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

MATH152 April 2017

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Question B 04 (a)

A model of love affairs is described by the following system of differential equations ${\begin{aligned}R'&=R+4J,\\J'&=-2R-3J,\end{aligned}}$ where $R(t)$ and $J(t)$ are, respectively, Romeo’s and Juliet’s amount of love for each other at time $t$. (a) Express the system in matrix form ${\frac {d{\mathbf {x}}}{dt}}=A{\mathbf {x}}$, where ${\mathbf {x}}(t)={\begin{bmatrix}R(t)&J(t)\end{bmatrix}}^{T}$ .

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
$A$ will be the 2x2 coefficient matrix of the system.

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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. We want to write the given system of differential equations as ${\frac {d{\mathbf {x}}}{dt}}=A{\mathbf {x}}$. Since ${\mathbf {x}}(t)={\begin{bmatrix}R\\J\end{bmatrix}}$, we have ${\frac {d{\mathbf {x}}}{dt}}={\begin{bmatrix}R'\\J'\end{bmatrix}}$. So, we need to find the matrix $A$, where ${\begin{bmatrix}R'\\J'\end{bmatrix}}={\begin{bmatrix}R+4J\\-2R-3J\end{bmatrix}}=A{\begin{bmatrix}R\\J\end{bmatrix}}$ Thus $A={\begin{bmatrix}\quad \!1&\quad \!4\\-2&-3\end{bmatrix}}$, and the answer is: $\color {blue}{\frac {d{\mathbf {x}}}{dt}}={\begin{bmatrix}\quad \!1&\quad \!4\\-2&-3\end{bmatrix}}{\mathbf {x}}$