Topics

Differential equations of the form y′ = f(x)

Covers AS91579

What a differential equation is

A differential equation is an equation containing a derivative.
$\dfrac{dy}{dx} = 6x^2$ is one; so is $\dfrac{d^2y}{dx^2} = \sin x$ and $\dfrac{dP}{dt} = 0.05P$ .
Solving one means finding the function that satisfies it — not a number.

The two kinds of solution

The general solution contains the arbitrary constant(s) and describes an entire family of curves.
The particular solution has the constant(s) pinned down by extra information — an initial condition or a point the curve passes through.
The standard restricts you to two forms:
- $y' = f(x)$ — solve by integrating once
- $y'' = f(x)$ — solve by integrating twice
- plus separable equations, covered on the next page

Solving $y' = f(x)$

Integrate both sides with respect to $x$ :

$y = \int f(x)\,dx + c$

Apply the initial condition by substituting the given values and solving for $c$ .
State the particular solution in full.
Worked through — solve $\dfrac{dy}{dx} = 3x^2 - 4$ given $y = 5$ when $x = 2$ :
- $y = x^3 - 4x + c$
- $5 = 8 - 8 + c$ , so $c = 5$
- $y = x^3 - 4x + 5$

Solving $y'' = f(x)$

Integrate twice, introducing a separate constant each time.
The first integration gives $y'$ ; the second gives $y$ .
Two constants need two pieces of information. These are usually a value of $y'$ and a value of $y$ , or the values of $y$ at two points.
Note the order of use. A condition about the gradient applies to $y'$ after the first integration; a condition about the curve applies to $y$ after the second. Using them in the wrong place is the standard error.
Worked through — solve $\dfrac{d^2y}{dx^2} = 12x$ , given $\dfrac{dy}{dx} = 5$ and $y = 3$ when $x = 1$ :
- First integration: $\dfrac{dy}{dx} = 6x^2 + c_1$ . At $x = 1$ : $5 = 6 + c_1$ , so $c_1 = -1$
- Second integration: $y = 2x^3 - x + c_2$ . At $x = 1$ : $3 = 2 - 1 + c_2$ , so $c_2 = 2$
- $y = 2x^3 - x + 2$

Notation you will meet

These all mean the same thing:
- $y'$ , $\dfrac{dy}{dx}$ , $f'(x)$ — the first derivative
- $y''$ , $\dfrac{d^2y}{dx^2}$ , $f''(x)$ — the second derivative
When the independent variable is time, $\dfrac{dP}{dt}$ and $\dfrac{dh}{dt}$ are used, and the initial condition is usually stated "when $t = 0$ ".

Interpreting the solution

A differential equation is a statement about how something changes; its solution says what that thing is.
$\dfrac{dV}{dt} = -3$ says a volume is falling at a constant 3 units per unit time. Integrating gives $V = -3t + c$ , and $c$ is the starting volume.
Always interpret the constant in context where the question is applied — it usually has a physical meaning, most often the initial value.

Reading the initial condition correctly

"When $t = 0$ , $P = 500$ " means substitute $t = 0$ and $P = 500$ .
"Initially" means at $t = 0$ .
"After 3 hours the amount is 200" means substitute $t = 3$ and the quantity $= 200$ — a second condition, often used to find a rate constant rather than the integration constant.

Tips and Tricks

One constant per integration, and they are different constants. Solving $y'' = f(x)$ needs $c_1$ and $c_2$ , found from two separate conditions. Using the same letter for both is a route to confusion.

Do not assume the constant is zero. The 2025 report describes candidates who assumed $c = 0$ because a quantity was zero at $t = 0$ , when the function was exponential and the constant was not zero. Substitute and solve, every time.

Match each condition to the right function. A statement about the gradient is a condition on $\frac{dy}{dx}$ ; a statement about the value is a condition on $y$ . Applying a gradient condition to $y$ produces a plausible-looking wrong answer.

The report lists using "the correct notation" for differential equations as an Excellence behaviour. Write $\frac{dy}{dx} = \ldots$ , then $y = \int \ldots dx$ , then the general solution, then the particular solution — as separate labelled steps.

