## MATH1013 Quizzes

Quiz 12: First order linear differential equations and predator prey models
Question 1 Questions
Two populations, $x=x\left(t\right)$ and $y=y\left(t\right)$, satisfy the following differential equations $\begin{array}{llll}\hfill {x}^{\prime }\left(t\right)& =2x\left(t\right)+3y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {y}^{\prime }\left(t\right)& =5x\left(t\right)-4y\left(t\right)\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Eliminate $y$ to find a second order linear differential equation in $x$ such that $x=x\left(t\right)$.
 a) ${x}^{″}\left(t\right)+2{x}^{\prime }\left(t\right)-23x\left(t\right)=0$ b) ${x}^{″}\left(t\right)-2{x}^{\prime }\left(t\right)+23x\left(t\right)=0$ c) ${x}^{″}\left(t\right)-\frac{2}{3}{x}^{\prime }\left(t\right)-\frac{7}{3}x\left(t\right)=0$ d) ${x}^{″}\left(t\right)-6{x}^{\prime }\left(t\right)-x\left(t\right)=0$

Choice (a) is correct!
$\begin{array}{llll}\hfill {x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)+3{y}^{\prime }\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)+15x\left(t\right)-12y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{\prime }\left(t\right)+15x\left(t\right)-4{x}^{\prime }\left(t\right)+8x\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-2{x}^{\prime }\left(t\right)+23x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{″}\left(t\right)& +2{x}^{\prime }\left(t\right)-23x\left(t\right)=0\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Two populations, $x=x\left(t\right)$ and $y=y\left(t\right)$, satisfy the following differential equations $\begin{array}{llll}\hfill \frac{dx}{dt}& =3x+2y,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{dy}{dt}& =x+y\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Eliminate $y$ to find a second order linear differential equation in $x$ such that $x=x\left(t\right)\phantom{\rule{0.3em}{0ex}}.$
 a) $\frac{{d}^{2}x}{d{t}^{2}}+4\frac{dx}{dt}-x=0$ b) $\frac{{d}^{2}x}{d{t}^{2}}-4\frac{dx}{dt}+5x=0$ c) $\frac{{d}^{2}x}{d{t}^{2}}-4\frac{dx}{dt}+x=0$ d) $\frac{{d}^{2}x}{d{t}^{2}}+4\frac{dx}{dt}-5x=0\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$\begin{array}{llll}\hfill \frac{{d}^{2}x}{d{t}^{2}}& =3\frac{dx}{dt}+2\frac{dy}{dt},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}x}{d{t}^{2}}& =3\frac{dx}{dt}+2x+2y,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}x}{d{t}^{2}}& =3\frac{dx}{dt}+2x+\frac{dx}{dt}-3x=0,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}x}{d{t}^{2}}& =4\frac{dx}{dt}-x,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}x}{d{t}^{2}}& -4\frac{dx}{dt}+x=0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (d) is incorrect
Let populations $A$ and $B$ be represented by $x\left(t\right)$ and $y\left(t\right)$ respectively at time $t\phantom{\rule{0.3em}{0ex}}.$ Population $A$’s growth is 4 times itself and it is decreased by 2 times the population of $B\phantom{\rule{0.3em}{0ex}}.$ Population $B$’s growth is 5 times population $A$ and it is decreased by 3 times its own population. Which of the systems below model these populations ?
 a) ${x}^{\prime }\left(t\right)=4x\left(t\right)-2y\left(t\right)$,${y}^{\prime }\left(t\right)=5y\left(t\right)-3x\left(t\right)$. b) ${x}^{\prime }\left(t\right)=4x\left(t\right)-2y\left(t\right)$,${y}^{\prime }\left(t\right)=5x\left(t\right)-3y\left(t\right)$. c) ${x}^{\prime }\left(t\right)=4x\left(t\right)+2y\left(t\right)$,${y}^{\prime }\left(t\right)=5x\left(t\right)+3y\left(t\right)$. d) ${x}^{\prime }\left(t\right)=4x\left(t\right)+2y\left(t\right)$,${y}^{\prime }\left(t\right)=5y\left(t\right)+3x\left(t\right)$.

