Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 4 (b)

MATH152 April 2015

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Question B 4 (b)

Consider the differential equation system $\mathbf {x} '=A\mathbf {x} ,$ where $A$ has the eigenvalues $\lambda _{1}=-1+i$ and $\lambda _{2}=-1-i$ with the corresponding eigenvectors $\mathbf {v} _{1}={\begin{bmatrix}1+i\\1-i\end{bmatrix}}$ and $\mathbf {v} _{2}={\begin{bmatrix}1-i\\1+i\end{bmatrix}}.$ (b) Find the solution that satisfies $\mathbf {x} (0)={\begin{bmatrix}1\\2\end{bmatrix}}$ in real form (no complex numbers or complex exponentials).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The solution to this initial value problem can be obtained, from the initial condition and the general solution from part (a), by substituting $t=0$ into your answer for part (a). To find the real-valued function, evaluate the complex valued function using Euler's formula then take the real part of the resulting function.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. From part (a), we have ${\mathbf {x}}(t)=c_{1}e^{(-1+i)t}{\begin{pmatrix}1+i\\1-i\end{pmatrix}}+c_{2}e^{(-1-i)t}{\begin{pmatrix}1-i\\1+i\end{pmatrix}}.$ Using the initial condition, ${\begin{pmatrix}1\\2\end{pmatrix}}={\mathbf {x}}(0)=c_{1}e^{(-1+i)\cdot 0}{\begin{pmatrix}1+i\\1-i\end{pmatrix}}+c_{2}e^{(-1-i)\cdot 0}{\begin{pmatrix}1-i\\1+i\end{pmatrix}}=c_{1}{\begin{pmatrix}1+i\\1-i\end{pmatrix}}+c_{2}{\begin{pmatrix}1-i\\1+i\end{pmatrix}}$ To find $c_{1}$ and $c_{2}$, so we first write the equation in the matrix form and then in augmented matrix; ${\begin{pmatrix}1+i&1-i\\1-i&1+i\end{pmatrix}}{\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}}={\begin{pmatrix}1\\2\end{pmatrix}}\implies {\begin{pmatrix}1+i&1-i&|&1\\1-i&1+i&|&2\end{pmatrix}}$ Using Gaussian elimination, we get the following. We have $(1+i)(1-i)=1^{2}-i+i-i^{2}=1^{2}-i^{2}=1-(-1)=2$ So multiplying the top and bottom rows of the above augmented matrix by $1-i$ and $1+i$ respectively gives ${\begin{pmatrix}2&(1-i)^{2}&|&1-i\\2&(1+i)^{2}&|&2+2i\end{pmatrix}}$ Since $(1-i)^{2}=1^{2}-2i+i^{2}=-2i$ and $(1+i)^{2}=1^{2}+2i+i^{2}=2i$ the above augmented matrix is ${\begin{pmatrix}2&-2i&|&1-i\\2&2i&|&2+2i\end{pmatrix}}$ Further row reduction now gives, ${\begin{pmatrix}2&-2i&|&1-i\\2&2i&|&2+2i\end{pmatrix}}\to {\begin{pmatrix}2&-2i&|&1-i\\0&-4i&|&-1-3i\end{pmatrix}}\to {\begin{pmatrix}4&-4i&|&2-2i\\0&4i&|&1+3i\end{pmatrix}}$ $\to {\begin{pmatrix}4&0&|&3+i\\0&4i&|&1+3i\end{pmatrix}}\to {\begin{pmatrix}1&0&|&(3+i)/4\\0&1&|&(1+3i)/(4i)\end{pmatrix}}$ So, as $1/i=-i$, $c_{1}={\frac {(3+i)}{4}}$ and $c_{2}={\frac {(3-i)}{4}}$. Plugging them back, ${\mathbf {x}}(t)={\frac {3+i}{4}}e^{(-1+i)t}{\begin{pmatrix}1+i\\1-i\end{pmatrix}}+{\frac {3-i}{4}}e^{(-1-i)t}{\begin{pmatrix}1-i\\1+i\end{pmatrix}}$ Since the first summand is the complex conjugate of the second summand, the solution ${\mathbf {x}}$ in the real form is ${\mathbf {x}}(t)=2{\text{Re}}\left({\frac {3+i}{4}}e^{(-1+i)t}{\begin{pmatrix}1+i\\1-i\end{pmatrix}}\right)=2{\text{Re}}\left({\frac {3+i}{4}}e^{-t}(\cos(t)+i\sin(t)){\begin{pmatrix}1+i\\1-i\end{pmatrix}}\right)=e^{-t}{\begin{pmatrix}\cos(t)-2\sin(t)\\2\cos(t)+\sin(t)\end{pmatrix}}$ Here, we used the Euler's formula: $e^{it}=\cos t+i\sin t$. Hence, $\color {blue}x(t)=e^{-t}{\begin{pmatrix}\cos(t)-2\sin(t)\\2\cos(t)+\sin(t)\end{pmatrix}}$