## MATH1013 Quizzes

Quiz 6: Systematic solution of first Order Linear Difference Equations
Question 1 Questions
Solve the first order linear difference equation
${X}_{n}-{X}_{n-1}=0.1{X}_{n-1}$
where ${X}_{1}^{}=100\phantom{\rule{0.3em}{0ex}}.$
 a) ${X}_{n}=100{\left(1.1\right)}^{n-1}\phantom{\rule{0.3em}{0ex}}.$ b) ${X}_{n}=100{\left(0.9\right)}^{n-1}$ c) ${X}_{n}=100{\left(1.1\right)}^{n}$ d) ${X}_{n}=0.9{\left(100\right)}^{n}$

Choice (a) is correct!
${X}_{n}=1.1{X}_{n-1}$ so the solution is ${X}_{n}={\left(1.1\right)}^{n-1}{X}_{1}^{}=100{\left(1.1\right)}^{n-1}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Solve the first order linear difference equation
${X}_{n}=\frac{4}{3}{X}_{n-1}-400$
where ${X}_{1}^{}=1000\phantom{\rule{0.3em}{0ex}}.$
 a) ${X}_{n}=\frac{2600}{3}{\left(\frac{4}{3}\right)}^{n-1}+\frac{400}{3}$ b) ${X}_{n}=\frac{2600}{3}{\left(\frac{4}{3}\right)}^{n-1}-\frac{400}{3}$ c) ${X}_{n}=200{\left(\frac{4}{3}\right)}^{n-1}-1200$ d) ${X}_{n}=1200-200{\left(\frac{4}{3}\right)}^{n-1}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\begin{array}{llll}\hfill {X}_{n}& ={\left(\frac{4}{3}\right)}^{n-1}{X}_{1}^{}-400\left(\sum _{k=0}^{n-2}{\left(\frac{4}{3}\right)}^{k}\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill Now\phantom{\rule{1em}{0ex}}\sum _{k=0}^{n-2}{\left(\frac{4}{3}\right)}^{k}& =\frac{{\left(\frac{4}{3}\right)}^{n-1}-1}{\frac{4}{3}-1}=3\left({\left(\frac{4}{3}\right)}^{n-1}-1\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill so\phantom{\rule{1em}{0ex}}{X}_{n}& =1000{\left(\frac{4}{3}\right)}^{n-1}-1200\left({\left(\frac{4}{3}\right)}^{n-1}-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1200-200{\left(\frac{4}{3}\right)}^{n-1}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
A disease spreads through a population in the following way. Each year 1000 new cases of the disease occur and half of the existing cases are cured. At the beginning of the year 2000 there were 1200 cases of the disease. Find the difference equation which describes this situation and solve it.
 a) The difference equation is ${X}_{n}=1.5{X}_{n-1}+100$ and the solution is ${X}_{n}=3200{\left(1.5\right)}^{n-1}-200\phantom{\rule{0.3em}{0ex}}.$ b) The difference equation is ${X}_{n}=1.5{X}_{n-1}+100$ and the solution is ${X}_{n}=2000-800{\left(1.5\right)}^{n-1}\phantom{\rule{0.3em}{0ex}}.$ c) The difference equation is ${X}_{n}=0.5{X}_{n-1}+100$ and the solution is ${X}_{n}=3200{\left(0.5\right)}^{n-1}-200\phantom{\rule{0.3em}{0ex}}.$ d) The difference equation is ${X}_{n}=0.5{X}_{n-1}+100$ and the solution is ${X}_{n}=2000-800{\left(1.5\right)}^{n-1}\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\begin{array}{llll}\hfill {X}_{n}& =0.5{X}_{n-1}+1000,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {X}_{n}& ={\left(0.5\right)}^{n-1}{X}_{1}^{}+1000\left(\sum _{k=0}^{n-2}{\left(0.5\right)}^{k}\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill Now\phantom{\rule{1em}{0ex}}\sum _{k=0}^{n-2}{\left(0.5\right)}^{k}& =\frac{{\left(0.5\right)}^{n-1}-1}{0.5-1}=-2\left({\left(0.5\right)}^{n-1}-1\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill so\phantom{\rule{1em}{0ex}}{X}_{n}& =1200{\left(0.5\right)}^{n-1}-2000\left({\left(0.5\right)}^{n-1}-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2000-800{\left(0.5\right)}^{n-1}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
A population increases by $20%$ per annum and every year 150 are culled. If the initial population size is 100, how long does it take for the population to reach 1500 ?
 a) At the beginning of the 6th year. b) At the beginning of the 7th year. c) At the beginning of the 9th year. d) Never, the population is decreasing.