Check your solution by differentiating it and confirming it satisfies both the differential equation and the initial condition. Two checks, both quick.

Worked ExampleA second-order equation in context

The acceleration of a particle is given by $\dfrac{d^2s}{dt^2} = 6t - 4$ m s⁻², where $s$ is displacement in metres and $t$ is time in seconds. When $t = 0$ the particle has velocity 5 m s⁻¹ and displacement 2 m. Find $s$ in terms of $t$ .

Step 1 — Identify what each derivative means

$\dfrac{d^2s}{dt^2}$ is acceleration
$\dfrac{ds}{dt}$ is velocity
$s$ is displacement

So we integrate twice, and the two conditions apply at different stages: the velocity condition after the first integration, the displacement condition after the second.

Step 2 — Integrate once to get velocity

$\frac{ds}{dt} = \int (6t - 4)\,dt = 3t^2 - 4t + c_1$

Step 3 — Apply the velocity condition

"When $t = 0$ the velocity is 5":

$5 = 3(0)^2 - 4(0) + c_1$

$c_1 = 5$

So the velocity function is:

$\frac{ds}{dt} = 3t^2 - 4t + 5$

Step 4 — Integrate again to get displacement

$s = \int \left(3t^2 - 4t + 5\right)dt = t^3 - 2t^2 + 5t + c_2$

Note this is a new and different constant.

Step 5 — Apply the displacement condition

"When $t = 0$ the displacement is 2":

$2 = 0 - 0 + 0 + c_2$

$c_2 = 2$

Step 6 — State the particular solution

$s = t^3 - 2t^2 + 5t + 2$

Step 7 — Check all three given facts

Differentiate once:

$\frac{ds}{dt} = 3t^2 - 4t + 5, \quad \text{and at } t = 0 \text{ this is } 5 \;\checkmark$

Differentiate again:

$\frac{d^2s}{dt^2} = 6t - 4 \;\checkmark$

Substitute $t = 0$ into $s$ :

$s = 2 \;\checkmark$

Differential equations of the form y′ = f(x)

Covers AS91579

What a differential equation is

A differential equation is an equation containing a derivative.
$\dfrac{dy}{dx} = 6x^2$ is one; so is $\dfrac{d^2y}{dx^2} = \sin x$ and $\dfrac{dP}{dt} = 0.05P$ .
Solving one means finding the function that satisfies it — not a number.

The two kinds of solution

The general solution contains the arbitrary constant(s) and describes an entire family of curves.
The particular solution has the constant(s) pinned down by extra information — an initial condition or a point the curve passes through.
The standard restricts you to two forms:
- $y' = f(x)$ — solve by integrating once
- $y'' = f(x)$ — solve by integrating twice
- plus separable equations, covered on the next page

Solving $y' = f(x)$

Integrate both sides with respect to $x$ :

$y = \int f(x)\,dx + c$

Apply the initial condition by substituting the given values and solving for $c$ .
State the particular solution in full.
Worked through — solve $\dfrac{dy}{dx} = 3x^2 - 4$ given $y = 5$ when $x = 2$ :
- $y = x^3 - 4x + c$
- $5 = 8 - 8 + c$ , so $c = 5$
- $y = x^3 - 4x + 5$

Solving $y'' = f(x)$

Integrate twice, introducing a separate constant each time.
The first integration gives $y'$ ; the second gives $y$ .
Two constants need two pieces of information. These are usually a value of $y'$ and a value of $y$ , or the values of $y$ at two points.
Note the order of use. A condition about the gradient applies to $y'$ after the first integration; a condition about the curve applies to $y$ after the second. Using them in the wrong place is the standard error.
Worked through — solve $\dfrac{d^2y}{dx^2} = 12x$ , given $\dfrac{dy}{dx} = 5$ and $y = 3$ when $x = 1$ :
- First integration: $\dfrac{dy}{dx} = 6x^2 + c_1$ . At $x = 1$ : $5 = 6 + c_1$ , so $c_1 = -1$
- Second integration: $y = 2x^3 - x + c_2$ . At $x = 1$ : $3 = 2 - 1 + c_2$ , so $c_2 = 2$
- $y = 2x^3 - x + 2$

Notation you will meet

These all mean the same thing:
- $y'$ , $\dfrac{dy}{dx}$ , $f'(x)$ — the first derivative
- $y''$ , $\dfrac{d^2y}{dx^2}$ , $f''(x)$ — the second derivative
When the independent variable is time, $\dfrac{dP}{dt}$ and $\dfrac{dh}{dt}$ are used, and the initial condition is usually stated "when $t = 0$ ".

