Mastering Differential Equations for Your Mathematics Exam

What the examiner is testing

The examiner is assessing your ability to identify the type of differential equation presented and apply the correct integration technique to find its general or particular solution. Marks are awarded for correct separation of variables, accurate integration, and the correct use of boundary conditions to determine constants.

The method

1. Identify the type of differential equation: Most commonly, these will be first-order separable differential equations of the form \( \frac{dy}{dx} = f(x)g(y) \).
2. Separate the variables: Rearrange the equation so that all terms involving \( y \) and \( dy \) are on one side, and all terms involving \( x \) and \( dx \) are on the other. This usually involves multiplying or dividing.
3. Integrate both sides: Integrate the \( y \)-side with respect to \( y \) and the \( x \)-side with respect to \( x \). Remember to include a single constant of integration, \( C \), on one side (conventionally the \( x \)-side).
4. Solve for \( y \) (if required): If the question asks for a general solution in the form \( y = f(x) \), rearrange your integrated equation to make \( y \) the subject.
5. Apply boundary conditions (if given): If specific values for \( x \) and \( y \) are provided (a particular solution is required), substitute these values into your integrated equation (before solving for \( y \) if possible) to find the value of \( C \).
6. Substitute \( C \) back: Replace \( C \) with its calculated value in your integrated equation to obtain the particular solution.

Worked example

A quantity \( P \) is increasing such that its rate of increase at time \( t \) is proportional to \( P \). Initially, \( P = 100 \) when \( t = 0 \). After 10 minutes, \( P = 200 \). Find an expression for \( P \) in terms of \( t \).

$$ \frac{dP}{dt} \propto P $$
$$ \frac{dP}{dt} = kP \quad \text{where } k \text{ is the constant of proportionality.} $$
Separate variables:
$$ \int \frac{1}{P} \, dP = \int k \, dt $$
Integrate both sides:
$$ \ln|P| = kt + C $$
Exponentiate to solve for \( P \):
$$ P = e^{kt+C} $$
$$ P = e^{kt}e^C $$
Let \( A = e^C \), where \( A \) is a positive constant:
$$ P = Ae^{kt} $$
Apply initial condition: \( P = 100 \) when \( t = 0 \):
$$ 100 = Ae^{k \times 0} $$
$$ 100 = Ae^0 $$
$$ 100 = A \times 1 $$
$$ A = 100 $$
So, the equation becomes:
$$ P = 100e^{kt} $$
Apply second condition: \( P = 200 \) when \( t = 10 \):
$$ 200 = 100e^{k \times 10} $$
Divide by 100:
$$ 2 = e^{10k} $$
Take natural logarithm of both sides:
$$ \ln 2 = 10k $$
$$ k = \frac{\ln 2}{10} $$
Substitute \( k \) back into the equation for \( P \):
$$ P = 100e^{\left(\frac{\ln 2}{10}\right)t} $$
This can also be written as:
$$ P = 100e^{\ln(2^{t/10})} $$
$$ P = 100 \times 2^{t/10} $$
Sanity check:
When \( t=0 \), \( P = 100 \times 2^0 = 100 \times 1 = 100 \). (Correct)
When \( t=10 \), \( P = 100 \times 2^{10/10} = 100 \times 2^1 = 200 \). (Correct)
The units for \( t \) are minutes, and \( P \) is a quantity, so no specific units are needed for \( P \).

Worked example: a harder one

A liquid is cooling in a room. The rate of decrease of its temperature \( \theta \) (in \( ^\circ C \)) with respect to time \( t \) (in minutes) is proportional to the difference between its temperature and the room temperature, which is \( 20^\circ C \). Initially, the liquid's temperature is \( 80^\circ C \). After 5 minutes, its temperature is \( 60^\circ C \). Find the time it takes for the liquid to reach \( 30^\circ C \).

The statement "rate of decrease of its temperature" means \( -\frac{d\theta}{dt} \).
"Proportional to the difference between its temperature and the room temperature" means \( k(\theta - 20) \).
So, \( -\frac{d\theta}{dt} = k(\theta - 20) \).
This can be written as \( \frac{d\theta}{dt} = -k(\theta - 20) \).

Why the obvious first move fails: If you try to integrate \( \int -k \, dt \) and \( \int \frac{1}{\theta - 20} \, d\theta \) directly without considering the negative sign, you might forget it or misplace it, leading to an incorrect constant or exponential decay. It's crucial to handle the negative sign correctly from the start.

Separate variables:
$$ \int \frac{1}{\theta - 20} \, d\theta = \int -k \, dt $$
Integrate both sides:
$$ \ln|\theta - 20| = -kt + C $$
Exponentiate to solve for \( \theta - 20 \):
$$ \theta - 20 = e^{-kt+C} $$
$$ \theta - 20 = e^{-kt}e^C $$
Let \( A = e^C \):
$$ \theta - 20 = Ae^{-kt} $$
$$ \theta = 20 + Ae^{-kt} $$
Apply initial condition: \( \theta = 80 \) when \( t = 0 \):
$$ 80 = 20 + Ae^{-k \times 0} $$
$$ 80 = 20 + A $$
$$ A = 60 $$
So, the equation is:
$$ \theta = 20 + 60e^{-kt} $$
Apply second condition: \( \theta = 60 \) when \( t = 5 \):
$$ 60 = 20 + 60e^{-k \times 5} $$
$$ 40 = 60e^{-5k} $$
$$ \frac{40}{60} = e^{-5k} $$
$$ \frac{2}{3} = e^{-5k} $$
Take natural logarithm of both sides:
$$ \ln\left(\frac{2}{3}\right) = -5k $$
$$ k = -\frac{1}{5}\ln\left(\frac{2}{3}\right) = \frac{1}{5}\ln\left(\frac{3}{2}\right) $$
Substitute \( k \) back into the equation for \( \theta \):
$$ \theta = 20 + 60e^{-\frac{1}{5}\ln\left(\frac{3}{2}\right)t} $$
Now, find the time \( t \) when \( \theta = 30^\circ C \):
$$ 30 = 20 + 60e^{-\frac{1}{5}\ln\left(\frac{3}{2}\right)t} $$
$$ 10 = 60e^{-\frac{1}{5}\ln\left(\frac{3}{2}\right)t} $$
$$ \frac{10}{60} = e^{-\frac{1}{5}\ln\left(\frac{3}{2}\right)t} $$
$$ \frac{1}{6} = e^{-\frac{1}{5}\ln\left(\frac{3}{2}\right)t} $$
Take natural logarithm of both sides:
$$ \ln\left(\frac{1}{6}\right) = -\frac{1}{5}\ln\left(\frac{3}{2}\right)t $$
$$ -\ln 6 = -\frac{1}{5}\ln\left(\frac{3}{2}\right)t $$
$$ \ln 6 = \frac{1}{5}\ln\left(\frac{3}{2}\right)t $$
$$ t = \frac{5 \ln 6}{\ln\left(\frac{3}{2}\right)} $$
$$ t \approx \frac{5 \times 1.79176}{0.40547} \approx 22.09 \text{ minutes} $$
The time it takes for the liquid to reach \( 30^\circ C \) is approximately 22.1 minutes (3 s.f.).

Practice

1. Find the general solution to the differential equation \( \frac{dy}{dx} = 2xy^2 \).
2. Solve the differential equation \( \frac{dy}{dx} = \frac{x^2}{y} \) given that \( y = 3 \) when \( x = 0 \).
3. The rate of change of the number of bacteria, \( N \), in a culture with respect to time, \( t \) (in hours), is given by \( \frac{dN}{dt} = 0.05N \). If initially there are 1000 bacteria, how many bacteria will there be after 24 hours? Give your answer to the nearest whole number.
4. A cup of coffee at \( 90^\circ C \) is placed in a room where the ambient temperature is \( 25^\circ C \). The rate at which the coffee cools is proportional to the difference between its temperature \( \theta \) and the ambient temperature. After 10 minutes, the coffee's temperature is \( 70^\circ C \). Calculate the time it takes for the coffee to cool to \( 40^\circ C \).

