1.8.11 Differential equations with separable variables (A-level only) - Revision Guide

What you'll learn

How to recognise a first-order separable differential equation.
How to rearrange and integrate it to find a general solution.
How factorising a common factor can reveal separable variables.
How to use an initial condition to find a particular solution.

Prerequisites

Before starting, you should be comfortable with:

rearranging algebraic expressions;
factorising by taking out a common factor;
integrating powers, exponentials and reciprocal functions;
using logarithm laws;
finding an arbitrary constant from a point on a curve.

In particular, remember:

dy=ln⁡∣y∣+C\int \frac{1}{y}\,dy=\ln|y|+C

The modulus signs matter because $y$ could be positive or negative.

What is a differential equation?

A differential equation is an equation containing a derivative, such as $dydx\frac{dy}{dx}$ . It describes how one variable changes in relation to another.

For example,

dydx=3xy\frac{dy}{dx}=3xy

states that the gradient of a solution curve depends on both $x$ and $y$ .

Definition

First-order differential equation

A differential equation is first order if its highest derivative is the first derivative, $dydx\frac{dy}{dx}$ .

A solution is a function relating $y$ and $x$ that satisfies the differential equation.

Separable variables

A differential equation has separable variables if it can be rearranged so that:

every expression involving $y$ is with $d y$ on one side;
every expression involving $x$ is with $d x$ on the other side.

The target form is:

dxf(y)\,dy=g(x)\,dx

You can then integrate both sides.

Key Idea

The central method

Rearrange to separate the variables, integrate both sides, and then use any given condition to determine the arbitrary constant.

Separating the variables

Consider:

dydx=3xy\frac{dy}{dx}=3xy

Divide by $y$ and multiply by $d x$ :

dx\frac{1}{y}\,dy=3x\,dx

The variables are now separated.

Example

Finding a general exponential solution

Solve:

dydx=3xy\frac{dy}{dx}=3xy

Separate the variables by dividing by $y$ and multiplying by $d x$ :
$dx\frac{1}{y}\,dy=3x\,dx$
Integrate both sides:
$dxln⁡∣y∣=32x2+C\begin{aligned} \int \frac{1}{y}\,dy&=\int 3x\,dx\\ \ln|y|&=\frac{3}{2}x^2+C \end{aligned}$
Exponentiate to make $y$ the subject:
$∣y∣=e32x2+C=eCe32x2|y|=e^{\frac{3}{2}x^2+C} =e^C e^{\frac{3}{2}x^2}$
The constant factor, including the possible sign of $y$ , can be replaced by a new arbitrary constant $A$ :
$y=Ae32x2y=Ae^{\frac{3}{2}x^2}$
This is the general solution.

Common Mistake

Treating dy/dx as an ordinary fraction

Writing $d y$ and $d x$ separately is a useful part of the standard method, but each rearrangement must correspond to valid algebra on the differential equation. Show the separation clearly before integrating.

General and particular solutions

Integrating introduces an arbitrary constant, usually written as $C$ . Different values of the constant give different solution curves.

Definition

General solution

A general solution contains an arbitrary constant and represents a family of functions satisfying the differential equation.

A condition such as $y = 3$ when $x = 0$ identifies one curve from this family. This is often called an initial condition, even when the variables do not represent time.

Definition

Particular solution

A particular solution is obtained when the arbitrary constant has been determined using a given condition.

The graph shows how the condition $(0, 3)$ selects one member of the family $y=Ae^{2x}$ .

Family of exponential solution curves with the particular solution through (0, 3)

Example

Using an initial condition

Given that

dydx=2y\frac{dy}{dx}=2y

and $y = 3$ when $x = 0$ , find the particular solution.

Separate and integrate:
$dxln⁡∣y∣=2x+C\begin{aligned} \frac{1}{y}\,dy&=2\,dx\\ \ln|y|&=2x+C \end{aligned}$
Substitute the condition $x = 0$ , $y = 3$ into the logarithmic form:
$ln⁡3=C\ln 3=C$
Substitute this value and exponentiate:
$ln⁡y=2x+ln⁡3y=e2x+ln⁡3y=3e2x\begin{aligned} \ln y&=2x+\ln 3\\ y&=e^{2x+\ln 3}\\ y&=3e^{2x} \end{aligned}$

Tip

Use the condition early

You can often substitute the initial condition immediately after integrating. This may be simpler than first rearranging the general solution into an explicit formula for $y$ .

Factorising before separation

A differential equation may not initially look separable. You may first need to take out a common factor.

For example:

dydx=xy+x\frac{dy}{dx}=xy+x

The right-hand side has a common factor of $x$ :

x y + x = x (y + 1)

The equation is now visibly separable.

