Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 4 (b)/Solution 1

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,

$\color {blue}x(t)=e^{-t}{\begin{pmatrix}\cos(t)-2\sin(t)\\2\cos(t)+\sin(t)\end{pmatrix}}$