Differential Equations — Word Problems

Mathematics · NEB / A-Level

Differential Equations

Language Based Questions — Word Problems & Solutions

NEB 2079

A water tank is filled in such a way that the rate at which the depth of water increases is proportional to the square root of the depth. Initially the depth is $n$ metres.

a)Write down a differential equation for $h$ .
b)Show that $\sqrth = kt/2 + \sqrtn$ where $k$ is a constant.
c)When $h = 16$ and $t = 6 hours$ , prove that $n = (4 - 3k)²$ .

Solution

Let the depth of water be $h$ .

Part a

According to the question, the rate of increase of depth is proportional to $\sqrth$ :

dh/dt \propto \sqrth ∴ dh/dt = k\sqrth

This is the required differential equation.

Part b

Separating variables:

dh / \sqrth = k dt

Integrating both sides:

\int h^(-1/2) dh = k \int dt h^(-1/2 + 1) / (-1/2 + 1) = kt + c 2\sqrth = kt + c

Applying initial condition: when $t = 0$ , $h = n$ :

2\sqrtn = 0 + c \to c = 2\sqrtn

Substituting back:

2\sqrth = kt + 2\sqrtn

√h = kt/2 + √n ✓ (proved)

Part c

Substituting $h = 16$ , $t = 6$ :

\sqrt16 = 6k/2 + \sqrtn 4 = 3k + \sqrtn \sqrtn = 4 - 3k

n = (4 − 3k)² ✓ (proved)

A college hostel accommodating 1000 students; one of them came from abroad with infection of corona virus, then the hostel was isolated. The rate at which the virus spreads is assumed to be proportional to the product of the number $N$ of infected students and number of non-infected students. The number of infected students is 50 after 4 days.

a)Express the above information in the form of a differential equation.
b)Solve the differential equation.
c)Show that more than 95% students will be infected after 10 days.

Solution

Total students = 1000, Infected = $N$ , Non-infected = $1000 - N$

Part a

dN/dt \propto N(1000 - N) ∴ dN/dt = kN(1000 - N)

Part b

Separating variables and using partial fractions:

dN / [N(1000-N)] = k dt (1/1000) \int (1/N + 1/(1000-N)) dN = kt + c' (1/1000) log[N/(1000-N)] = kt + c' log[N/(1000-N)] = 1000kt + c

When $t = 0$ , $N = 1$ :

c = log(1/999)

Substituting:

log[999N/(1000-N)] = 1000kt ...(i)

When $t = 4$ , $N = 50$ :

k = (1/4000) log(999/19)

Final solution:

4 log[999N/(1000−N)] = log(999/19) · t

Part c — t = 10 days

log[999N/(1000-N)] = log(999/19)^(5/2) 999N/(1000-N) = (999/19)^(5/2) N = 952

Infection % = 952/1000 × 100 = 95.2% > 95% ✓

Water is drained from a vertical cylindrical tank by opening a valve at the base. The rate at which the water level drops is proportional to the square root of water depth $y$ , with constant of proportionality $k = 1/15$ . Find the time taken to empty the tank if water is $4 m$ deep at the beginning.

Solution

-dy/dt = k\sqrty (rate of drop)

Separating and integrating with limits $y: 4\to0$ , $t: 0\tot$ :

\int₄⁰ dy/\sqrty = -k \int₀ᵗ dt [2\sqrty]₄⁰ = -kt 0 - 2\sqrt4 = -kt 4 = kt

t = 4/k = 4/(1/15) = 60 minutes

Water flows out of a tank through a hole in the bottom. At time $t$ minutes, the depth is $x$ metres. The rate at which depth decreases is proportional to the square root of the depth.

a)Write down a differential equation modelling this situation.
b)When $t = 0, x = 2$ ; when $t = 5, x = 1$ . Find $t$ when $x = 0.5$ (correct to 1 d.p.).

Solution

Part a

-dx/dt \propto \sqrtx ∴ dx/dt = -k\sqrtx

Part b

Separating variables and integrating:

\int dx/\sqrtx = -k \int dt 2\sqrtx = -kt + c

When $t = 0, x = 2$ : $c = 2\sqrt2$

When $t = 5, x = 1$ :

2 = -5k + 2\sqrt2 k = (2\sqrt2 - 2)/5

When $x = 0.5$ :

2\sqrt0.5 = -[(2\sqrt2 - 2)/5] \cdot t + 2\sqrt2

t = 8.5 minutes

The rate at which body temperature $T$ falls is proportional to the difference between the body temperature $T$ and the surrounding temperature $T₀$ . Find a differential equation relating body temperature $T$ and time $t$ .