Example

Separating after factorisation

Solve:

dydx=xy+x\frac{dy}{dx}=xy+x

Factorise the right-hand side:
$dydx=x(y+1)\frac{dy}{dx}=x(y+1)$
Separate the variables:
$dx\frac{1}{y+1}\,dy=x\,dx$
Integrate both sides:
$dxln⁡∣y+1∣=x22+C\begin{aligned} \int\frac{1}{y+1}\,dy&=\int x\,dx\\ \ln|y+1|&=\frac{x^2}{2}+C \end{aligned}$
Exponentiate and absorb the sign and $e^C$ into a constant $A$ :
$y+1=Aex2/2y=Aex2/2−1\begin{aligned} y+1&=Ae^{x^2/2}\\ y&=Ae^{x^2/2}-1 \end{aligned}$

Common Mistake

Missing the common factor

The expression $x y + x$ is not a sum that should be separated term by term. Factorise it as $x (y + 1)$ before rearranging.

Check for solutions lost by division

When you divide by an expression involving $y$ , check whether that expression could equal zero.

In the previous equation, separation involved division by $y + 1$ . If $y + 1 = 0$ , then $y = - 1$ . Substituting this into the original equation gives:

dydx=0\frac{dy}{dx}=0

and

x y + x = x (- 1) + x = 0

Therefore, $y = - 1$ is also a solution. It is called an equilibrium solution because its value remains constant.

Common Mistake

Division can remove a solution

Before dividing by $f (y)$ , test the values satisfying $f (y) = 0$ in the original differential equation. They may give constant solutions that must be stated separately.

In the form $y=Ae^{x^2/2}-1$ , the equilibrium solution is recovered by allowing $A = 0$ , although the logarithmic working itself assumed $y+1≠0y+1\neq0$ .

Solutions involving powers

Not every separable equation produces logarithms. The result depends on the expression you integrate.

Example

Finding a solution involving a square

Solve

dydx=xy\frac{dy}{dx}=\frac{x}{y}

given that $y = 2$ when $x = 0$ .

Multiply by $dxy\,dx$ to separate the variables:
$dxy\,dy=x\,dx$
Integrate both sides:
$dxy22=x22+C\begin{aligned} \int y\,dy&=\int x\,dx\\ \frac{y^2}{2}&=\frac{x^2}{2}+C \end{aligned}$
Substitute $x = 0$ and $y = 2$ :
$222=C\frac{2^2}{2}=C$
so $C = 2$ .
Rearrange and choose the branch consistent with the condition $y = 2$ :
$y2=x2+4y=x2+4\begin{aligned} y^2&=x^2+4\\ y&=\sqrt{x^2+4} \end{aligned}$

Common Mistake

Forgetting the correct square-root branch

From $y^2=x^2+4$ , the algebraic possibilities are $y=±x2+4y=\pm\sqrt{x^2+4}$ . The condition $y = 2$ when $x = 0$ selects the positive branch.

Checking your solution

Differentiate your final answer and compare it with the original differential equation.

For example, if

y=3e^{2x},

then

dydx=6e2x.\frac{dy}{dx}=6e^{2x}.

Also,

2y=2(3e2x)=6e2x.2y=2\left(3e^{2x}\right)=6e^{2x}.

Therefore, $dydx=2y\frac{dy}{dx}=2y$ , as required.

Tip

A reliable final check

Verify both parts: your function must satisfy the differential equation and, for a particular solution, it must satisfy the given condition.

Exam technique

In the exam

Factorise first if this allows the equation to be written as a product of an $x$ -expression and a $y$ -expression.
Separate completely before integrating, keeping $d y$ with the $y$ -expression and $d x$ with the $x$ -expression.
Include an arbitrary constant once, then use any given condition to find it.
Check values excluded when you divide by an expression involving $y$ .
Differentiate your final answer to confirm that it satisfies the original equation.

Self review

Check yourself

Can you solve $dydx=x(1+y)\frac{dy}{dx}=x(1+y)$ and state any equilibrium solution?
How would you use the condition $y = 4$ when $x = 1$ to obtain a particular solution?
Why must you be careful before dividing a differential equation by $y$ or by $y + 1$ ?

What you'll learn

How to recognise a first-order separable differential equation.
How to rearrange and integrate it to find a general solution.
How factorising a common factor can reveal separable variables.
How to use an initial condition to find a particular solution.

Prerequisites

Before starting, you should be comfortable with:

rearranging algebraic expressions;
factorising by taking out a common factor;
integrating powers, exponentials and reciprocal functions;
using logarithm laws;
finding an arbitrary constant from a point on a curve.

In particular, remember:

dy=ln⁡∣y∣+C\int \frac{1}{y}\,dy=\ln|y|+C

The modulus signs matter because $y$ could be positive or negative.

What is a differential equation?

