Questions 8 (Hints and solutions start on page .)

Q 4.1   Solve these differential equations together with the given conditions. Remember that you can always check your solution by plugging it back into the equation.

y' = $${\frac{x^2}{y^2}}$$        y(1) = 1,                y' = y²(x + 1)        y(0) = 1

xy' = y²(1 + x)        y(1) = - 1,                y' = y(1 - y)        y(0) = $${\textstyle\frac{1}{2}}$$

$${\frac{dr}{d\theta}}$$ = r²sin$$\theta$$        r(0) = 1,                $${\frac{du}{dv}}$$ = e^{u + v}        u(0) = 0

Q 4.2   Find the solution to y'''(x) = x that satisfies y(0) = 0, y(1) = 1 and y(2) = 2.

Q 4.3   Newton's Law of cooling says that, under certain circumstances, a body cools (loses temperature) at a rate proportional to the difference between the temperature of the body and that of the surrounding air. So if the temperature of the body at time t is T(t) and the temperature of the air is T₀ then $\dot{T}$ = - $\kappa$(T - T₀), where $\kappa$ is a positive constant.

Solve this equation (on the assumption that T₀ is constant and T(0) > T₀). Suppose that T₀ = 20^o and that T(0) = 100^o. If the temperature of the body after 10 seconds is 80^o what is the temperature after 20 seconds. When does the temperature drop to 30^o?

Q 4.4   Pareto's Law in Economics says that if y(x) represents the number of people in a stable economy whose incomes are greater than x then the rate of decrease of y with respect to x is proportional to y and inversely proportional to x. Show that this leads to a differential equation of the form

$${\frac{dy}{dx}}$$ = - c$${\frac{y}{x}}$$

Find the general solution to this equation (Pareto's Law).

*Q 4.5   Consider the differential equation x²y' = xy + y² You probably can't solve this as it stands. Now introduce a new function v(x) by y(x) = xv(x). Rewrite the equation in terms of v and show that you get xv' = v². Solve this and hence solve the original equation.

*Q 4.6   An infectious disease breaks out in a population. At any time thereafter let N(t) be the number of people who have not caught the disease yet, let D(t) be the number of people who currently have the disease and let R(t) be the number of people who are out of it for some reason (immune, isolated, recovered or dead). If P is the total population (assumed constant if we count the dead) then N + D + R = P at all times.

The rate at which susceptible people catch the disease depends on how likely they are to meet somebody who has the disease. The chances of one susceptible person meeting a diseased person in the time interval from t to t + dt is assumed to be proportional to (D/P)dt. This leads to the equation

$$\dot{N} \alpha {\frac{ND}{P}}$$ (4.4)

where $\alpha$ is a positive constant.

Suppose that people `recover' from the disease at a fixed rate $\beta$. Then, taking into account the people who are catching the disease, we have the equation

$$\dot{D} \alpha {\frac{ND}{P}} \beta$$ (4.5)

Consequently, to keep the total population constant, we have $\dot{R}$ = $\beta$D.

Now, accepting these three equations, we want to know what is going to happen. If we give the disease to a small number of people in the population we want to know things like (a) how many people will eventually catch the disease, (b) how many people have the disease at its peak, (c) how long the epidemic lasts, (d) is there actually an epidemic or does the disease fail to take hold in the population?

Divide equation (4.5) by (4.4) and show that you get

$${\frac{dD}{dN}}$$ = $${\frac{\beta P}{\alpha N}}$$ - 1

Solve this under the condition that N(0) = P and D(0) = 0. (This is meant to idealise the situation of having a very small number of diseased people in the population. If you don't like this then set N(0) = P - $\epsilon$ and D(0) = $\epsilon$--it doesn't make a significant difference.)

You should get

$${\frac{D}{P}}$$ = $${\frac{\beta}{\alpha}}$$ln($${\frac{N}{P}}$$) + 1 - $${\frac{N}{P}}$$

Suppose that when the epidemic finally ends 50% of the population have had the disease. What is the value of $\alpha$/$\beta$ and what percentage of the population had the disease when it was at its peak?

Suppose that, at the peak of the epidemic, 10% of the population have the disease. What percentage of the population will have had the disease by the end of the epidemic? (You probably cannot solve the equations you get, but use your calculator to try to estimate the answer roughly.)

Can you show that the condition for an `epidemic' to occur, as opposed to just a handful of people getting the disease, is that $\beta$ should be (significantly) less than $\alpha$? In other words, what happens if $\beta$ is greater than $\alpha$?