What is sign of the damping exponent?

Decaying oscillations need p < 0 in e^px; a positive p means the amplitude grows.

Find the integrating factor for dydx + 2y = e^x. [2 marks]

Find a particular integral of d^2ydx^2 + y = 6. [3 marks]

Pearson Edexcel Edexcel 2019-style practice: Solve the differential equation (dy)/(dx) + 2y = e^-x, given that y = 3 when x = 0. Express y in terms of x.

Use an integrating factor since the equation is first order linear. Here P = 2, so the integrating factor is I = e^int 2\,dx = e^2x (M1 A1). Multiplying through, (d)/(dx)(y e^2x) = e^2xe^-x = e^x (M1). Integrating gives y e^2x = e^x + c (A1). So y = e^-x + c e^-2x. Applying y = 3 at x = 0: 3 = 1 + c, so c = 2 (M1). Hence y = e^-x + 2e^-2x (A1).

EnglandFurther MathsSyllabus dot point

How do you solve first and second order differential equations, and model oscillations with them?

First order linear equations by integrating factor, second order constant-coefficient equations using the auxiliary equation, complementary function and particular integral, and modelling damped and forced oscillations and coupled systems.

A focused answer to the Edexcel A-Level Further Mathematics differential equations content, covering first order linear equations by integrating factor, second order constant-coefficient equations via the auxiliary equation, the complementary function and particular integral, and modelling simple harmonic, damped and forced systems.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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Quick answer

A first order linear equation $\frac{dy}{dx} + P(x)y = Q(x)$ is solved by the integrating factor $I = e^{\int P\,dx}$ , after which the left side becomes $\frac{d}{dx}(Iy)$ and you integrate $IQ$ . A second order equation $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$ is solved by the auxiliary equation $am^2 + bm + c = 0$ giving the complementary function, plus a particular integral matching the form of $f(x)$ . Distinct real roots give exponentials (overdamped); a repeated root gives $(A + Bx)e^{mx}$ (critically damped); complex roots $p \pm qi$ give $e^{px}(A\cos qx + B\sin qx)$ (oscillatory, decaying when $p < 0$ ).

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What this dot point is asking
First order linear equations
Second order constant-coefficient equations
Particular integrals and forcing
Examples in context
Try this

What this dot point is asking

Edexcel wants you to solve first order linear differential equations using an integrating factor, solve second order constant-coefficient equations through the auxiliary equation (complementary function plus particular integral), apply boundary or initial conditions to fix the arbitrary constants, and interpret solutions as models of simple harmonic, damped and forced motion. These appear in every Core Pure paper, often in a multi-part modelling context.

First order linear equations

A first order equation is linear when it can be written in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ . The trick is to multiply by a factor $I(x)$ that turns the whole left side into the derivative of a product. The right factor is $I = e^{\int P\,dx}$ , because then $\frac{d}{dx}(Iy) = I\frac{dy}{dx} + I P y$ , which is exactly $I$ times the left side.

Solve a first order linear equation

Solve $\dfrac{dy}{dx} + \dfrac{2}{x}y = x$ for $x > 0$ .

Step 1: Identify $P(x)$ and compute the integrating factor

The equation is already in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P = \frac{2}{x}$ . The integrating factor turns the left-hand side into an exact derivative, and is chosen as $I = e^{\int P\,dx}$ .

\int P\,dx = \int \frac{2}{x}\,dx = 2\ln x = \ln x^2, \qquad \text{so } I = e^{\ln x^2} = x^2.

Step 2: Multiply through and recognise the product rule

Multiplying the entire equation by $x^2$ makes the left-hand side the derivative of $x^2 y$ by the product rule. This is the whole point of the integrating factor: we convert a messy equation into one we can integrate directly.

\frac{d}{dx}(x^2 y) = x^2 \cdot x = x^3.

Step 3: Integrate both sides and solve for $y$

Integrate the right-hand side with respect to $x$ , remembering to include the arbitrary constant $c$ here (not in the integrating factor step).

x^2 y = \frac{x^4}{4} + c, \qquad \text{so} \quad y = \frac{x^2}{4} + \frac{c}{x^2}.

Final answer: $y = \dfrac{x^2}{4} + \dfrac{c}{x^2}$ .

Second order constant-coefficient equations

For $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$ the general solution is the sum of two parts: the complementary function $y_c$ (the general solution of the homogeneous equation with $f(x) = 0$ ) and a particular integral $y_p$ (any single solution of the full equation). You read off $y_c$ from the auxiliary equation $am^2 + bm + c = 0$ .

Solve a homogeneous second order equation

Solve $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$ .

Step 1: Write and solve the auxiliary equation

For any second order constant-coefficient homogeneous equation we substitute the trial solution $y = e^{mx}$ , which produces the auxiliary equation. The roots $m$ tell us the form of the general solution. Here the auxiliary equation is obtained by replacing $y''$ by $m^2$ , $y'$ by $m$ , and $y$ by $1$ .

m^2 - 5m + 6 = 0

This factorises as $(m - 2)(m - 3) = 0$ , giving $m = 2$ or $m = 3$ .

Step 2: Write the general solution from the roots

Two distinct real roots $m_1 = 2$ and $m_2 = 3$ mean the two linearly independent solutions are $e^{2x}$ and $e^{3x}$ . The general solution is their linear combination, with arbitrary constants $A$ and $B$ fixed by any boundary or initial conditions given.

y = Ae^{2x} + Be^{3x}.

Final answer: $y = Ae^{2x} + Be^{3x}$ , where $A$ and $B$ are determined by boundary conditions.