A differential equation is an equation containing a derivative, such as $dydx\frac{dy}{dx}$ . It describes how one variable changes in relation to another.

For example,

dydx=3xy\frac{dy}{dx}=3xy

states that the gradient of a solution curve depends on both $x$ and $y$ .

Definition

First-order differential equation

A differential equation is first order if its highest derivative is the first derivative, $dydx\frac{dy}{dx}$ .

A solution is a function relating $y$ and $x$ that satisfies the differential equation.

Separable variables

A differential equation has separable variables if it can be rearranged so that:

every expression involving $y$ is with $d y$ on one side;
every expression involving $x$ is with $d x$ on the other side.

The target form is:

dxf(y)\,dy=g(x)\,dx

You can then integrate both sides.

Key Idea

The central method

Rearrange to separate the variables, integrate both sides, and then use any given condition to determine the arbitrary constant.

Separating the variables

Consider:

dydx=3xy\frac{dy}{dx}=3xy

Divide by $y$ and multiply by $d x$ :

dx\frac{1}{y}\,dy=3x\,dx

The variables are now separated.

Example

Finding a general exponential solution

Solve:

dydx=3xy\frac{dy}{dx}=3xy

Separate the variables by dividing by $y$ and multiplying by $d x$ :
$dx\frac{1}{y}\,dy=3x\,dx$
Integrate both sides:
$dxln⁡∣y∣=32x2+C\begin{aligned} \int \frac{1}{y}\,dy&=\int 3x\,dx\\ \ln|y|&=\frac{3}{2}x^2+C \end{aligned}$
Exponentiate to make $y$ the subject:
$∣y∣=e32x2+C=eCe32x2|y|=e^{\frac{3}{2}x^2+C} =e^C e^{\frac{3}{2}x^2}$
The constant factor, including the possible sign of $y$ , can be replaced by a new arbitrary constant $A$ :
$y=Ae32x2y=Ae^{\frac{3}{2}x^2}$
This is the general solution.

Common Mistake

Treating dy/dx as an ordinary fraction

General and particular solutions

Integrating introduces an arbitrary constant, usually written as $C$ . Different values of the constant give different solution curves.

Definition

General solution

A general solution contains an arbitrary constant and represents a family of functions satisfying the differential equation.

A condition such as $y = 3$ when $x = 0$ identifies one curve from this family. This is often called an initial condition, even when the variables do not represent time.

Definition

Particular solution

A particular solution is obtained when the arbitrary constant has been determined using a given condition.

The graph shows how the condition $(0, 3)$ selects one member of the family $y=Ae^{2x}$ .

Family of exponential solution curves with the particular solution through (0, 3)

Example

Using an initial condition

Given that

dydx=2y\frac{dy}{dx}=2y

and $y = 3$ when $x = 0$ , find the particular solution.

Separate and integrate:
$dxln⁡∣y∣=2x+C\begin{aligned} \frac{1}{y}\,dy&=2\,dx\\ \ln|y|&=2x+C \end{aligned}$
Substitute the condition $x = 0$ , $y = 3$ into the logarithmic form:
$ln⁡3=C\ln 3=C$
Substitute this value and exponentiate:
$ln⁡y=2x+ln⁡3y=e2x+ln⁡3y=3e2x\begin{aligned} \ln y&=2x+\ln 3\\ y&=e^{2x+\ln 3}\\ y&=3e^{2x} \end{aligned}$

Tip

Use the condition early

You can often substitute the initial condition immediately after integrating. This may be simpler than first rearranging the general solution into an explicit formula for $y$ .

Factorising before separation

A differential equation may not initially look separable. You may first need to take out a common factor.

For example:

dydx=xy+x\frac{dy}{dx}=xy+x

The right-hand side has a common factor of $x$ :

x y + x = x (y + 1)

The equation is now visibly separable.

Example

Separating after factorisation

Solve:

dydx=xy+x\frac{dy}{dx}=xy+x

Factorise the right-hand side:
$dydx=x(y+1)\frac{dy}{dx}=x(y+1)$
Separate the variables:
$dx\frac{1}{y+1}\,dy=x\,dx$
Integrate both sides:
$dxln⁡∣y+1∣=x22+C\begin{aligned} \int\frac{1}{y+1}\,dy&=\int x\,dx\\ \ln|y+1|&=\frac{x^2}{2}+C \end{aligned}$
Exponentiate and absorb the sign and $e^C$ into a constant $A$ :
$y+1=Aex2/2y=Aex2/2−1\begin{aligned} y+1&=Ae^{x^2/2}\\ y&=Ae^{x^2/2}-1 \end{aligned}$

Common Mistake

Missing the common factor

The expression $x y + x$ is not a sum that should be separated term by term. Factorise it as $x (y + 1)$ before rearranging.