Choice (a) is incorrect
Choice (b) is correct!
This is a classic predator-prey relationship where population $B$ is the predator and population $A$ is the prey.
Choice (c) is incorrect
Choice (d) is incorrect
Let populations $A$ and $B$ be represented by $x\left(t\right)$ and $y\left(t\right)$ respectively at time $t\phantom{\rule{0.3em}{0ex}}.$ Population $A$s growth is 2 times itself and it is increased by 3 times the population of $B\phantom{\rule{0.3em}{0ex}}.$ Population $B$s growth is 3 times itself and it is increased by 4 times population $A\phantom{\rule{0.3em}{0ex}}.$ Which of the systems below model these populations ?
 a) ${x}^{\prime }\left(t\right)=2x\left(t\right)-3y\left(t\right)$,${y}^{\prime }\left(t\right)=3y\left(t\right)-4x\left(t\right)$. b) ${x}^{\prime }\left(t\right)=2x\left(t\right)+3y\left(t\right)$,${y}^{\prime }\left(t\right)=3x\left(t\right)+4y\left(t\right)$. c) ${x}^{\prime }\left(t\right)=2x\left(t\right)-3y\left(t\right)$,${y}^{\prime }\left(t\right)=4x\left(t\right)-3y\left(t\right)$. d) ${x}^{\prime }\left(t\right)=2x\left(t\right)+3y\left(t\right)$,${y}^{\prime }\left(t\right)=4x\left(t\right)+3y\left(t\right)$.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
This is a classic symbiotic relationship where each population depends on the other for growth.
Find the general solution to the system of linear differential equations $\begin{array}{llll}\hfill {x}^{\prime }\left(t\right)& =4x\left(t\right)-2y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {y}^{\prime }\left(t\right)& =5x\left(t\right)-3y\left(t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
 a) $x\left(t\right)=A{e}^{2t}+B{e}^{t}$,$y\left(t\right)=3A{e}^{2t}+\frac{3}{2}B{e}^{t}.$ b) $x\left(t\right)=A{e}^{-2t}+B{e}^{-t}$,$y\left(t\right)=5A{e}^{2t}+\frac{5}{2}B{e}^{t}$. c) $x\left(t\right)=A{e}^{-2t}+B{e}^{t}$,$y\left(t\right)=6A{e}^{-2t}+\frac{3}{2}B{e}^{t}.$ d) $x\left(t\right)=A{e}^{2t}+B{e}^{-t},$$y\left(t\right)=A{e}^{2t}+\frac{5}{2}B{e}^{-t}.$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
First find a differential equation involving $x\left(t\right)$ only : $\begin{array}{llll}\hfill {x}^{″}\left(t\right)& =4{x}^{\prime }\left(t\right)-2{y}^{\prime }\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{″}\left(t\right)& =4{x}^{\prime }\left(t\right)-10x\left(t\right)+6y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4{x}^{\prime }\left(t\right)-10x\left(t\right)+12{x}^{\prime }\left(t\right)-3x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{\prime }\left(t\right)+2x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{″}\left(t\right)& -{x}^{\prime }\left(t\right)-2x\left(t\right)=0\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ The auxiliary equation is ${k}^{2}-k-2=\left(k-2\right)\left(k+1\right)=0$ and the general solution is $\begin{array}{llll}\hfill x\left(t\right)& =A{e}^{2t}+B{e}^{-t},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{\prime }\left(t\right)& =2A{e}^{2t}-B{e}^{-t},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y\left(t\right)& =2x\left(t\right)-\frac{1}{2}{x}^{\prime }\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2A{e}^{2t}+2B{e}^{-t}-A{e}^{2t}+\frac{1}{2}B{e}^{-t},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =A{e}^{2t}+\frac{5}{2}B{e}^{-t}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Two