Product substitution and a logistic epidemic

(a) By means of the substitution z=xyz=xy, show that the differential equation 2xdydx=1x−y2x\frac{dy}{dx}=\frac1x-y can be reduced to the form 21+zdzdx=1x\frac2{1+z}\frac{dz}{dx}=\frac1x. marks

Hence, solve the differential equation, giving your answer in the form f(x,y)=Axf(x,y)=Ax, where f(x,y)f(x,y) is a function in x,yx,y and AA is an arbitrary constant. marks

(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 xx denote the number of infected cats (in hundreds) in tt days. Show that the rate of increase of infected cats, dxdt\frac{dx}{dt} can be expressed as dxdt=kx(15−x)\frac{dx}{dt}=kx(15-x), where kk is an exact constant to be determined. marks

Find xx in terms of tt. marks

[2017/RVHS/Promo/10]

Total marks: 11

Solution

(a) substitution

Substitution

Use z=xyz=xy to reduce 2xdydx=1x−y2x\dfrac{dy}{dx}=\dfrac1x-y to 21+zdzdx=1x\dfrac2{1+z}\dfrac{dz}{dx}=\dfrac1x.

Hint

  1. z′=y+xy′z'=y+xy' gives xy′=z′−zxxy'=z'-\frac zx.

Answer · (a) substitution

2z′/(1+z)=1/x2z\prime/(1+z)=1/x.

(a) solution

Separation of variables

Hence solve the equation, giving the answer in the form f(x,y)=Axf(x,y)=Ax.

Explore this idea

Hint

  1. ∫21+zdz=∫1xdx\int\frac2{1+z}dz=\int\frac1xdx.

Answer · (a) solution

(1+xy)2=Ax(1+xy)^2=Ax.

x≠0x\ne0; A=0A=0 includes y=−1/xy=-1/x lost by division.

Explore the idea

Solutions of 2xy' = 1/x − y

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of y at x = 10
Read the solution at x2
Try an example
y at x = 2
0.207
Rate dy/dx there
0.073
  • Solution that starts at the purple dot
Key idea

Each solution is a branch of (1+xy)2=Ax(1+xy)^2=Ax, defined for xx of one sign (for A≠0A\neq0).

Ideas to try
Try the other starting values: every solution satisfies (1+xy)2=Ax(1+xy)^2=Ax.
(b) differential equation

Logistic model

Show that dxdt=kx(15−x)\dfrac{dx}{dt}=kx(15-x) and find kk (infected cats xx in hundreds of a population of 1500).

Hint

  1. The rate is proportional to infected ×\times not infected: x(15−x)x(15-x) in hundreds.

  2. At x=1x=1, dxdt=0.15\frac{dx}{dt}=0.15.

Answer · (b) differential equation

x′=kx(15−x)x\prime=kx(15-x); k=3/280k=3/280.

(b) solution

Logistic solution

Find xx in terms of tt.

Explore this idea

Hint

  1. Partial fractions: 1x(15−x)=115(1x+115−x)\frac1{x(15-x)}=\frac1{15}\left(\frac1x+\frac1{15-x}\right).

Answer · (b) solution

x=15/(1+29e−9t/56)x=15/(1+29e^{-9t/56}) (hundreds).

Explore the idea

The epidemic

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of x at t = 00.5
Read the solution at t21
Try an example
x at t = 21
7.428
Rate dx/dt there
0.602
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The curve is S-shaped and tends to 15 hundred: the whole population.

Ideas to try
  1. Read when half the cats (7.5 hundred) are infected.
  2. Start above 15: what happens?