Product substitution and a logistic epidemic · source reference

Printed question

(a) By means of the substitution $z=xy$, show that the differential equation $2x\frac{dy}{dx}=\frac1x-y$ can be reduced to the form $\frac2{1+z}\frac{dz}{dx}=\frac1x$. [2]

Hence, solve the differential equation, giving your answer in the form $f(x,y)=Ax$, where $f(x,y)$ is a function in $x,y$ and $A$ is an arbitrary constant. [3]

(b) A small town has a population of 1500 cats. During an epidemic, 50 cats are initially infected with this disease and these infected cats in the town spread the disease to other cats at a rate proportional to the product of the number of infected cats and the number of cats that are not infected. It is also given that the number of infected cats increase at a rate of 15 cats per day when the number of infected cats is 100. Let $x$ denote the number of infected cats (in hundreds) in $t$ days. Show that the rate of increase of infected cats, $\frac{dx}{dt}$ can be expressed as $\frac{dx}{dt}=kx(15-x)$, where $k$ is an exact constant to be determined. [2]

Find $x$ in terms of $t$. [4]

[2017/RVHS/Promo/10]

Printed working

Part (a)

$z=xy$ (1); $z'=y+xy'$ (2).

$2xy'=1/x-y$; $2(z'-z/x)=1/x-z/x$; $2z'=(1+z)/x$; $2z'/(1+z)=1/x$.

$\int2/(1+z)dz=\int1/x\,dx$; $2\ln|1+z|=\ln|x|+c$; $\ln|(1+z)^2/x|=c$.

$|(1+z)^2/x|=e^c$; $(1+z)^2/x=A$, $A=\pm e^c$; $(1+z)^2=Ax$; $(1+xy)^2=Ax$.

Part (b)

$x'\propto x(15-x)$; $x'=kx(15-x)$. At $x=1$, $x'=0.15$, $0.15=k(1)(15-1)$; $k=0.15/14=3/280$.

$\int dx/[x(15-x)]=\int(3/280)dt$.

$\frac1{15}\int[1/x+1/(15-x)]dx=\int(3/280)dt$; $\int[1/x+1/(15-x)]dx=\int(9/56)dt$.

$\ln|x|-\ln|15-x|=9t/56+C$; $\ln|x/(15-x)|=9t/56+C$.

$x/(15-x)=\pm e^{9t/56+C}=Ae^{9t/56}$, $A=\pm e^C$.

$x=Ae^{9t/56}(15-x)$; $x(1+Ae^{9t/56})=15Ae^{9t/56}$; $x=15Ae^{9t/56}/(1+Ae^{9t/56})$.

At 0, $x=0.5$: $0.5=15A/(1+A)\Rightarrow1+A=30A\Rightarrow A=1/29$.

$x=15e^{9t/56}/(29+e^{9t/56})$.

Printed answer key

(a) $(1+xy)^2=Ax$. (b) $x=15e^{9t/56}/(29+e^{9t/56})$.

Teacher note 1

Method 1: partial fractions. $1/[x(15-x)]=A/x+B/(15-x)$. Thus $1=A(15-x)+Bx$. Let $x=0$: $A=1/15$; let $x=15$: $B=1/15$. Therefore $1/[x(15-x)]=(1/15)[1/x+1/(15-x)]$.

Mathematical review

(a) $2(z'-z/x)=1/x-z/x$ gives $2z'=(1+z)/x$. For z≠−1, integrate $2\ln|1+z|=\ln|x|+C$, so $(1+xy)^2=Ax$ on a connected interval with x≠0. The separately lost solution z=−1, or y=−1/x, is included by A=0. For A≠0 the implicit relation has branches $xy=-1\pm\sqrt{Ax}$ with Ax>0.

(b) $x'=kx(15-x)$. At x=1, x′=0.15, so $k=0.15/14=3/280$. Separation using $1/[x(15-x)]=(1/15)(1/x+1/(15-x))$ gives $\ln[x/(15-x)]=(9/56)t+C$. Since x(0)=1/2, the odds are 1/29. Thus $x=15/(1+29e^{-9t/56})=15e^{9t/56}/(29+e^{9t/56})$. Differentiation verifies the logistic equation; x rises from 0.5 toward 15.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 5. Working page(s): 20, 21, 22. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.