populations $x=x\left(t\right)$ and $y=y\left(t\right)$ satisfy the following differential equations $\begin{array}{llll}\hfill {x}^{\prime }\left(t\right)& =2x\left(t\right)+3y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {y}^{\prime }\left(t\right)& =4x\left(t\right)+3y\left(t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ What is the solution to this system if $x\left(0\right)=3000$ and $y\left(0\right)=500\phantom{\rule{0.3em}{0ex}}?$
 a) $x\left(t\right)=3500{e}^{2t}-500{e}^{3t}$,$y\left(t\right)=500{e}^{3t}\phantom{\rule{0.3em}{0ex}}.$ b) $x\left(t\right)=\frac{9500}{3}{e}^{2t}-\frac{500}{3}{e}^{3t}$,$y\left(t\right)=\frac{500}{3}{e}^{3t}\phantom{\rule{0.3em}{0ex}}.$ c) $x\left(t\right)=1500{e}^{-6t}+1500{e}^{t},$$y\left(t\right)=2000{e}^{-6t}+1500{e}^{t}\phantom{\rule{0.3em}{0ex}}.$ d) $x\left(t\right)=1500{e}^{6t}+1500{e}^{-t},$$y\left(t\right)=2000{e}^{6t}-1500{e}^{-t}\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
We first find the differential equation for $x\left(t\right)$ : $\begin{array}{llll}\hfill {x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)+3{y}^{\prime }\left(t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)+12x\left(t\right)+9y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{\prime }\left(t\right)+12x\left(t\right)+3{x}^{\prime }\left(t\right)-6x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5{x}^{\prime }\left(t\right)+6x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& -5x\left(t\right)-6x\left(t\right)=0\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ It follows that the general solution is $x\left(t\right)=A{e}^{6t}+B{e}^{-t}$ and $y\left(t\right)=\frac{4}{3}A{e}^{6t}-B{e}^{-t}\phantom{\rule{0.3em}{0ex}}.$ Substituting the initial conditions gives $A+B=3000$ and $\frac{4}{3}A-B=500$ which gives $A=B=1500\phantom{\rule{0.3em}{0ex}}.$
Two populations $x=x\left(t\right)$ and $y=y\left(t\right)$ satisfy the following differential equations $\begin{array}{llll}\hfill \frac{dx}{dt}& =3x+2y,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{dy}{dt}& =4x+y\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Find the general solution to this system.
 a) $x\left(t\right)=A{e}^{\left(2-\sqrt{15}\right)t}+B{e}^{\left(2+\sqrt{15}\right)t}$ $y\left(t\right)=\frac{5-\sqrt{15}}{2}A{e}^{\left(2-\sqrt{15}\right)t}+\frac{5+\sqrt{15}}{2}A{e}^{\left(2+\sqrt{15}\right)t}$ b) $x\left(t\right)=A{e}^{\left(-2-\sqrt{15}\right)t}+B{e}^{\left(-2+\sqrt{15}\right)t}$,$y\left(t\right)=\frac{1-\sqrt{15}}{2}A{e}^{\left(-2-\sqrt{15}\right)t}+\frac{1+\sqrt{15}}{2}A{e}^{\left(-2+\sqrt{15}\right)t}$. c) $x\left(t\right)=A{e}^{5t}+B{e}^{-t}$,$y\left(t\right)=A{e}^{5t}+2B{e}^{-t}.$ d) $x\left(t\right)=A{e}^{-5t}+B{e}^{t},$$y\left(t\right)=4A{e}^{-5t}-B{e}^{t}.$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
