Differential Equations Explorer

Exercise 10.8 - Complete Solutions with Mathematical Beauty

1

Bacteria Growth

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours?

Solution

1
Differential Equation
\[ \frac{dN}{dt} = kN \]

where \( N \) is the number of bacteria at time \( t \), and \( k \) is the constant of proportionality.

2
Solution
\[ \int \frac{dN}{N} = \int k \, dt \] \[ \ln|N| = kt + C \] \[ N = e^{kt + C} = N_0 e^{kt} \]
3
Apply Conditions

Triples in 5 hours:

\[ N(5) = 3N_0 = N_0 e^{5k} \] \[ 3 = e^{5k} \] \[ k = \frac{\ln 3}{5} \]
4
Find at t=10
\[ N(10) = N_0 e^{10k} = N_0 e^{10 \cdot \frac{\ln 3}{5}} = N_0 e^{2 \ln 3} = N_0 \cdot 3^2 = 9N_0 \]
After 10 hours, the number of bacteria will be 9 times the initial population
2

Population Growth

Find the population of a city at any time \( t \), given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 3,00,000 to 4,00,000.

Solution

1
Differential Equation
\[ \frac{dP}{dt} = kP \]
2
Solution
\[ P(t) = P_0 e^{kt} \]
3
Apply Conditions
\[ P(40) = 400000 = 300000 \cdot e^{40k} \] \[ \frac{4}{3} = e^{40k} \] \[ k = \frac{\ln(4/3)}{40} \]
4
General Solution
\[ P(t) = 300000 \cdot e^{\frac{\ln(4/3)}{40} t} = 300000 \cdot \left( \frac{4}{3} \right)^{t/40} \]
Population at time t: \( P(t) = 300000 \cdot \left( \frac{4}{3} \right)^{t/40} \)
3

Electric Circuit

The equation of electromotive force for an electric circuit containing resistance and self-inductance is \( E = Ri + L \frac{di}{dt} \), where \( E \) is the electromotive force is given to the circuit, \( R \) the resistance and \( L \), the coefficient of induction. Find the current \( i \) at time \( t \) when \( E = 0 \).

Solution

1
Equation with E=0
\[ 0 = Ri + L \frac{di}{dt} \]
2
Rearrange
\[ L \frac{di}{dt} = -Ri \] \[ \frac{di}{i} = -\frac{R}{L} dt \]
3
Integrate
\[ \int \frac{di}{i} = -\frac{R}{L} \int dt \] \[ \ln|i| = -\frac{R}{L} t + C \]
4
Solve for i
\[ i = e^C e^{-(R/L)t} \] \[ i = i_0 e^{-(R/L)t} \]
Current at time t: \( i = i_0 e^{-(R/L)t} \)
4

Motor Boat Velocity

The engine of a motor boat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to the velocity at that time. Find the velocity after 2 seconds of switching off the engine.

Solution

1
Set up Equation
\[ \text{Retardation} = -\frac{dv}{dt} = v \] \[ \frac{dv}{dt} = -v \]
2
Solve
\[ \frac{dv}{v} = -dt \] \[ \int \frac{dv}{v} = -\int dt \] \[ \ln|v| = -t + C \]
3
Apply Initial Condition
\[ \text{At } t=0, v=10 \] \[ \ln 10 = C \] \[ \ln v = -t + \ln 10 \] \[ v = 10 e^{-t} \]
4
Find at t=2
\[ v(2) = 10 e^{-2} \approx 1.353 \text{ m/s} \]
Velocity after 2 seconds: 1.353 m/s
5

Continuous Compounding

Suppose a person deposits ₹10,000 in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?

Solution

1
Differential Equation
\[ \frac{dA}{dt} = rA \]

where r = 0.05 (5%)

2
Solution
\[ A(t) = A_0 e^{rt} \]
3
Parameters
\[ A_0 = 10000, \quad r = 0.05, \quad t = 1.5 \text{ years} \]
4
Calculate
\[ A(1.5) = 10000 \cdot e^{0.05 \times 1.5} = 10000 \cdot e^{0.075} \approx 10000 \times 1.07788 \]
Amount after 18 months: ₹10,778.80
6

Radioactive Decay

Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of the original radioactive nuclei will remain after 1000 years?

Solution

1
Differential Equation
\[ \frac{dN}{dt} = -kN \]
2
Solution
\[ N(t) = N_0 e^{-kt} \]
3
Apply Condition
\[ \text{After 100 years, 90% remain:} \] \[ 0.9N_0 = N_0 e^{-k \cdot 100} \] \[ 0.9 = e^{-100k} \] \[ k = -\frac{\ln 0.9}{100} \]
4
Find at t=1000
\[ N(1000) = N_0 e^{-k \cdot 1000} = N_0 e^{\ln(0.9) \cdot 10} = N_0 (0.9)^{10} \]
Percentage remaining: \( (0.9)^{10} \approx 34.87\% \)
7

Cooling Water

Water at temperature 100°C cools in 10 minutes to 80°C in a room temperature of 25°C. Find (i) The temperature of water after 20 minutes (ii) The time when the temperature is 40°C

