Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 06 (a)

MATH152 April 2016

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Question B 06 (a)

Consider the three-component differential equation ${\textbf {x}}^{\prime }=A{\textbf {x}}$. The $3\times 3$ matrix $A$ has real entries. It has an eigenvalue $\lambda _{1}=-2$ and an eigenvalue $\lambda _{2}=-1+i$ with corresponding eigenvectors ${\textbf {v}}_{1}=\left[{\begin{array}{c}1\\1\\1\end{array}}\right]$ and ${\textbf {v}}_{2}=\left[{\begin{array}{c}0\\1+i\\1\end{array}}\right].$ (a) Write the general solution to the differential equation.

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Hint
Suppose a matrix $A$ has real entries, and $\lambda$ is an eigenvalue of $A$ with associated eigenvector ${\textbf {v}}$. Then their complex conjugates ${\bar {\lambda }}$ and ${\bar {\textbf {v}}}$ are also an eigenvalue and eigenvector pair of the matrix $A$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. One eigenvalue and eigenvector pair, $\lambda _{1}=-2$ and ${\textbf {v}}_{1}=\left[{\begin{array}{c}1\\1\\1\end{array}}\right]$ is real. The other eigenvalue and eigenvector pair, $\lambda _{2}=-1+i$ and ${\textbf {v}}_{2}=\left[{\begin{array}{c}0\\1+i\\1\end{array}}\right]$, is complex. Because the matrix $A$ has real entries, the complex conjugates of $\lambda _{2}$ and ${\textbf {v}}_{2}$ must also be an eigenvalue and eigenvector pair, $\lambda _{3}=-1-i$ and ${\textbf {v}}_{3}=\left[{\begin{array}{c}0\\1-i\\1\end{array}}\right]$. Since the matrix $A$ is a $3\times 3$ matrix, it can have at most 3 eigenvalues. So we have found all of them and they are distinct. In this case, the general solution to the differential equation is given by ${\textbf {x}}(t)=C_{1}e^{\lambda _{1}t}{\textbf {v}}_{1}+C_{2}e^{\lambda _{2}t}{\textbf {v}}_{2}+C_{3}e^{\lambda _{3}t}{\textbf {v}}_{3}.$ Substituting the eigenvalues and eigenvectors that we found, the general solution to the differential equation is ${\textbf {x}}(t)=C_{1}e^{-2t}\left[{\begin{array}{c}1\\1\\1\end{array}}\right]+C_{2}e^{(-1+i)t}\left[{\begin{array}{c}0\\1+i\\1\end{array}}\right]+C_{3}e^{(-1-i)t}\left[{\begin{array}{c}0\\1-i\\1\end{array}}\right].$ To put this in real form, we replace the complex term $e^{(-1+i)t}\left[{\begin{array}{c}0\\(1+i)\\1\end{array}}\right]$ and its conjugate by the real and imaginary parts of the former. Since $e^{(-1+i)t}=e^{-t}(\cos t+i\sin t)=e^{-t}\cos t+ie^{-t}\sin t$ and $e^{(-1+i)t}(1+i)=e^{-t}(\cos t+i\sin t)(1+i)=e^{-t}(\cos t-\sin t)+ie^{-t}(\cos t+\sin t)$ we then have that the general solution in real form is: $\color {blue}{\textbf {x}}(t)=C_{1}e^{-2t}\left[{\begin{array}{c}1\\1\\1\end{array}}\right]+C_{2}\left[{\begin{array}{c}0\\e^{-t}(\cos t-\sin t)\\e^{-t}\cos t\end{array}}\right]+C_{3}\left[{\begin{array}{c}0\\e^{-t}(\cos t+\sin t)\\e^{-t}\sin t\end{array}}\right].$