Choice (a) is incorrect
Choice (b) is correct!
$\begin{array}{llll}\hfill {X}_{n}& =1.2{X}_{n-1}-150,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {X}_{n}& ={\left(1.2\right)}^{n-1}{X}_{1}^{}-150\left(\sum _{k=0}^{n-2}{\left(1.2\right)}^{k}\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill Now\phantom{\rule{1em}{0ex}}\sum _{k=0}^{n-2}{\left(1.2\right)}^{k}& =\frac{{\left(1.2\right)}^{n-1}-1}{1.2-1}=5\left({\left(1.2\right)}^{n-1}-1\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill so\phantom{\rule{1em}{0ex}}{X}_{n}& =1000{\left(1.2\right)}^{n-1}-750\left({\left(1.2\right)}^{n-1}-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =750+250{\left(1.2\right)}^{n-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill Now\phantom{\rule{1em}{0ex}}solving\phantom{\rule{1em}{0ex}}1500& =750+250{\left(1.2\right)}^{n-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ⇒250{\left(1.2\right)}^{n-1}& =750,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(1.2\right)}^{n-1}& =\frac{750}{250}=3,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n-1& =\frac{ln3}{ln1.2}=6.02\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ⇒n& =7.02.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Therefore the population will be 1500 at the beginning of the 7th year.
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the difference equation
${x}_{n+1}-9{x}_{n-1}=0.$
 a) ${x}_{n}={9}^{n-1}{x}_{1}^{}$ b) ${x}_{n}=A{3}^{n}+An{3}^{n}$ c) ${x}_{n}=\left(A+B{\left(-1\right)}^{n}\right){3}^{n}$ d) ${x}_{n}=A{3}^{n}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The auxiliary equation is ${\lambda }^{2}-9=\left(\lambda -3\right)\left(\lambda +3\right)=0\phantom{\rule{0.3em}{0ex}},$ so the solution is
${x}_{n}=A{3}^{n}+B{\left(-3\right)}^{n}=\left(A+B{\left(-1\right)}^{n}\right){3}^{n}\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Find the general solution to the difference equation
${x}_{n+1}=4{x}_{n-1}+21{x}_{n}.$
 a) ${x}_{n}=A\left(-{7}^{n}\right)+B\left(-{3}^{n}\right)$ b) ${x}_{n}=A\left({7}^{n}\right)+B\left({3}^{n}\right)$ c) ${x}_{n}=A\left({7}^{n}\right)+B\left(-{3}^{n}\right)$ d) ${x}_{n}=A\left(-{7}^{n}\right)+B\left({3}^{n}\right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The difference equation is ${x}_{n+1}-4{x}_{n-1}-21{x}_{n}=0$. The auxiliary equation is ${\lambda }^{2}-4\lambda -21=\left(\lambda -7\right)\left(\lambda +3\right)=0\phantom{\rule{0.3em}{0ex}},$ so the solution is
${x}_{n}=A\left({7}^{n}\right)+B{\left(-3\right)}^{n}.$
Choice (d) is incorrect
Find the solution to the second order linear difference equation
${x}_{n+2}-3{x}_{n+1}-4{x}_{n}=0,$
where ${x}_{0}^{}=1$ and ${x}_{1}^{}=2\phantom{\rule{0.3em}{0ex}}.$
 a) ${x}_{n}=A{4}^{n}+B{\left(-1\right)}^{n}$ b) ${x}_{n}=A{\left(-4\right)}^{n}+B$ c) ${x}_{n}=\left(0.6\right){4}^{n}+\left(0.4\right){\left(-1\right)}^{n}$ d) ${x}_{n}=\left(-0.2\right){\left(-4\right)}^{n}+1.2$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The auxiliary equation is ${\lambda }^{2}-3\lambda -4=\left(\lambda -4\right)\left(\lambda +1\right)=0,$ so the solution is
${x}_{n}=A{4}^{n}+B{\left(-1\right)}^{n}.$
Using the initial conditions $\begin{array}{llll}\hfill A+B& =1\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}4A-B=2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{1em}{0ex}}implies\phantom{\rule{1em}{0ex}}A& =0.6\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}B=0.4\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (d) is incorrect
A population increases in such a way that its growth in any year is three times its growth in the previous year. Find the second order linear difference equation that models this population.
 a) ${x}_{n+2}-4{x}_{n+1}+3{x}_{n}=0$ b) ${x}_{n+2}+2{x}_{n+1}+3{x}_{n}=0$ c) ${x}_{n+2}-3{x}_{n}=0$ d) ${x}_{n+2}-2{x}_{n+1}+3{x}_{n}=0$

