Second order differential equations

Edexcel International A-Level Further Mathematics· FP2 5.1-5.2 (2018 Spec Issue 3)· 45 min read

1. 1. Complementary Functions from the Auxiliary Equation★★☆☆☆⏱ 15 min

✓ Calculator OK

The standard form of a linear constant-coefficient second order ODE is $a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = f (x)$ where $a, b, c$ are real constants. The general solution is the sum of two components: the complementary function (CF), which solves the homogeneous equation $a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = 0$ , and a particular integral (PI), which solves the full non-homogeneous equation.

📘 Definition

Complementary Function

The general solution to the homogeneous version of the ODE, containing two arbitrary constants (A and B) for a second order equation.

Example:

For $\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = e^{x}$ , the CF solves $\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 0$ .

🔬 Derivation

Goal:

Derive the auxiliary equation for homogeneous ODEs

Starting from:

$a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = 0$

1
Assume a trial solution of the form $y = e^{m x}$ where $m$ is a constant to be found.
2
$\frac{dy}{dx} = m e^{mx}, \quad \frac{d^2y}{dx^2} = m^2 e^{mx}$
3
Substitute into the homogeneous ODE:
4
$a m^2 e^{mx} + b m e^{mx} + c e^{mx} = 0$
5
Factor out $e^{m x}$ (which is never zero):
6
$a m^2 + b m + c = 0$

Result:

This quadratic equation in $m$ is the auxiliary equation. Its roots determine the form of the complementary function.

Root type	Auxiliary equation property	Form of complementary function
Distinct real roots $m_{1}, m_{2}$	$b^{2} - 4 a c > 0$	$y = A e^{m_{1} x} + B e^{m_{2} x}$
Equal real roots $m$	$b^{2} - 4 a c = 0$	$y = (A + B x) e^{m x}$
Complex conjugate roots $α \pm β i$	$b^{2} - 4 a c < 0$	$y = e^{α x} (A cos β x + B sin β x)$

📐 Worked Example

Find the complementary function for the ODE $\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = 3 cos x$

1
First write the auxiliary equation for the homogeneous ODE $\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = 0$ :
2
$m^2 + 2m +5 =0$
3
Solve the quadratic using the quadratic formula:
4
$m = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i$
5
These are complex conjugate roots with $α = - 1$ and $β = 2$
6
Write the complementary function using the complex root form:
7
$y_{CF} = e^{-x}(A \cos 2x + B \sin 2x)$

Exam tip:

Always write the auxiliary equation correctly first: copy coefficients directly from the ODE, preserving all signs. A sign error here will make your entire CF incorrect.

2. 2. Particular Integrals for Non-Homogeneous ODEs★★★☆☆⏱ 20 min

✓ Calculator OK

The particular integral (PI) is a specific solution to the full non-homogeneous ODE. You find it by choosing a trial function that matches the form of $f (x)$ , substituting it into the ODE, and solving for unknown coefficients.

📘 Definition

Particular Integral

A single, constant-free solution to the full non-homogeneous ODE, added to the complementary function to form the general solution.

Form of f(x)	Standard trial PI
$k e^{p x}$	$λ e^{p x}$
Linear polynomial $A + B x$	$α + β x$
Quadratic polynomial $p + q x + r x^{2}$	$α + β x + γ x^{2}$
$m cos ω x + n sin ω x$	$α cos ω x + β sin ω x$