Solution

Since temperature is falling, the rate is negative:

-dT/dt \propto (T - T₀)

dT/dt = −k(T − T₀)

This is Newton's Law of Cooling.

In studying the spread of a disease, a scientist thinks that the rate of infection is proportional to the product of the number of people infected and the number of people uninfected. If $N$ is the number infected at time $t$ and $P$ is the total population, form a differential equation to summarise the scientist's theory.

Solution

Total = $P$ , Infected = $N$ , Non-infected = $P - N$ .

dN/dt \propto N(P - N)

dN/dt = kN(P − N)

The rate of increase of a population is proportional to its size at the time. Write down a differential equation to describe this situation. When the population was 2 million, the rate of increase was 1,40,000 per day. Find the constant of proportionality.

Solution

dP/dt \propto P ∴ dP/dt = kP

When $P = 2,000,000$ and $dP/dt = 140,000$ :

140,000 = k \times 2,000,000

k = 0.07

A tank is draining such that when height of water is $h cm$ , it decreases at rate $0.5\sqrth cm/s$ . Initially height is $25 cm$ .

a)Write down a differential equation describing this situation.
b)Solve the differential equation to find $h$ as a function of time.
c)What is the height after 10 seconds?
d)How long to reach a height of 5 cm?

Solution

Part a

dh/dt = −0.5√h

Part b

dh/\sqrth = -0.5 dt \int dh/\sqrth = -0.5 \int dt 2\sqrth = -0.5t + c

When $t = 0, h = 25$ : $c = 2\sqrt25 = 10$

2\sqrth = -0.5t + 10

h = (−0.25t + 5)²

Part c — t = 10 s

h = (-0.25\times10 + 5)² = (2.5)²

h = 6.25 cm

Part d — h = 5 cm

5 = (-0.25t + 5)² \sqrt5 = -0.25t + 5 t = (\sqrt5 - 5)/(-0.25)

t ≈ 11.05 seconds

A stone falls through the air from rest and, $t$ seconds after it was dropped, its speed $v$ satisfies the equation $dv/dt = 10 - 0.2v$ .

a)Show that $v = 50(1 - e^(-0.2t))$ .
b)Calculate the time at which the stone reaches a speed of $20 ms⁻¹$ .

Solution

Part a

dv/(10 - 0.2v) = dt

Integrating with limits $v: 0\tov$ , $t: 0\tot$ :

(1/-0.2)[log(10 - 0.2v)]₀ᵛ = t -5[log(10-0.2v) - log 10] = t log[(10-0.2v)/10] = -0.2t (10-0.2v)/10 = e^(-0.2t) 0.2v = 10 - 10e^(-0.2t)

v = 50(1 − e^(−0.2t)) ✓ (shown)

Part b — v = 20

20 = 50(1 - e^(-0.2t)) e^(-0.2t) = 1 - 2/5 = 3/5 -0.2t = log(3/5)

t = log(3/5)/(−0.2) ≈ 2.55 seconds

The spread of a disease occurs at a rate proportional to the product of the number of people infected and the number not infected. Initially 50 out of a population of 1050 are infected, and the disease is spreading at a rate of 10 new cases per day.

a)If $n$ is the number infected after $t$ days, show that $dn/dt = n(1050-n)/5000$ .
b)Solve the differential equation to find the number infected after $t$ days.
c)How long will it take for 250 people to be infected?
d)Explain why everyone in the population will eventually be infected.

Solution

Part a

dn/dt = kn(1050 - n)

When $t = 0, n = 50, dn/dt = 10$ :

10 = k \times 50 \times 1000 \to k = 1/5000

dn/dt = n(1050 − n)/5000 ✓

Part b

\int dn/[n(1050-n)] = \int dt/5000 (1/1050)\int(1/n + 1/(1050-n))dn = t/5000 + c log[n/(1050-n)] = 0.21t + c

When $t = 0, n = 50$ : $c = log(50/1000) = -2.995$

n/(1050 − n) = (1/20)e^(0.21t)

Part c — n = 250

250/800 = (1/20)e^(0.21t) e^(0.21t) = 25/4 0.21t = log(25/4) = 1.833

t = 1.833/0.21 ≈ 8.73 days

Part d

The number of infected people increases exponentially with time. As $t \to \infty$ , the ratio $n/(1050-n) \to \infty$ , which means $n \to 1050$ . Therefore, given sufficient time, the entire population will eventually be infected.