Check for solutions lost by division

When you divide by an expression involving $y$ , check whether that expression could equal zero.

In the previous equation, separation involved division by $y + 1$ . If $y + 1 = 0$ , then $y = - 1$ . Substituting this into the original equation gives:

dydx=0\frac{dy}{dx}=0

and

x y + x = x (- 1) + x = 0

Therefore, $y = - 1$ is also a solution. It is called an equilibrium solution because its value remains constant.

Common Mistake

Division can remove a solution

Before dividing by $f (y)$ , test the values satisfying $f (y) = 0$ in the original differential equation. They may give constant solutions that must be stated separately.

In the form $y=Ae^{x^2/2}-1$ , the equilibrium solution is recovered by allowing $A = 0$ , although the logarithmic working itself assumed $y+1≠0y+1\neq0$ .

Solutions involving powers

Not every separable equation produces logarithms. The result depends on the expression you integrate.

Example

Finding a solution involving a square

Solve

dydx=xy\frac{dy}{dx}=\frac{x}{y}

given that $y = 2$ when $x = 0$ .

Multiply by $dxy\,dx$ to separate the variables:
$dxy\,dy=x\,dx$
Integrate both sides:
$dxy22=x22+C\begin{aligned} \int y\,dy&=\int x\,dx\\ \frac{y^2}{2}&=\frac{x^2}{2}+C \end{aligned}$
Substitute $x = 0$ and $y = 2$ :
$222=C\frac{2^2}{2}=C$
so $C = 2$ .
Rearrange and choose the branch consistent with the condition $y = 2$ :
$y2=x2+4y=x2+4\begin{aligned} y^2&=x^2+4\\ y&=\sqrt{x^2+4} \end{aligned}$

Common Mistake

Forgetting the correct square-root branch

From $y^2=x^2+4$ , the algebraic possibilities are $y=±x2+4y=\pm\sqrt{x^2+4}$ . The condition $y = 2$ when $x = 0$ selects the positive branch.

Checking your solution

Differentiate your final answer and compare it with the original differential equation.

For example, if

y=3e^{2x},

then

dydx=6e2x.\frac{dy}{dx}=6e^{2x}.

Also,

2y=2(3e2x)=6e2x.2y=2\left(3e^{2x}\right)=6e^{2x}.

Therefore, $dydx=2y\frac{dy}{dx}=2y$ , as required.

Tip

A reliable final check

Verify both parts: your function must satisfy the differential equation and, for a particular solution, it must satisfy the given condition.

Exam technique

In the exam

Factorise first if this allows the equation to be written as a product of an $x$ -expression and a $y$ -expression.
Separate completely before integrating, keeping $d y$ with the $y$ -expression and $d x$ with the $x$ -expression.
Include an arbitrary constant once, then use any given condition to find it.
Check values excluded when you divide by an expression involving $y$ .
Differentiate your final answer to confirm that it satisfies the original equation.

Self review

Check yourself

Can you solve $dydx=x(1+y)\frac{dy}{dx}=x(1+y)$ and state any equilibrium solution?
How would you use the condition $y = 4$ when $x = 1$ to obtain a particular solution?
Why must you be careful before dividing a differential equation by $y$ or by $y + 1$ ?

1.11.7 Differential equations with separable variables (A-level only)

What you'll learn

Prerequisites

What is a differential equation?

Separable variables

Separating the variables

General and particular solutions

Factorising before separation

Check for solutions lost by division

Solutions involving powers

Checking your solution

Review 1.8.11 Differential equations with separable variables (A-level only) by teaching Genie

1.11.7 Differential equations with separable variables (A-level only)

What you'll learn

Prerequisites

What is a differential equation?

Separable variables

Separating the variables

General and particular solutions

Factorising before separation

Check for solutions lost by division

Solutions involving powers

Checking your solution

Review 1.8.11 Differential equations with separable variables (A-level only) by teaching Genie

1.11.7 Differential equations with separable variables (A-level only)

What you'll learn

Prerequisites

What is a differential equation?

Separable variables

Separating the variables

General and particular solutions

Factorising before separation

Check for solutions lost by division

Solutions involving powers

Checking your solution

Review 1.8.11 Differential equations with separable variables (A-level only) by teaching Genie

1.8.11 Differential equations with separable variables (A-level only) Revision Guide

1.11.7 Differential equations with separable variables (A-level only)

What you'll learn

Prerequisites

What is a differential equation?

Separable variables

Separating the variables

General and particular solutions

Factorising before separation

Check for solutions lost by division

Solutions involving powers

Checking your solution

Review 1.8.11 Differential equations with separable variables (A-level only) by teaching Genie

1.8.11 Differential equations with separable variables (A-level only) Revision Guide