📐 Worked Example

Find the general solution to $\frac{d ^{2} y}{d x ^{2}} + 4 y = sin 2 x$

1
First find the complementary function by solving the homogeneous ODE:
2
$m^2 +4 =0 \implies m = \pm 2i$
3
So CF is $y_{C F} = A cos 2 x + B sin 2 x$
4
Notice that $f (x) = sin 2 x$ matches a term in the CF, so we have a resonance case. Multiply the standard trial PI by $x$ :
5
Trial PI: $y_{P I} = x (α cos 2 x + β sin 2 x)$
6
Compute first and second derivatives of $y_{P I}$ using the product rule:
7
$\frac{dy_{PI}}{dx} = (\alpha \cos 2x + \beta \sin 2x) + x(-2\alpha \sin 2x + 2\beta \cos 2x)$
8
$\frac{d^2 y_{PI}}{dx^2} = -4\alpha \sin 2x + 4\beta \cos 2x -4x(\alpha \cos 2x + \beta \sin 2x)$
9
Substitute $y_{P I}$ and its second derivative into the original ODE:
10
$(-4\alpha \sin 2x + 4\beta \cos 2x -4x(\alpha \cos 2x + \beta \sin 2x)) + 4(x(\alpha \cos 2x + \beta \sin 2x)) = \sin 2x$
11
Simplify: the $x$ terms cancel out, leaving:
12
$-4\alpha \sin 2x + 4\beta \cos 2x = \sin 2x$
13
Equate coefficients of $cos 2 x$ and $sin 2 x$ :
14
Coefficient of $cos 2 x$ : $4 β = 0 ⟹ β = 0$
15
Coefficient of $sin 2 x$ : $- 4 α = 1 ⟹ α = - 1/4$
16
So $y_{P I} = - \frac{1}{4} x cos 2 x$
17
General solution is CF + PI:
18
$y = A \cos 2x + B \sin 2x - \frac{1}{4}x \cos 2x$

Exam tip:

When calculating derivatives for resonance cases, write out each product rule step explicitly to avoid arithmetic errors, which are common for these questions.

3. 3. Finding Particular Solutions from Conditions★★☆☆☆⏱ 10 min

✓ Calculator OK

Once you have the general solution (CF + PI), you can use given initial or boundary conditions to find the values of the arbitrary constants A and B, producing a single particular solution. Always substitute the conditions into the full general solution, not just the complementary function.

📐 Worked Example

For the ODE in the previous example, find the particular solution given that when $x = 0$ , $y = 0$ and $\frac{d y}{d x} = 0$ .

1
Start with the general solution:
2
$y = A \cos 2x + B \sin 2x - \frac{1}{4}x \cos 2x$
3
Apply first condition: $x = 0$ , $y = 0$ :
4
$0 = A \cos 0 + B \sin 0 - 0 \implies A = 0$
5
Now simplify the solution with A=0:
6
$y = B \sin 2x - \frac{1}{4}x \cos 2x$
7
Compute the first derivative to apply the second condition:
8
$\frac{dy}{dx} = 2B \cos 2x - \frac{1}{4}\cos 2x + \frac{1}{2}x \sin 2x$
9
Apply $\frac{d y}{d x} = 0$ when $x = 0$ :
10
$0 = 2B \cos 0 - \frac{1}{4}\cos 0 + 0 \implies 2B = \frac{1}{4} \implies B = \frac{1}{8}$
11
Write the final particular solution:
12
$y = \frac{1}{8}\sin 2x - \frac{1}{4}x \cos 2x$

Exam tip:

Never substitute constants before finding both the CF and PI. Fixing A and B using only the homogeneous CF will produce a solution that does not satisfy the non-homogeneous ODE.

4. 4. Solving Reducible ODEs with Given Substitutions★★★★☆⏱ 20 min

✓ Calculator OK

Some non-standard second order ODEs can be converted into the linear constant-coefficient form you already know how to solve, using a substitution that is always given to you in the exam question. You only need to apply the provided substitution, rewrite the ODE in terms of the new variable, solve the resulting ODE, then convert back to the original variables.

📐 Worked Example

Use the substitution $z = y e^{2 x}$ to reduce the ODE $\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 4 y = 3 e^{- 2 x}$ to a simpler ODE, hence find its general solution.