Interpreting the solution

A differential equation is a statement about how something changes; its solution says what that thing is.
$\dfrac{dV}{dt} = -3$ says a volume is falling at a constant 3 units per unit time. Integrating gives $V = -3t + c$ , and $c$ is the starting volume.
Always interpret the constant in context where the question is applied — it usually has a physical meaning, most often the initial value.

Reading the initial condition correctly

"When $t = 0$ , $P = 500$ " means substitute $t = 0$ and $P = 500$ .
"Initially" means at $t = 0$ .
"After 3 hours the amount is 200" means substitute $t = 3$ and the quantity $= 200$ — a second condition, often used to find a rate constant rather than the integration constant.

Tips and Tricks

Check your solution by differentiating it and confirming it satisfies both the differential equation and the initial condition. Two checks, both quick.

Worked ExampleA second-order equation in context

Step 1 — Identify what each derivative means

$\dfrac{d^2s}{dt^2}$ is acceleration
$\dfrac{ds}{dt}$ is velocity
$s$ is displacement

So we integrate twice, and the two conditions apply at different stages: the velocity condition after the first integration, the displacement condition after the second.

Step 2 — Integrate once to get velocity

$\frac{ds}{dt} = \int (6t - 4)\,dt = 3t^2 - 4t + c_1$

Step 3 — Apply the velocity condition

"When $t = 0$ the velocity is 5":

$5 = 3(0)^2 - 4(0) + c_1$

$c_1 = 5$

So the velocity function is:

$\frac{ds}{dt} = 3t^2 - 4t + 5$

Step 4 — Integrate again to get displacement

$s = \int \left(3t^2 - 4t + 5\right)dt = t^3 - 2t^2 + 5t + c_2$

Note this is a new and different constant.

Step 5 — Apply the displacement condition

"When $t = 0$ the displacement is 2":

$2 = 0 - 0 + 0 + c_2$

$c_2 = 2$

Step 6 — State the particular solution

$s = t^3 - 2t^2 + 5t + 2$

Step 7 — Check all three given facts

Differentiate once:

$\frac{ds}{dt} = 3t^2 - 4t + 5, \quad \text{and at } t = 0 \text{ this is } 5 \;\checkmark$

Differentiate again:

$\frac{d^2s}{dt^2} = 6t - 4 \;\checkmark$

Substitute $t = 0$ into $s$ :

$s = 2 \;\checkmark$

Differential equations of the form y′ = f(x)

What a differential equation is

The two kinds of solution

Solving y′=f(x)y' = f(x)y′=f(x)

Solving y′′=f(x)y'' = f(x)y′′=f(x)

Notation you will meet

Interpreting the solution

Reading the initial condition correctly

Step 1 — Identify what each derivative means

Step 2 — Integrate once to get velocity

Step 3 — Apply the velocity condition

Step 4 — Integrate again to get displacement

Step 5 — Apply the displacement condition

Step 6 — State the particular solution

Step 7 — Check all three given facts

Differential equations of the form y′ = f(x)

What a differential equation is

The two kinds of solution

Solving y′=f(x)y' = f(x)y′=f(x)

Solving y′′=f(x)y'' = f(x)y′′=f(x)

Notation you will meet

Interpreting the solution

Reading the initial condition correctly

Step 1 — Identify what each derivative means

Step 2 — Integrate once to get velocity

Step 3 — Apply the velocity condition

Step 4 — Integrate again to get displacement

Step 5 — Apply the displacement condition

Step 6 — State the particular solution

Step 7 — Check all three given facts

Solving $y' = f(x)$

Solving $y'' = f(x)$

Solving $y' = f(x)$

Solving $y'' = f(x)$