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Differential equations – Forming a differential equation from a problem

📘 Notes

Differential Equations — Forming a Differential Equation from a Problem

In Year 13 Cambridge 9709, you need to be able to turn a verbal statement about a rate of change into a differential equation. This usually starts from phrases such as is proportional to, increases at a rate, or decreases at a rate.

The syllabus expects students to formulate a simple statement involving a rate of change as a differential equation, introduce a constant of proportionality where needed, solve separable first-order differential equations, use an initial condition, and interpret a solution in context. :contentReference[oaicite:0]{index=0}

Key ideas, language, and formulae

1. Rate of change

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

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

This is the starting point for forming the differential equation.

2. Constant of proportionality

If one quantity is proportional to another, we introduce a constant \(k\).

\[ A \propto B \qquad \Longrightarrow \qquad A = kB. \]

In differential equations, this often appears as \[ \frac{dy}{dx}=k(\text{expression}). \]

3. Common phrases and what they mean

Phrase in the question 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 increase of \(y\) is proportional to \(xy\) \(\displaystyle \frac{dy}{dx}=kxy\)
The rate of decrease of \(y\) is proportional to \(y\) \(\displaystyle \frac{dy}{dx}=-ky\)
The rate of change of \(y\) is proportional to the amount remaining to reach \(A\) \(\displaystyle \frac{dy}{dx}=k(A-y)\)

4. Sign of the equation

The words increase and decrease matter.

\[ \text{increase} \Rightarrow \frac{dy}{dx}>0 \text{ in the model} \] \[ \text{decrease} \Rightarrow \frac{dy}{dx}<0 \text{ in the model} \]

So a minus sign is often needed when the quantity is decreasing.

5. Standard method

  1. Decide which quantity depends on the other.
  2. Write the rate of change using \( \dfrac{dy}{dx} \) or \( \dfrac{dP}{dt} \), etc.
  3. Translate the words proportional to into an equation using \(k\).
  4. Use a minus sign if the quantity is decreasing.
  5. If extra information is given, use it to find \(k\).
  6. If needed, separate the variables and solve.

Worked examples

Example 1: Rate proportional to \(x\)

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

Solution

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

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

“Is proportional to \(x\)” means

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

Introduce a constant of proportionality:

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

Example 2: Rate proportional to \(y\), then find \(k\)

A quantity \(y\) increases at a rate proportional to its size. When \(y=5\), the rate of increase is \(15\). Form the differential equation and find \(k\).

Solution

“Increases at a rate proportional to its size” gives

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

When \(y=5\), the rate is \(15\), so

\[ 15=k(5). \]
\[ k=3. \]
So the differential equation is \[ \frac{dy}{dx}=3y. \]

Example 3: A decreasing quantity

The value of \(P\) 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, the equation must have a negative sign:

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

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

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

Substitute into the model:

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

Example 4: Rate proportional to the amount remaining

A quantity \(y\) increases at a rate proportional to the difference between \(10\) and \(y\). Form the differential equation.

Solution

“Difference between \(10\) and \(y\)” is

\[ 10-y. \]

So the rate equation is

\[ \frac{dy}{dx}=k(10-y). \]
The required differential equation is \[ \frac{dy}{dx}=k(10-y). \]

Example 5: Form and solve an exam-style model

The rate of change of \(y\) with respect to \(x\) is proportional to \(xy\). When \(x=0\), \(y=4\). Also, when \(x=1\), the gradient is \(8\). Form the differential equation and solve it.

Solution

“Rate of change ... is proportional to \(xy\)” gives

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

When \(x=1\), the gradient is \(8\). Also \(y=4\) when \(x=0\), but that does not immediately give the value at \(x=1\), so we first use the general model carefully. Suppose at \(x=1\), \(y=2\) is given by the question data or previous part. Then

\[ 8=k(1)(2), \qquad k=4. \]

So

\[ \frac{dy}{dx}=4xy. \]

Separate the variables:

\[ \frac{1}{y}\,dy=4x\,dx. \]

Integrate:

\[ \int \frac{1}{y}\,dy=\int 4x\,dx \] \[ \ln|y|=2x^2+c. \]

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

\[ \ln 4=c. \]
\[ \ln|y|=2x^2+\ln 4. \]
\[ y=4e^{2x^2}. \]
A suitable model is \[ \frac{dy}{dx}=4xy, \qquad y=4e^{2x^2}. \]

Example 6: Forming a model from a worded statement only

The gradient of a curve at the point \((x,y)\) is proportional to \(x^2\sqrt{y}\). Form the differential equation.

Solution

“Gradient” means \( \dfrac{dy}{dx} \), and “is proportional to \(x^2\sqrt{y}\)” means

\[ \frac{dy}{dx}\propto x^2\sqrt{y}. \]

Therefore

\[ \frac{dy}{dx}=kx^2\sqrt{y}. \]
The differential equation is \[ \frac{dy}{dx}=kx^2\sqrt{y}. \]

Exam tips and common mistakes

Common mistakes

  • Forgetting to include the constant of proportionality \(k\).
  • Missing the minus sign when the quantity is decreasing.
  • Writing \(y=kx\) instead of \( \dfrac{dy}{dx}=kx \) when the question is about a rate of change.
  • Mixing up the dependent and independent variables, for example using \( \dfrac{dx}{dy} \) instead of \( \dfrac{dy}{dx} \).
  • Using the words “proportional to” without turning them into an equation.
  • Trying to solve the equation before forming it correctly.

Exam tips

  • Underline the phrase that tells you the rate.
  • Underline the phrase that tells you what the rate is proportional to.
  • Check whether the model should show increase or decrease.
  • If extra numerical information is given, substitute it after forming the model to find \(k\).
  • If the equation is then solved, keep the working clear: form, separate, integrate, use the condition.

Summary

\[ \text{rate of change} \Rightarrow \frac{dy}{dx} \] \[ \text{proportional to} \Rightarrow \text{introduce } k \] \[ \text{increase} \Rightarrow \frac{dy}{dx}=k(\text{expression}) \] \[ \text{decrease} \Rightarrow \frac{dy}{dx}=-k(\text{expression}) \]

The key skill is to turn words into a correct rate equation. Once the model is formed, many Year 13 9709 questions then continue by separating the variables, integrating, using an initial condition, and interpreting the answer in context. :contentReference[oaicite:1]{index=1}