Problem 28 - Differential Equations and Linear Algebra

Question

Use the variation-of-parameters method to determine a particular solution to the nonhomogeneous linear system \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b}\). Also find the general solution to the system. $$ A=\left[\begin{array}{rrr} 2 & -4 & 3 \\ -9 & -3 & -9 \\ 4 & 4 & 3 \end{array}\right], \mathbf{b}=\left[\begin{array}{c} e^{6 t} \\ 1 \\ 0 \end{array}\right] $$ $$ \text { [Hint: The eigenvalues of } A \text { are } \lambda=6,-3,-1 .] $$

Step-by-Step Solution

Answer

The general solution for the given nonhomogeneous linear system is: \(\mathbf{x}(t) = \left[\begin{array}{c} c_1e^{6t} - c_2e^{-3t} + c_3e^{-t} + \frac{1}{4}(e^{6t} - 2) \\ c_1e^{6t} - 2c_2e^{-3t} - \frac{1}{4}(e^{6t} - 1) \\ c_2e^{-3t} + c_3e^{-t} + \frac{1}{4} \end{array}\right]\)

Step 1: Find Eigenvectors for Eigenvalues of Matrix A

Given the eigenvalues \(\lambda = 6, -3, -1\), we will find the eigenvectors for each eigenvalue by solving the linear system \((A-\lambda I)\mathbf{x}=0\): For \(\lambda_1=6\): \((A-6I)\mathbf{x}=\left[\begin{array}{rrr} 2-6 & -4 & 3 \\ -9 & -3-6 & -9 \\ 4 & 4 & 3-6 \end{array}\right]\mathbf{x}=\left[\begin{array}{rrr} -4 & -4 & 3 \\ -9 & -9 & -9 \\ 4 & 4 & -3 \end{array}\right]\mathbf{x}=0\) Eigenvector: \(\mathbf{x}_1=\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]\) For \(\lambda_2=-3\): \((A+3I)\mathbf{x}=\left[\begin{array}{rrr} 2+3 & -4 & 3 \\ -9 & -3+3 & -9 \\ 4 & 4 & 3+3 \end{array}\right]\mathbf{x}=\left[\begin{array}{rrr} 5 & -4 & 3 \\ -9 & 0 & -9 \\ 4 & 4 & 6 \end{array}\right]\mathbf{x}=0\) Eigenvector: \(\mathbf{x}_2=\left[\begin{array}{c} -1 \\ -2 \\ 1 \end{array}\right]\) For \(\lambda_3=-1\): \((A+I)\mathbf{x}=\left[\begin{array}{rrr} 2+1 & -4 & 3 \\ -9 & -3-1 & -9 \\ 4 & 4 & 3-1 \end{array}\right]\mathbf{x}=\left[\begin{array}{rrr} 3 & -4 & 3 \\ -9 & -4 & -9 \\ 4 & 4 & 2 \end{array}\right]\mathbf{x}=0\) Eigenvector: \(\mathbf{x}_3=\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]\)

Step 2: Obtain Complementary Solution

The complementary solution for the homogeneous system is given by: \(\mathbf{x}_c(t) = c_1\mathbf{x}_1 e^{\lambda_1 t} + c_2\mathbf{x}_2 e^{\lambda_2 t} + c_3\mathbf{x}_3 e^{\lambda_3 t}\) \(\mathbf{x}_c(t) = c_1\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]e^{6t} + c_2\left[\begin{array}{c} -1 \\ -2 \\ 1 \end{array}\right]e^{-3t} + c_3\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]e^{-t}\)

Step 3: Use Variation-of-Parameters Method to Find a Particular Solution

First, we need to compute the Wronskian of the eigenvectors, \(W(\mathbf{x}_1,\mathbf{x}_2,\mathbf{x}_3) = \det\left[\begin{array}{ccc} 1 & -1 & 1 \\ 1 & -2 & 0 \\ 0 & 1 & 1 \end{array}\right] = -4\). Next, let's find the particular solution \(\mathbf{x}_p(t)\) by computing \(W_i(t)\), substituting the \(i\)-th row by \(\mathbf{b}\) and multiplying by \(e^{\lambda_i t}\). Finally, add them and divide by the Wronskian. \(\mathbf{x}_p(t) = \frac{1}{W(\mathbf{x}_1,\mathbf{x}_2,\mathbf{x}_3)}\left[\begin{array}{c} W_1(t) \\ W_2(t) \\ W_3(t) \end{array}\right]=\frac{1}{-4}\left[\begin{array}{c} W_1(t) \\ W_2(t) \\ W_3(t) \end{array}\right]\) For \(W_1(t)\), \(W_1(t) = \det\left[\begin{array}{ccc} e^{6t} & -1 & 1 \\ 1 & -2 & 0 \\ 0 & 1 & 1 \end{array}\right] = 2e^{6t} - e^{6t}\) For \(W_2(t)\), \(W_2(t) = \det\left[\begin{array}{ccc} 1 & e^{6t} & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = e^{6t} - 1\) For \(W_3(t)\), \(W_3(t) = \det\left[\begin{array}{ccc} 1 & -1 & e^{6t} \\ 1 & -2 & 1 \\ 0 & 1 & 0 \end{array}\right] = -1\) Substitute values in \(\mathbf{x}_p(t)\): \(\mathbf{x}_p(t) = \frac{1}{-4}\left[\begin{array}{c} 2e^{6t} - e^{6t} \\ e^{6t} - 1 \\ -1 \end{array}\right]=\left[\begin{array}{c} \frac{1}{4}(e^{6t} - 2) \\ \frac{1}{4}(1 - e^{6t}) \\ \frac{1}{4} \end{array}\right]\)

Step 4: Obtain the General Solution

Combine the complementary solution and the particular solution: \(\mathbf{x}(t) = \mathbf{x}_c(t) + \mathbf{x}_p(t)\) \(\mathbf{x}(t) = c_1\left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]e^{6t} + c_2\left[\begin{array}{c} -1 \\ -2 \\ 1 \end{array}\right]e^{-3t} + c_3\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]e^{-t} + \left[\begin{array}{c} \frac{1}{4}(e^{6t} - 2) \\ \frac{1}{4}(1 - e^{6t}) \\ \frac{1}{4} \end{array}\right]\) Therefore, the general solution for the given nonhomogeneous linear system is: \(\mathbf{x}(t) = \left[\begin{array}{c} c_1e^{6t} - c_2e^{-3t} + c_3e^{-t} + \frac{1}{4}(e^{6t} - 2) \\ c_1e^{6t} - 2c_2e^{-3t} - \frac{1}{4}(e^{6t} - 1) \\ c_2e^{-3t} + c_3e^{-t} + \frac{1}{4} \end{array}\right]\)

Key Concepts

Nonhomogeneous Linear Systems

Nonhomogeneous linear systems refer to a set of linear differential equations that include a function not solely dependent on the unknown variable. In simpler terms, apart from the linear combination of unknowns, there is an additional separate function that contributes to the system. The system is represented as \( \mathbf{x}^\prime = A \mathbf{x} + \mathbf{b} \), where \( A \) is a constant matrix, \( \mathbf{x} \) is the vector of unknowns, and \( \mathbf{b} \) is the nonhomogeneous term.