, specifically \( \mathbf{y}_{c}(t) = \operatorname{Re}[C_1 e^{i t}\mathbf{v}_1 + C_2 e^{-i t}\mathbf{v}_2] \), where \( C_1 \) and \( C_2 \) are arbitrary constants. This form captures the natural behavior of the system without external influence.

Particular Solution

The particular solution accounts for the non-homogeneous part of the differential equation, expressed as \( \mathbf{g}(t) \). To find it, we use a method of undetermined coefficients, where we assume a form for \( \mathbf{y}_p(t) \) and adjust parameters within this form to match the equation. In this problem, the suggested form is \( \mathbf{y}_p(t) = e^{2t}\mathbf{a} + t\mathbf{b} + \mathbf{c} \).

Firstly, differentiate \( \mathbf{y}_p(t) \) to get \( \mathbf{y}^{\prime}_p(t) = 2e^{2t}\mathbf{a} + \mathbf{b} \). This derivative accounts for the rate of change expressed in the original equation.

Next, plug these expressions into the equation \( \mathbf{y}^{\prime} = A\mathbf{y} + \mathbf{g}(t) \), and equate it respectively to find \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \). Solving this system results in:\

• \( \mathbf{a} = \left[\begin{array}{c} 0 \ -\frac{1}{2}\end{array}\right] \)
• \( \mathbf{b} = \left[\begin{array}{c} -1 \ -1\end{array}\right] \)
• \( \mathbf{c} = \left[\begin{array}{c} 0 \ 1\end{array}\right] \)

The particular solution \( \mathbf{y}_p(t) \) thus becomes \( e^{2t}\left[\begin{array}{c} 0 \ -\frac{1}{2}\end{array}\right] + t\left[\begin{array}{c} -1 \ -1\end{array}\right] + \left[\begin{array}{c} 0 \ 1\end{array}\right] \).

Eigenvectors and Eigenvalues

Eigenvalues and eigenvectors are crucial for understanding systems of linear differential equations. They simplify the processes of finding solutions, especially for homogeneous equations.

Given a square matrix \( A \), eigenvalues \( \lambda \) are values that satisfy the characteristic equation \( \det(A - \lambda I) = 0 \). In our exercise, this results in the quadratic equation \( \lambda^2 + 1 = 0 \), yielding complex eigenvalues \( \lambda_1 = i \) and \( \lambda_2 = -i \).

Eigenvectors are non-zero vectors that, when multiplied by their matrix \( A \), only scale by the associated eigenvalue. Calculating eigenvectors involves solving \( (A - \lambda I)\mathbf{v} = 0 \). For \( \lambda_1 = i \), we have the eigenvector \( \mathbf{v}_{1} = \left[\begin{array}{c}1 \ i\end{array}\right] \) and for \( \lambda_2 = -i \), \( \mathbf{v}_{2} = \left[\begin{array}{c}1 \ -i\end{array}\right] \).

These eigenvectors play an instrumental role in forming the complementary solution, as they define the basis that describe the intrinsic behavior of the system.

General Solution

The general solution of a differential equation combines both the complementary and particular solutions. This approach captures both the natural dynamics of the homogeneous part and the forced dynamics introduced by the non-homogeneous function.

To construct this solution, you sum the complementary solution \( \mathbf{y}_c(t) \) and the particular solution \( \mathbf{y}_p(t) \). Continuing from our solutions:

• Complementary Solution: \( \mathbf{y}_{c}(t) = \operatorname{Re}[C_1 e^{i t}\mathbf{v}_1 + C_2 e^{-i t}\mathbf{v}_2] \)
• Particular Solution: \( \mathbf{y}_{p}(t) = e^{2t}\left[\begin{array}{c} 0 \ -\frac{1}{2}\end{array}\right] + t\left[\begin{array}{c} -1 \ -1\end{array}\right] + \left[\begin{array}{c} 0 \ 1\end{array}\right] \)

Combining these results, the general solution is given by \( \mathbf{y}(t) = \mathbf{y}_c(t) + \mathbf{y}_p(t) \). This combination fully describes the system's response under given conditions, such as an initial value \( \mathbf{y}(0) \), leading to expressions for constants \( C_1 \) and \( C_2 \). Thus, the general solution solves the initial value problem completely.