Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 04 (d)

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MATH152 April 2017

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Question B 04 (d)

A model of love affairs is described by the following system of differential equations $\begin{array}{rl}{R}^{\prime }& =R+4J,\\ J^{\prime }& =-2R-3J,\end{array}$ where $R(t)$ and $J(t)$ are, respectively, Romeo’s and Juliet’s amount of love for each other at time $t$. (d) [2] Assuming that initially the two have equal love for each other, i.e. $\mathbf{𝐱}(0)={\left[\begin{array}{cc}1& 1\end{array}\right]}^T$, find $\mathbf{𝐱}(t)$ in its real form (involving no complex numbers).

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
Use Euler's formula to re-write the complex solution from part c) into a real form. Afterwards, use the initial condition to obtained the real-valued solution.

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Solution

We know from part c) that the general form of the solution is $c_1\left[\begin{array}{c}1\\ \frac{-1+i}{2}\\ \end{array}\right]e^{(-1+2i)t}+c_2\left[\begin{array}{c}1\\ \frac{-1-i}{2}\\ \end{array}\right]e^{(-1-2i)t}$ with arbitrary constants $c_1,c_2.$ We first turn the solution into a real form, and then use the given initial condition to replace the arbitrary constants with specific numbers. To turn the solution into real form, we replace the complex terms $\mathbf{𝐯}{e}^{\lambda t}$ and $\overline{\mathbf{v}}{e}^{\overline{\lambda }t}$ appearing in the general solution with $\text{Re}(\mathbf{𝐯}{e}^{\lambda t})$ and $\text{Im}(\mathbf{𝐯}{e}^{\lambda t})$. Using the Euler relation $e^{it}=\cos t+i\sin t,$ we have: $e^{(-1+2i)t}=e^{-t}{e}^{2it}=e^{-t}(\cos (2t)+i\sin (2t))=e^{-t}\cos (2t)+i{e}^{-t}\sin (2t)$ Also: $\begin{array}{rl}\left(\frac{-1+i}{2}\right)e^{(-1+2i)t}& =(\frac{-1}{2}+\frac{i}{2})(e^{-t}\cos (2t)+i{e}^{-t}\sin (2t))\\ & =(-\frac{e^{-t}}{2}\cos (2t)-\frac{e^{-t}}{2}\sin (2t))+i(\frac{e^{-t}}{2}\cos (2t)-\frac{e^{-t}}{2}\sin (2t))\end{array}$ So the complex term can be written as $\begin{array}{rl}\left[\begin{array}{c}1\\ \frac{-1+i}{2}\\ \end{array}\right]e^{(-1+2i)t}& =\left[\begin{array}{c}\cos (2t)+i\sin (2t)\\ (-\frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t))+i(\frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t))\\ \end{array}\right]e^{-t}\\ & =\left[\begin{array}{c}\cos (2t)\\ -\frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t)\\ \end{array}\right]e^{-t}+i\left[\begin{array}{c}\sin (2t)\\ \frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t)\\ \end{array}\right]e^{-t}\end{array}$ Thus the real form of the general solution is: $\mathbf{𝐱}(t)=A\left[\begin{array}{c}\cos (2t)\\ -\frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t)\\ \end{array}\right]e^{-t}+B\left[\begin{array}{c}\sin (2t)\\ \frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t)\\ \end{array}\right]e^{-t},$ for arbitrary real constants $A$ and $B$. Now we use the given initial data $\mathbf{𝐱}(0)=\left[\begin{array}{c}1\\ 1\\ \end{array}\right]$ to solve for $A$ and $B$. Remember that $e^0=1,\cos 0=1,$ and $\sin 0=0.$ Putting the initial data at time $t=0$ into the general solution gives: $\mathbf{𝐱}(0)=\left[\begin{array}{c}1\\ 1\\ \end{array}\right]=A\left[\begin{array}{c}1\\ -\frac{1}{2}\\ \end{array}\right]+B\left[\begin{array}{c}0\\ \frac{1}{2}\\ \end{array}\right],$ This gives the two equations $1=A$ and $1=\frac{1}{2}(-A+B)$ which has solution $A=1,B=3.$ So the solution satisfying the initial data is $\begin{array}{rl}\mathbf{𝐱}(t)& =\left[\begin{array}{c}\cos (2t)\\ -\frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t)\\ \end{array}\right]e^{-t}+3\left[\begin{array}{c}\sin (2t)\\ \frac{1}{2}\cos (2t)-\frac{1}{2}\sin (2t)\\ \end{array}\right]e^{-t}\\ & =\left[\begin{array}{c}\cos (2t)+3\sin (2t)\\ \cos (2t)-2\sin (2t)\\ \end{array}\right]e^{-t}\end{array}$

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