Choice (a) is correct!
${x}_{n+2}-{x}_{n+1}$ is the growth in year $n+2$. ${x}_{n+1}-{x}_{n}$ is the growth in the previous year i.e. year $n+1$. Hence ${x}_{n+2}-{x}_{n+1}=3\left({x}_{n+1}-{x}_{n}\right)$ and ${x}_{n+2}-4{x}_{n+1}+3{x}_{n}=0\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
A population increases in such a way that its growth in any year is three times its growth in the previous year. Find the population in year $n$ if the initial population was 1000 and the population the next year was 1400.
 a) ${x}_{n}=600\left({3}^{n}\right)+400{\left(-1\right)}^{n}$ b) ${x}_{n}=200\left({3}^{n}\right)+800$ c) ${x}_{n}=\frac{400}{3}\left({4}^{n}\right)+\frac{2600}{3}$ d) ${x}_{n}=80\left({4}^{n}\right)+920{\left(-1\right)}^{n}$

Choice (a) is incorrect
Choice (b) is correct!
From the previous question we know that the difference equation is ${x}_{n+2}-4{x}_{n+1}+3{x}_{n}=0\phantom{\rule{0.3em}{0ex}}.$ Hence the auxiliary equation is ${\lambda }^{2}-4\lambda +3=\left(\lambda -3\right)\left(\lambda -1\right)=0\phantom{\rule{0.3em}{0ex}}.$ Therefore the solution is
${x}_{n}=A{\left(3\right)}^{n}+B{\left(1\right)}^{n}.$
Using initial conditions $\begin{array}{llll}\hfill A+B& =1000\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}3A+B=1400\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill implies\phantom{\rule{1em}{0ex}}A& =200\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}B=800.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Hence ${x}_{n}=200{\left(3\right)}^{n}+800$.
Choice (c) is incorrect
Choice (d) is incorrect
During an epidemic the number of new cases of a disease occurring during the $n$th week is equal to twice the number of cases that existed at the end of the $n-2$th week. Find the number of cases in the $n$th week if ${x}_{0}^{}={x}_{1}^{}=1\phantom{\rule{0.3em}{0ex}}.$
 a) ${x}_{n}=\left(\frac{1}{2}+\frac{\sqrt{2}}{4}\right){\left(\sqrt{2}\right)}^{n}+\left(\frac{1}{2}-\frac{\sqrt{2}}{4}\right){\left(-\sqrt{2}\right)}^{n}$ b) ${x}_{n}=\left(\frac{1}{2}+\frac{\sqrt{2}}{4}\right){\left(\sqrt{2}\right)}^{n}+\left(\frac{1}{2}-\frac{\sqrt{2}}{4}\right){\left(\sqrt{-2}\right)}^{n}$ c) ${x}_{n}=\frac{2}{3}\left({2}^{n}\right)+\frac{1}{3}{\left(-1\right)}^{n}$ d) ${x}_{n}=\frac{1}{2}\left({2}^{n}\right)+\frac{1}{2}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Since ${x}_{n}-{x}_{n-1}$ is the number of new cases occurring in week $n$ we have
${x}_{n}-{x}_{n-1}=2{x}_{n-2}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{x}_{n}-{x}_{n-1}-2{x}_{n-2}=0.$
Hence the auxiliary equation is ${\lambda }^{2}-\lambda -2=\left(\lambda -2\right)\left(\lambda +1\right)=0$ so the solution is
${x}_{n}=A{\left(2\right)}^{n}+B{\left(-1\right)}^{n}.$
Using initial conditions $\begin{array}{llll}\hfill A+B& =1\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}2A-B=1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill implies\phantom{\rule{1em}{0ex}}A& =\frac{2}{3}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}B=\frac{1}{3}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Hence ${x}_{n}=\frac{2}{3}{\left(3\right)}^{n}+\frac{1}{3}{\left(-1\right)}^{n}\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect