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⬅ Forming a differential equation from a problem
The technique of separating the variables Forming a differential equation from a problem Review

Differential equations — Review 39 problems

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Differential Equations — Review

In Year 13 Cambridge 9709, differential equations are used to model a rate of change and then solve for the function involved. The main ideas are forming a differential equation from words, separating the variables, integrating, and using an initial condition.

In exam questions, it is important to write the differential equation correctly first, then show the separation step clearly, integrate carefully, and use any given condition accurately.

Key definitions and formulae

1. Rate of change

If \(y\) changes with respect to \(x\), then the rate of change is

\[ \frac{dy}{dx}. \]

2. Forming a differential equation from words

Phrases involving proportional to lead to a constant of proportionality \(k\).

\[ A \propto B \qquad \Longrightarrow \qquad A=kB. \]
Statement Differential equation
The rate of change of \(y\) with respect to \(x\) is proportional to \(x\) \(\displaystyle \frac{dy}{dx}=kx\)
The rate of change of \(y\) with respect to \(x\) is proportional to \(y\) \(\displaystyle \frac{dy}{dx}=ky\)
The rate of decrease of \(y\) is proportional to \(y\) \(\displaystyle \frac{dy}{dx}=-ky\)
The rate of change of \(y\) is proportional to \(xy\) \(\displaystyle \frac{dy}{dx}=kxy\)
The rate of change of \(y\) is proportional to \(A-y\) \(\displaystyle \frac{dy}{dx}=k(A-y)\)

3. Separable differential equations

A differential equation is separable if it can be written in the form

\[ \frac{dy}{dx}=f(x)g(y). \]

Then we separate the variables:

\[ \frac{1}{g(y)}\,dy=f(x)\,dx. \]

4. Integration step

After separating,

\[ \int \frac{1}{g(y)}\,dy=\int f(x)\,dx. \]

Do not forget the constant of integration.

5. General solution and particular solution

After integrating, the answer with the constant \(c\) is the general solution.

If a condition such as \(y=y_0\) when \(x=x_0\) is given, use it to find \(c\). This gives the particular solution.

\[ y=y_0 \quad \text{when} \quad x=x_0. \]

6. Standard method

  1. Form the differential equation from the words in the problem.
  2. Check whether it is separable.
  3. Separate the variables.
  4. Integrate both sides.
  5. Add the constant of integration.
  6. Use any initial condition to find the constant.
  7. Rearrange if needed to make \(y\) the subject.

Worked examples

Example 1: Form a differential equation from a statement

The rate of change of \(y\) with respect to \(x\) is proportional to \(xy\). Form the differential equation.

Solution

“Rate of change of \(y\) with respect to \(x\)” means

\[ \frac{dy}{dx}. \]

“Is proportional to \(xy\)” means

\[ \frac{dy}{dx}\propto xy. \]

Introduce a constant of proportionality:

\[ \frac{dy}{dx}=kxy. \]
The differential equation is \[ \frac{dy}{dx}=kxy. \]

Example 2: Separate the variables and solve

Solve

\[ \frac{dy}{dx}=xy. \]

Solution

Separate the variables:

\[ \frac{1}{y}\,dy=x\,dx. \]

Integrate both sides:

\[ \int \frac{1}{y}\,dy=\int x\,dx \] \[ \ln|y|=\frac{1}{2}x^2+c. \]
The general solution is \[ \ln|y|=\frac{1}{2}x^2+c. \]

Example 3: Solve and use an initial condition

Solve

\[ \frac{dy}{dx}=2xy \]

given that \(y=3\) when \(x=0\).

Solution

Separate the variables:

\[ \frac{1}{y}\,dy=2x\,dx. \]

Integrate:

\[ \int \frac{1}{y}\,dy=\int 2x\,dx \] \[ \ln|y|=x^2+c. \]

Use \(y=3\) when \(x=0\):

\[ \ln 3=c. \]

So

\[ \ln|y|=x^2+\ln 3. \]

Exponentiate:

\[ y=3e^{x^2}. \]
The particular solution is \[ y=3e^{x^2}. \]

Example 4: A decreasing quantity

A quantity \(P\) decreases at a rate proportional to \(P\). When \(P=80\), the rate of decrease is \(20\). Form the differential equation.

Solution

Since \(P\) is decreasing,

\[ \frac{dP}{dt}=-kP. \]

When \(P=80\), the rate of decrease is \(20\), so

\[ \frac{dP}{dt}=-20. \]

Substitute:

\[ -20=-k(80) \] \[ k=\frac{1}{4}. \]
The differential equation is \[ \frac{dP}{dt}=-\frac{1}{4}P. \]

Example 5: Rate proportional to the amount remaining

A quantity \(y\) increases at a rate proportional to \(10-y\). Given that \(y=4\) when \(x=0\), solve

\[ \frac{dy}{dx}=10-y. \]

Solution

Separate the variables:

\[ \frac{1}{10-y}\,dy=dx. \]

It is usually easier to write

\[ \int \frac{1}{10-y}\,dy=\int dx. \]

Integrate carefully:

\[ -\ln|10-y|=x+c. \]

Use \(y=4\) when \(x=0\):

\[ -\ln 6=c. \]

So

\[ -\ln|10-y|=x-\ln 6. \]

Rearrange:

\[ \ln|10-y|=\ln 6-x \] \[ 10-y=6e^{-x}. \]
\[ y=10-6e^{-x}. \]
The particular solution is \[ y=10-6e^{-x}. \]

Common mistakes and exam tips

Common mistakes

  • Forgetting to introduce the constant of proportionality \(k\).
  • Missing the negative sign when the quantity is decreasing.
  • Writing \(y=kx\) instead of a rate equation such as \(\dfrac{dy}{dx}=kx\).
  • Not separating the variables correctly before integrating.
  • Forgetting the constant of integration.
  • Writing \(\ln y\) instead of \(\ln|y|\) when integrating \(\dfrac{1}{y}\).
  • Using the initial condition incorrectly or too early.

Exam tips

  • Identify the dependent variable and the independent variable first.
  • Underline the words rate of change, proportional to, increase, and decrease.
  • Show the separation step clearly, for example \[ \frac{1}{y}\,dy=x\,dx. \]
  • Use the initial condition immediately after integrating.
  • Leave the answer in implicit form if that is simpler, unless the question asks for \(y\) explicitly.
  • Check that your final solution satisfies the given condition.

Summary

\[ \text{rate of change} \Rightarrow \frac{dy}{dx} \] \[ \text{proportional to} \Rightarrow \text{introduce } k \] \[ \frac{dy}{dx}=f(x)g(y) \] \[ \frac{1}{g(y)}\,dy=f(x)\,dx \] \[ \int \frac{1}{g(y)}\,dy=\int f(x)\,dx \]

The main Year 13 differential equation skills are forming the equation from a problem, separating the variables, integrating, and using a condition to find the particular solution.

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