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Let us solve the following system of ODEs \begin{align*} \mathbf{x}' = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1 \end{pmatrix} \mathbf{x} \end{align*}

Solution

Let \(\mathbf{A}\) be the matrix above, so let us find the characteristic polynomial of our system

Substitute the eigenvalues above to find the corresponding eigenvectors \(\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}\)

\begin{align*} \lambda = \lambda_1 \implies \begin{pmatrix} 1 - \lambda_1 & -1 & 4 \\ 3 & 2 - \lambda_1 & -1 \\ 2 & 1 & -1 - \lambda_1 \end{pmatrix} \mathbf{v}_1 \implies \mathbf{v}_1 = \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix} \end{align*}

\begin{align*} \lambda = \lambda_2 \implies \begin{pmatrix} 1 - \lambda_2 & -1 & 4 \\ 3 & 2 - \lambda_2 & -1 \\ 2 & 1 & -1 - \lambda_2 \end{pmatrix} \mathbf{v}_2 \implies \mathbf{v}_2 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \end{align*}

\begin{align*} \lambda = \lambda_3 \implies \begin{pmatrix} 1 - \lambda_3 & -1 & 4 \\ 3 & 2 - \lambda_3 & -1 \\ 2 & 1 & -1 - \lambda_3 \end{pmatrix} \mathbf{v}_3 \implies \mathbf{v}_3 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \end{align*}

Therefore the final solution has the form

\begin{align*} \mathbf{x}(t) & = C_1 e^{\lambda_1 t} \mathbf{v_1} + C_2 e^{\lambda_2 t} \mathbf{v_2} + C_3 e^{\lambda_3 t} \mathbf{v_3}\\\\ & = C_1 e^{t} \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + C_3 e^{3t} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \end{align*}

The solution strategy is the same as Problem 9. We find that eigenvalues are \(\lambda_1 = 4\) and \(\lambda_2 = 2\) with their corresponding eigenvectors \(\mathbf{v}_1 = \begin{pmatrix}1 \\ 1 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix}1 \\ 3 \end{pmatrix}\), such that the solution is

\begin{align*} \mathbf{x}(t) = C_1 e^{4t} \begin{pmatrix}1 \\ 1 \end{pmatrix} + C_2 e^{2t} \begin{pmatrix}1 \\ 3 \end{pmatrix} \end{align*}

By knowing the initial value

\(\mathbf{x}(0) = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\), we solve the system

\begin{align*} \begin{cases} C_1 + C_2 = 2 \\ C_1 + 3 C_2 = -1 \end{cases} \implies \begin{cases} C_1 = \frac{7}{2} \\ C_2 = -\frac{3}{2} \end{cases} \end{align*}

Finally, the solution to the initial value problem is

\begin{align*} \mathbf{x}(t) = \frac{7}{2} e^{4t} \begin{pmatrix}1 \\ 1 \end{pmatrix} + - \frac{3}{2} e^{2t} \begin{pmatrix}1 \\ 3 \end{pmatrix} \end{align*}

Recall the eigenvalues and their eigenvectors we found in Problem 10. Applying Problem 13, the solution is

\begin{align*} \mathbf{x}(t) = C_1 t^4 \begin{pmatrix} 1\\ 1\end{pmatrix} + C_2 t^2 \begin{pmatrix} 1\\ 3\end{pmatrix} \end{align*}

Similarly to all problems above, we find eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = -2\), where their respective eigenvectors are \(\mathbf{v}_1 = \begin{pmatrix} 3 \\ 4\end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 2\end{pmatrix}\). Applying Problem 13, we have the solution

\begin{align*} \mathbf{x}(t) = C_1 \begin{pmatrix} 3\\ 4\end{pmatrix} + C_2 t^{-2} \begin{pmatrix} 1\\ 2\end{pmatrix} \end{align*}

Solutions are omitted for the sake of expediting the grading before Midterm II.