Particular integrals and forcing

To handle a non-zero right-hand side $f(x)$ , try a particular integral of the same form: a constant for a constant, a linear polynomial for a linear $f$ , $k e^{\lambda x}$ for an exponential, and $p\cos\omega x + q\sin\omega x$ for trigonometric forcing. Substitute the trial form into the full equation and match coefficients. The one trap is when your trial form already appears in the complementary function: then multiply the trial by $x$ (or $x^2$ for a repeated root) before substituting.

Find a particular integral

Find a particular integral of $\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + 2y = 4x$ .

Step 1: Choose a trial function matching the right-hand side

The right-hand side is $4x$ , a polynomial of degree one, so we try a particular integral of the same form: $y_p = ax + b$ . The constants $a$ and $b$ are to be determined by substitution. (If $x$ already appeared in the complementary function, we would multiply by $x$ , but it does not here.)

y_p = ax + b, \quad y_p' = a, \quad y_p'' = 0.

Step 2: Substitute into the equation and collect terms

Replace $y$ , $y'$ and $y''$ in the original equation with the trial expressions. The result is an identity in $x$ , and we group the coefficient of $x$ and the constant separately.

0 - 3a + 2(ax + b) = 4x \implies 2ax + (2b - 3a) = 4x.

Step 3: Match coefficients to find $a$ and $b$

For the equation to hold for all $x$ , the coefficient of $x$ on the left must equal $4$ and the constant term must equal $0$ .

Coefficients of $x$ : $2a = 4$ , so $a = 2$ . Constants: $2b - 3a = 0$ , so $2b = 6$ and $b = 3$ .

Step 4: Write the general solution

The particular integral is $y_p = 2x + 3$ . The general solution combines the complementary function (found from the auxiliary equation $m^2 - 3m + 2 = (m-1)(m-2) = 0$ , roots $m = 1, 2$ ) with the particular integral.

y = Ae^{x} + Be^{2x} + 2x + 3.

Final answer: $y = Ae^{x} + Be^{2x} + 2x + 3$ .

Examples in context

Differential equations are where complex numbers, calculus and mechanics meet. The complex auxiliary roots $p \pm qi$ produce oscillatory solutions through Euler's formula, the same identity that drives the modulus-argument form of complex numbers. Damped harmonic motion (a mass on a spring with resistance) is modelled by $m\ddot{x} + c\dot{x} + kx = 0$ , and the three damping regimes correspond exactly to the three root cases. Forced oscillations add a periodic right-hand side and can produce resonance when the forcing frequency matches the natural frequency. Coupled first order systems, expressed as $\dot{\mathbf{x}} = A\mathbf{x}$ , link directly to eigenvalues of the matrix $A$ , tying this dot point back to matrices.

Try this

Q1. Find the integrating factor for $\dfrac{dy}{dx} + 2y = e^x$ . [2 marks]

Cue. $I = e^{\int 2\,dx} = e^{2x}$ .

Q2. Write the complementary function when the auxiliary roots are $-1 \pm 2i$ . [2 marks]

Cue. $y = e^{-x}(A\cos 2x + B\sin 2x)$ .

Q3. Find a particular integral of $\dfrac{d^2y}{dx^2} + y = 6$ . [3 marks]

Cue. Try $y_p = k$ (constant): $k = 6$ , so the particular integral is $y_p = 6$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksSolve the differential equation

\frac{dy}{dx} + 2y = e^{-x}

, given that

y = 3

when

x = 0

. Express

y

in terms of

x

Show worked answer →

Use an integrating factor since the equation is first order linear.

Here $P = 2$ , so the integrating factor is $I = e^{\int 2\,dx} = e^{2x}$ (M1 A1).

Multiplying through, $\frac{d}{dx}(y e^{2x}) = e^{2x}e^{-x} = e^{x}$ (M1). Integrating gives $y e^{2x} = e^{x} + c$ (A1).

So $y = e^{-x} + c e^{-2x}$ . Applying $y = 3$ at $x = 0$ : $3 = 1 + c$ , so $c = 2$ (M1). Hence $y = e^{-x} + 2e^{-2x}$ (A1).

Edexcel 20218 marksFind the general solution of

\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 13y = 26

. Hence describe the long-term behaviour of

y

x \to \infty

Show worked answer →

Find the complementary function from the auxiliary equation, then a particular integral.

Auxiliary equation: $m^2 - 4m + 13 = 0$ , so $m = \frac{4 \pm \sqrt{16 - 52}}{2} = 2 \pm 3i$ (M1 A1).

Complex roots $p \pm qi$ with $p = 2$ , $q = 3$ give the complementary function $y_c = e^{2x}(A\cos 3x + B\sin 3x)$ (A1).

For the particular integral, try $y = k$ (constant): then $13k = 26$ , so $k = 2$ (M1 A1).

General solution: $y = e^{2x}(A\cos 3x + B\sin 3x) + 2$ (A1). Since $p = 2 > 0$ , the oscillation grows without bound, so $y$ diverges and oscillates with ever-increasing amplitude as $x \to \infty$ (M1 A1 for the reasoned conclusion).

Edexcel 20235 marksA second order equation has auxiliary equation

m^2 + 6m + 9 = 0

. Show that the system is critically damped and write down the form of the complementary function.

Show worked answer →

Solve the auxiliary equation and interpret a repeated root.

$m^2 + 6m + 9 = (m + 3)^2 = 0$ , so $m = -3$ is a repeated (double) root (M1 A1).

A repeated real root corresponds to critical damping, the boundary between overdamped (two distinct negative real roots) and underdamped (complex roots giving oscillation) behaviour (B1 for the interpretation).

The complementary function for a repeated root $m$ is $y = (A + Bx)e^{mx}$ , so here $y = (A + Bx)e^{-3x}$ (M1 A1). Since the exponent is negative, $y \to 0$ as $x \to \infty$ with no oscillation.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)