Answers:

1. \( y = -\frac{1}{x^2 + C} \)
2. \( y = \sqrt{\frac{2}{3}x^3 + 9} \)
3. \( N = 1000e^{0.05t} \). When \( t=24 \), \( N = 1000e^{0.05 \times 24} = 1000e^{1.2} \approx 3320 \) bacteria.
4. Let the temperature of the coffee be \( \theta \) and the ambient temperature be \( T_a = 25^\circ C \).
   The differential equation is \( \frac{d\theta}{dt} = -k(\theta - T_a) \), so \( \frac{d\theta}{dt} = -k(\theta - 25) \).
   Separate variables: \( \int \frac{1}{\theta - 25} \, d\theta = \int -k \, dt \)
   Integrate: \( \ln|\theta - 25| = -kt + C \)
   Exponentiate: \( \theta - 25 = Ae^{-kt} \implies \theta = 25 + Ae^{-kt} \)
   Initial condition: \( \theta = 90 \) when \( t = 0 \).
   \( 90 = 25 + Ae^0 \implies A = 65 \).
   So, \( \theta = 25 + 65e^{-kt} \).
   Second condition: \( \theta = 70 \) when \( t = 10 \).
   \( 70 = 25 + 65e^{-10k} \)
   \( 45 = 65e^{-10k} \)
   \( \frac{45}{65} = e^{-10k} \implies \frac{9}{13} = e^{-10k} \)
   \( \ln\left(\frac{9}{13}\right) = -10k \implies k = -\frac{1}{10}\ln\left(\frac{9}{13}\right) = \frac{1}{10}\ln\left(\frac{13}{9}\right) \).
   The equation for \( \theta \) is \( \theta = 25 + 65e^{-\frac{1}{10}\ln\left(\frac{13}{9}\right)t} \).
   Find \( t \) when \( \theta = 40^\circ C \):
   \( 40 = 25 + 65e^{-\frac{1}{10}\ln\left(\frac{13}{9}\right)t} \)
   \( 15 = 65e^{-\frac{1}{10}\ln\left(\frac{13}{9}\right)t} \)
   \( \frac{15}{65} = e^{-\frac{1}{10}\ln\left(\frac{13}{9}\right)t} \implies \frac{3}{13} = e^{-\frac{1}{10}\ln\left(\frac{13}{9}\right)t} \)
   \( \ln\left(\frac{3}{13}\right) = -\frac{1}{10}\ln\left(\frac{13}{9}\right)t \)
   \( t = \frac{10 \ln\left(\frac{3}{13}\right)}{-\ln\left(\frac{13}{9}\right)} = \frac{10 (-\ln\left(\frac{13}{3}\right))}{-\ln\left(\frac{13}{9}\right)} = \frac{10 \ln\left(\frac{13}{3}\right)}{\ln\left(\frac{13}{9}\right)} \)
   \( t \approx \frac{10 \times 1.4663}{0.3677} \approx 39.88 \text{ minutes} \).
   It takes approximately 39.9 minutes (3 s.f.) for the coffee to cool to \( 40^\circ C \).

The three mistakes that lose marks

1. Forgetting the constant of integration \( C \): This is a fundamental error. When integrating, if you omit \( C \), you will only find a specific solution, not the general one, or you won't be able to use boundary conditions to find the particular solution.
   Wrong answer example: For \( \int \frac{1}{y} \, dy = \int x \, dx \), writing \( \ln|y| = \frac{1}{2}x^2 \) instead of \( \ln|y| = \frac{1}{2}x^2 + C \). This means you cannot apply boundary conditions to find a unique solution.

2. Incorrect separation of variables: Rearranging the equation incorrectly before integration will lead to an unsolvable or incorrect integral. This often happens with terms that are products or sums.
   Wrong answer example: For \( \frac{dy}{dx} = x+y \), attempting to write \( \int \frac{1}{y} \, dy = \int x \, dx \) or \( \int dy = \int (x+y) \, dx \). This equation is not separable in this form.

3. Incorrectly applying logarithms/exponentials: Errors in manipulating \( \ln \) or \( e \) can lead to incorrect solutions, especially when dealing with the constant of integration. A common mistake is \( e^{kt+C} = e^{kt} + e^C \).
   Wrong answer example: From \( \ln|P| = kt + C \), incorrectly writing \( P = e^{kt} + C \) instead of \( P = e^{kt}e^C = Ae^{kt} \). This fundamentally changes the nature of the solution.

30-second recap

Differential equations describe rates of change. For separable first-order equations, rearrange to get all \( y \) terms with \( dy \) and all \( x \) terms with \( dx \). Integrate both sides, remembering the constant of integration. Use given conditions to find the constant and any other unknown parameters.