1.08k–lPureStretch

Differential Equations

A biologist observes that a population's rate of growth is proportional to its current size, but has no formula for the population itself — only an equation connecting a rate of change to the quantity. Solving that equation, using integration, recovers the actual model.

34 min Video by TLMaths Integration

The big picture

A $\textbf{differential equation}$ describes how a quantity CHANGES, typically as $\dfrac{dy}{dx}=$ some expression, rather than giving $y$ directly. This chapter's technique — $\textbf{separating the variables}$ — collects every $y$ term on one side and every $x$ term on the other, so that integrating both sides recovers the actual relationship between $y$ and $x$ . This is precisely how real-world rates of change (population growth, cooling, chemical reactions) are converted into usable models.

A-Level Maths H7-01: Differential Equations: Introducing Differential EquationsCurated by Kepler, not produced by us. Watch it, then work through the notes below.

Open on YouTube

What you'll be able to do

Recognise a first-order differential equation of the form $\dfrac{dy}{dx}=f(x)g(y)$
Separate the variables and integrate both sides
Find the general solution and use a given condition to find the particular solution
Interpret differential equation models in real-world contexts

Separating the variables

When $\dfrac{dy}{dx}$ can be written as a function of $x$ multiplied by a function of $y$ , rearrange algebraically so every $y$ -term (and $dy$ ) is on one side and every $x$ -term (and $dx$ ) is on the other, then integrate each side independently.

\dfrac{dy}{dx} = f(x)g(y) \;\Rightarrow\; \int\dfrac{1}{g(y)}\,dy = \int f(x)\,dx

Separate, then integrate both sides.

Example 1Solve

\dfrac{dy}{dx}=2xy

1Separate variables:

\dfrac{1}{y}\,dy = 2x\,dx

2Integrate both sides:

\int\dfrac1y\,dy = \int2x\,dx

\ln|y| = x^2+c

Answer

\ln|y|=x^2+c

, or equivalently

y=Ae^{x^2}

(where

A=\pm e^c

)

Tip — Only ONE constant of integration is needed overall — combine both sides' constants into a single $c$ (usually written on the side that gives the neater equation).

From general to particular solution

The result of integration is the $\textbf{general solution}$ — a whole family of curves, one for every value of the constant. A given initial condition (a specific point the solution must pass through) pins down the exact constant, giving the $\textbf{particular solution}$ .

Example 1Solve

\dfrac{dy}{dx}=\dfrac{y}{x}

given that

y=6

when

x=2

1Separate:

\dfrac1y\,dy=\dfrac1x\,dx

2Integrate:

\ln|y|=\ln|x|+c

3Rewrite using laws of logs:

\ln|y|-\ln|x|=c \Rightarrow \ln\left|\dfrac{y}{x}\right|=c \Rightarrow \dfrac{y}{x}=A

(where

A=e^c

4Use the condition:

\dfrac{6}{2}=A

, so

A=3

Answer

y=3x

Why this works

Converting a logarithmic general solution into the form $y=A\times(\text{function of }x)$ using the exponential/logarithm laws is a standard final step — it makes the constant much easier to interpret and solve for.

Modelling with differential equations

Real contexts describe a rate of change in words ("the rate of decrease is proportional to the amount present") — translating this into a differential equation, then separating and integrating, recovers the full model.

Example 1The mass

m

of a substance decreases at a rate proportional to

m

itself. Given

m=50

t=0

and

m=25

t=10

, find

m

in terms of

t

1"Rate proportional to

m

" means

\dfrac{dm}{dt}=-km

(negative since it decreases), for some constant

k>0

2Separate and integrate:

\int\dfrac1m\,dm=\int-k\,dt \Rightarrow \ln m = -kt+c \Rightarrow m=Ae^{-kt}

3Use

t=0

m=50

A=50

4Use

t=10

m=25

25=50e^{-10k} \Rightarrow e^{-10k}=0.5 \Rightarrow -10k=\ln0.5 \Rightarrow k=\dfrac{\ln2}{10}

Answer

m=50e^{-(\ln2/10)t}

(exponential decay, halving every 10 units of

t

)

Think like an examiner

Always check the differential equation CAN be separated (every $x$ term must be separable from every $y$ term by multiplication/division only) before attempting this method.
Combine the two integration constants into a single constant as early as possible, then convert a logarithmic solution into an exponential form using the laws of logs where it makes the answer clearer.
Use the given initial condition as the LAST step, substituting into the general solution to find the specific constant.

Common misconceptions

Writing two separate constants of integration, one on each side.

Combine them into a SINGLE constant

c

(or

A

) once both sides have been integrated — two separate constants are mathematically redundant.

Forgetting the modulus signs when integrating

\int\dfrac1y\,dy=\ln|y|+c

Strictly, integrating

\dfrac1y

gives

\ln|y|

, though in modelling contexts with

y>0

throughout, the modulus is often dropped for readability.

Applying the initial condition before fully solving for the general solution.

Complete the separation and integration FIRST to get the general solution, then substitute the given point to find the constant.

Separating variables

\int\dfrac{1}{g(y)}\,dy = \int f(x)\,dx

Separate then integrate both sides.

\dfrac{dy}{dx}=ky \Rightarrow y=Ae^{kx}

A very common exponential growth/decay result.

Stretch yourself

Solve the differential equation $\dfrac{dy}{dx} = \dfrac{x}{y^2}$ , given that $y=2$ when $x=0$ .

Hint — Separate so all y-terms (including $y^2\,dy$ ) are on one side and all x-terms on the other, then integrate both sides before applying the initial condition.

Questions students ask

What does it mean to "separate the variables"?

It means rearranging the differential equation algebraically so every term involving

y

(and

dy

) is on one side of the equation, and every term involving

x

(and

dx

) is on the other — then each side can be integrated independently.

What is the difference between a general and particular solution?

The general solution is the result of integration, containing an unknown constant — it represents a whole family of solutions. The particular solution uses a given point (initial condition) to find the exact value of that constant.

Why do so many models end up as

y=Ae^{kx}

Because "rate of change proportional to current amount" is an extremely common real-world description (population growth, radioactive decay, cooling), and separating variables on

\dfrac{dy}{dx}=ky

always produces exactly this exponential form.

Key takeaways

A differential equation of the form $\dfrac{dy}{dx}=f(x)g(y)$ can be separated: $\int\dfrac1{g(y)}\,dy=\int f(x)\,dx$ .
Integrating both sides gives the general solution, with one combined constant.
An initial condition (a known point) gives the particular solution by finding that constant.
Many real models follow $\dfrac{dy}{dx}=ky$ , giving the exponential solution $y=Ae^{kx}$ .