We first find the differential equation for $x\left(t\right)$ : $\begin{array}{llll}\hfill \frac{{d}^{2}x}{d{t}^{2}}& =3\frac{dx}{dt}+2\frac{dy}{dt},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}x}{d{t}^{2}}& -3\frac{dx}{dt}-8x-2y=0,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}x}{d{t}^{2}}& -3\frac{dx}{dt}-8x-\frac{dx}{dt}+3x=0,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}x}{d{t}^{2}}& -4\frac{dx}{dt}-5x=0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$ The auxiliary equation is ${k}^{2}-4k-5=\left(k-5\right)\left(k+1\right)=0$ hence the general solution is $\begin{array}{llll}\hfill x\left(t\right)& =A{e}^{5t}+B{e}^{-t}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}\frac{dx}{dt}& =5A{e}^{5t}-B{e}^{-t},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y\left(t\right)& =\frac{1}{2}\frac{dx}{dt}-\frac{3}{2}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5}{2}A{e}^{5t}-\frac{1}{2}B{e}^{-t}-\frac{3}{2}A{e}^{5t}-\frac{3}{2}B{e}^{-t}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =A{e}^{5t}-2B{e}^{-t}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (d) is incorrect
Find the particular solution to the system of linear differential equations $\begin{array}{llll}\hfill {x}^{\prime }\left(t\right)& =7x\left(t\right)-12y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {y}^{\prime }\left(t\right)& =4x\left(t\right)-7y\left(t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
 a) $x\left(t\right)=3{e}^{-t}-2{e}^{t}$,$y\left(t\right)=2{e}^{-t}-{e}^{t}.$ b) $x\left(t\right)=2{e}^{-t}-{e}^{t}$,$y\left(t\right)=\frac{2}{3}{e}^{-t}+\frac{1}{3}{e}^{t}$. c) $x\left(t\right)=\frac{11}{7}sint+cost$,$y\left(t\right)=\frac{5}{6}sint+cost$. d) $x\left(t\right)=\frac{1}{7}{e}^{7t}+\frac{6}{7}{e}^{-7t}$,$y\left(t\right)={e}^{-7t}$.

Choice (a) is correct!
We find the differential equation for $x\left(t\right)$ : $\begin{array}{llll}\hfill {x}^{″}\left(t\right)& =7{x}^{\prime }\left(t\right)-12{y}^{\prime }\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& =7{x}^{\prime }\left(t\right)-48x\left(t\right)+84y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =7{x}^{\prime }\left(t\right)-48x\left(t\right)-7{x}^{\prime }\left(t\right)+49x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{″}\left(t\right)& -x\left(t\right)=0\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ So $x\left(t\right)=A{e}^{t}+B{e}^{-t}$ and $y\left(t\right)=\frac{1}{2}A{e}^{t}+\frac{2}{3}B{e}^{-t}\phantom{\rule{0.3em}{0ex}}.$ Hence $A+B=1$ and $\frac{1}{2}A+\frac{2}{3}B=1$ which gives $A=-2$ and $B=3\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the system of linear differential equations $\begin{array}{llll}\hfill {x}^{\prime }\left(t\right)& =2x\left(t\right)-3y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {y}^{\prime }\left(t\right)& =3x\left(t\right)+2y\left(t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
 a) $x\left(t\right)=A{e}^{\left(2+\sqrt{17}\right)t}+B{e}^{\left(2-\sqrt{17}\right)t},$$y\left(t\right)=3A{e}^{2t}+\frac{3}{2}B{e}^{t}$. b) $x\left(t\right)=A{e}^{\left(-2+\sqrt{17}\right)t}+B{e}^{\left(-2-\sqrt{17}\right)t},$$y\left(t\right)=-\frac{1}{3}A{e}^{\left(-2+\sqrt{17}\right)t}+-\frac{1}{3}B{e}^{\left(-2-\sqrt{17}\right)t}$. c) $x\left(t\right)={e}^{-2t}\left(Asin3t+Bcos3t\right)$,$y\left(t\right)={e}^{-2t}\left(\left(\frac{4}{3}A+B\right)sin3t-\left(A-\frac{4}{3}B\right)cos3t\right)$. d) $x\left(t\right)={e}^{2t}\left(Asin3t+Bcos3t\right)$,$y\left(t\right)={e}^{2t}\left(Bsin3t-Acos3t\right)$.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