Solution

1
Newton's Law of Cooling
\[ \frac{dT}{dt} = -k(T - T_s) \] \[ T_s = 25^\circ C \]
2
Solution
\[ T - T_s = (T_0 - T_s) e^{-kt} \] \[ T - 25 = 75 e^{-kt} \]
3
Find k
\[ \text{At } t=10, T=80: \] \[ 80 - 25 = 75 e^{-10k} \] \[ 55 = 75 e^{-10k} \] \[ k = -\frac{1}{10} \ln \left( \frac{11}{15} \right) \approx 0.03101 \]
4
(i) At t=20
\[ T - 25 = 75 e^{-0.03101 \times 20} \approx 75 \times 0.5376 = 40.32 \] \[ T \approx 65.32^\circ C \]
5
(ii) When T=40°C
\[ 40 - 25 = 75 e^{-0.03101 t} \] \[ 15 = 75 e^{-0.03101 t} \] \[ e^{-0.03101 t} = \frac{1}{5} \] \[ t = \frac{\ln 5}{0.03101} \approx 51.9 \text{ minutes} \]
(i) 65.32°C after 20 min
(ii) 51.9 minutes to reach 40°C
8

Cooling Coffee

At 10.00 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby Kitchen counter to cool. At this instant the temperature of the coffee was 180°F, and 10 minutes later it was 160°F. Assume that constant temperature of the kitchen was 70°F. (i) What was the temperature of the coffee at 10.15A.M.? (ii) Between what times should she have drunk the coffee?

Solution

1
Newton's Law of Cooling
\[ \frac{dT}{dt} = -k(T - 70) \]
2
Solution
\[ T - 70 = (T_0 - 70) e^{-kt} \] \[ T_0 = 180 \]
3
Apply Condition at t=10
\[ 160 - 70 = (180 - 70) e^{-10k} \] \[ 90 = 110 e^{-10k} \] \[ e^{-10k} = \frac{90}{110} = \frac{9}{11} \] \[ k = -\frac{1}{10} \ln \left( \frac{9}{11} \right) \approx 0.02006 \]
4
(i) At 10:15 AM (t=15 min)
\[ T - 70 = 110 e^{-0.02006 \times 15} \approx 110 \times 0.740 = 81.4 \] \[ T \approx 70 + 81.4 = 151.4^\circ F \]
5
(ii) Time when T=130°F and T=140°F
\[ \text{For } T=130: \] \[ 130 - 70 = 110 e^{-0.02006 t} \] \[ 60 = 110 e^{-0.02006 t} \] \[ e^{-0.02006 t} = \frac{6}{11} \] \[ t = -\frac{1}{0.02006} \ln \left( \frac{6}{11} \right) \approx 30.2 \text{ minutes} \] \[ \text{For } T=140: \] \[ 140 - 70 = 110 e^{-0.02006 t} \] \[ 70 = 110 e^{-0.02006 t} \] \[ e^{-0.02006 t} = \frac{7}{11} \] \[ t = -\frac{1}{0.02006} \ln \left( \frac{7}{11} \right) \approx 22.5 \text{ minutes} \]
(i) At 10:15 AM: 151.4°F
(ii) Drink between 10:22.5 and 10:30.2 AM
9

Kitchen Temperature

A pot of boiling water at 100°C is removed from a stove at time \( t = 0 \) and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80°C, and another 5 minutes later it has dropped to 65°C. Determine the temperature of the kitchen.

Solution

1
Newton's Law of Cooling
\[ \frac{dT}{dt} = -k(T - T_s) \]

\( T_s \) is the kitchen temperature.

2
Solution
\[ T - T_s = (T_0 - T_s) e^{-kt} \] \[ T_0 = 100^\circ C \]
3
Apply Conditions
\[ \text{At } t=5, T=80: \] \[ 80 - T_s = (100 - T_s) e^{-5k} \quad (1) \] \[ \text{At } t=10, T=65: \] \[ 65 - T_s = (100 - T_s) e^{-10k} \quad (2) \]
4
Divide Equation (2) by (1)
\[ \frac{65 - T_s}{80 - T_s} = \frac{(100 - T_s) e^{-10k}}{(100 - T_s) e^{-5k}} = e^{-5k} \]
5
From Equation (1)
\[ e^{-5k} = \frac{80 - T_s}{100 - T_s} \] \[ \text{So: } \frac{65 - T_s}{80 - T_s} = \frac{80 - T_s}{100 - T_s} \]
6
Solve for \( T_s \)
\[ (65 - T_s)(100 - T_s) = (80 - T_s)^2 \] \[ 6500 - 165T_s + T_s^2 = 6400 - 160T_s + T_s^2 \] \[ 6500 - 165T_s = 6400 - 160T_s \] \[ 100 = 5T_s \] \[ T_s = 20^\circ C \]
Kitchen temperature is 20°C
10

Salt in Tank

A tank initially contains 50 litres of pure water. Starting at time \( t = 0 \) a brine containing with 2 grams of dissolved salt per litre flows into the tank at the rate of 3 litres per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time \( t > 0 \).

Solution

1
Set up Equation
\[ \frac{dx}{dt} = \text{rate in} - \text{rate out} \] \[ = (2 \times 3) - \left( \frac{x}{50} \times 3 \right) \] \[ = 6 - \frac{3x}{50} \]
2
Rearrange
\[ \frac{dx}{dt} + \frac{3x}{50} = 6 \]
3
Integrating Factor
\[ \mu(t) = e^{\int \frac{3}{50} dt} = e^{3t/50} \]
4
Solve
\[ e^{3t/50} \frac{dx}{dt} + \frac{3}{50} e^{3t/50} x = 6 e^{3t/50} \] \[ \frac{d}{dt} \left( x e^{3t/50} \right) = 6 e^{3t/50} \] \[ x e^{3t/50} = 6 \cdot \frac{50}{3} e^{3t/50} + C \] \[ x e^{3t/50} = 100 e^{3t/50} + C \]
5
Initial Condition
\[ \text{At } t=0, x=0: \] \[ 0 = 100 + C \implies C = -100 \] \[ x = 100 \left(1 - e^{-3t/50}\right) \]
Amount of salt: \( x(t) = 100 \left(1 - e^{-0.06t}\right) \) grams