1
Start with the given substitution: $z = y e^{2 x}$ , so rearrange to get $y = z e^{- 2 x}$
2
Compute first and second derivatives of y with respect to x using the product rule:
3
$\frac{dy}{dx} = \frac{dz}{dx} e^{-2x} - 2 z e^{-2x} = e^{-2x}\left(\frac{dz}{dx} - 2z\right)$
4
$\frac{d^2y}{dx^2} = e^{-2x}\left(\frac{d^2z}{dx^2} -4\frac{dz}{dx} +4z\right)$
5
Substitute y, dy/dx, d²y/dx² into the original ODE:
6
$e^{-2x}\left(\frac{d^2z}{dx^2} -4\frac{dz}{dx} +4z\right) + 4 e^{-2x}\left(\frac{dz}{dx} - 2z\right) + 4 z e^{-2x} = 3 e^{-2x}$
7
Divide both sides by $e^{- 2 x}$ (non-zero) to simplify:
8
$\frac{d^2z}{dx^2} -4\frac{dz}{dx} +4z +4\frac{dz}{dx} -8z +4z = 3$
9
$\frac{d^2z}{dx^2} = 3$
10
Integrate twice to solve for z:
11
$\frac{dz}{dx} = 3x + A$
12
$z = \frac{3}{2}x^2 + A x + B$
13
Substitute back $z = y e^{2 x}$ to get the general solution for y:
14
$y = e^{-2x}\left(\frac{3}{2}x^2 + A x + B\right)$

Exam tip:

Double-check your substitution step before solving the reduced ODE: a single differentiation error here will make the rest of your solution incorrect.

5. Common Pitfalls

Wrong move:

Using wrong sign in auxiliary equation coefficients

Why:

Mismatching coefficients between the ODE and auxiliary equation leads to incorrect roots and wrong CF

Correct move:

Copy coefficients of d²y/dx², dy/dx, y directly into the auxiliary equation, preserving all signs

Wrong move:

Forgetting to multiply trial PI by x in resonance cases

Why:

If the trial PI duplicates a CF term, substituting it will give 0 on the left-hand side, so you cannot solve for coefficients

Correct move:

Always check if your standard trial PI matches any term in the CF before substitution; if yes, multiply the trial PI by x

Wrong move:

Substituting initial conditions into the CF before finding the PI

Why:

The constants in the CF apply to the full general solution, not just the homogeneous part

Correct move:

Form the full general solution (CF + PI) first, then substitute initial/boundary conditions to find A and B

Wrong move:

Misapplying product rule for resonance trial PIs or substitution problems

Why:

Trial PIs for resonance and substitution variables have multiplied functions of x, requiring careful product rule use

Correct move:

Write out each derivative step explicitly, separating terms before combining like terms

Wrong move:

Mixing up CF forms for different root types

Why:

Using the wrong CF form leads to a solution that does not satisfy the homogeneous ODE

Correct move:

Memorize the three CF forms for distinct real, equal real, and complex conjugate roots, and match them to your auxiliary equation roots

6. Quick Reference Cheatsheet

Component	Rule/Form
Auxiliary Equation	$a m^{2} + bm + c = 0$ for $a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = f (x)$
CF: Distinct real roots $m_{1}, m_{2}$	$A e^{m_{1} x} + B e^{m_{2} x}$
CF: Equal real roots $m$	$(A + B x) e^{m x}$
CF: Complex roots $α \pm β i$	$e^{α x} (A cos β x + B sin β x)$
PI Trial: $k e^{p x}$	$λ e^{p x}$ (multiply by x if matches CF)
PI Trial: $p + q x + r x^{2}$	$α + β x + γ x^{2}$ (multiply by x if matches CF)
PI Trial: $m cos ω x + n sin ω x$	$α cos ω x + β sin ω x$ (multiply by x if matches CF)
General Solution	Complementary Function + Particular Integral
Reducible ODEs	Apply given substitution first, solve reduced ODE, convert back to original variables

Going deeper

formula_sheetEdexcel IAL Further Maths FP2 Formula ListNo ODE-specific formulas are provided for this topic; all forms must be recalled

What's Next

Now that you have mastered second order differential equations for Edexcel IAL FP2, you are ready to apply these skills to more advanced topics in further mathematics. The ability to solve these ODEs is foundational for studying series solutions of ODEs, which is covered later in the FP2 unit, as well as for applications in mechanics and differential equations modules in further study. Practise past paper questions to build speed and accuracy, particularly for resonance cases and substitution problems, which are frequently tested in FP2 exams. Make sure you can recall all CF and PI forms without reference, as no formula sheet is provided for this topic.