We find the differential equation for $x\left(t\right)$ : $\begin{array}{llll}\hfill {x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)-3{y}^{\prime }\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)-9x\left(t\right)-6y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{\prime }\left(t\right)-9x\left(t\right)+2{x}^{\prime }\left(t\right)-4x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={4}^{\prime }\left(t\right)-13x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& -4{x}^{\prime }\left(t\right)+13x\left(t\right)=0\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ So the auxiliary equation ${k}^{2}-4k+13=0$ and its roots are $k=2±3\sqrt{-1}$ and the general solution is $\begin{array}{llll}\hfill x\left(t\right)& ={e}^{2t}\left(Asin3t+Bcos3t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{\prime }\left(t\right)& =2{e}^{2t}\left(Asin3t+Bcos3t\right)+{e}^{2t}\left(3Acos3t-3Bsin3t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y\left(t\right)& =\frac{2}{3}x\left(t\right)-\frac{1}{3}{x}^{\prime }\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{2t}\left(Bsin3t-Acos3t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Find the particular solution to the system of linear differential equations $\begin{array}{llll}\hfill {x}^{\prime }\left(t\right)& =x\left(t\right)-2y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {y}^{\prime }\left(t\right)& =4x\left(t\right)-3y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ where $y\left(0\right)=1$ and $y\left(\pi \right)=2\phantom{\rule{0.3em}{0ex}}.$
 a) $x\left(t\right)={e}^{-t}\left(\frac{1}{2}sin2t-\frac{5}{2}cos2t\right)$,$y\left(t\right)={e}^{2t}\left(2sin2t+cos2t\right)$. b) $x\left(t\right)={e}^{-t}\left(\frac{1}{2}sin2t+\frac{3}{2}cos2t\right)$,$y\left(t\right)={e}^{2t}\left(2sin2t+cos2t\right)$. c) $x\left(t\right)=A{e}^{\left(2+\sqrt{17}\right)t}+B{e}^{\left(2-\sqrt{17}\right)t}$,$y\left(t\right)=3A{e}^{2t}+\frac{3}{2}B{e}^{t}$. d) $x\left(t\right)=A{e}^{\left(-2+\sqrt{17}\right)t}+B{e}^{\left(-2-\sqrt{17}\right)t}$,$y\left(t\right)=-\frac{1}{3}A{e}^{\left(-2+\sqrt{17}\right)t}+-\frac{1}{3}B{e}^{\left(-2-\sqrt{17}\right)t}$.

Choice (a) is incorrect
Choice (b) is correct!
$\begin{array}{llll}\hfill {x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)-3{y}^{\prime }\left(t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& =2{x}^{\prime }\left(t\right)-9x\left(t\right)-6y\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{\prime }\left(t\right)-9x\left(t\right)+2{x}^{\prime }\left(t\right)-4x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={4}^{\prime }\left(t\right)-13x\left(t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{″}\left(t\right)& =4{x}^{\prime }\left(t\right)+13x\left(t\right)=0\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ The auxiliary equation is thus ${k}^{2}-4k-13=0$ and its roots are $k=2±3\sqrt{-1}$. The general solution is $\begin{array}{llll}\hfill x\left(t\right)& ={e}^{2t}\left(Asin3t+Bcos3t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{therefore,}\phantom{\rule{0.3em}{0ex}}{x}^{\prime }\left(t\right)& =2{e}^{2t}\left(Asin3t-Bcos3t\right)+{e}^{2t}\left(3Acos3t-3Bsin3t\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y\left(t\right)& =\frac{2}{3}x\left(t\right)-\frac{1}{3}{x}^{\prime }\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{e}^{2t}\left(Bsin3t-Acos3t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (c) is incorrect
Choice (d) is incorrect