# Second order differential equations

> Edexcel International A-Level Further Mathematics · FP2
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp2-second-order-differential-equations/

This guide teaches you to solve linear constant-coefficient second order differential equations for Edexcel IAL Further Maths FP2, including complementary functions, particular integrals, resonance cases, and ODE reduction via given substitutions.

**Prerequisites:** [First order differential equations (FP2)](https://www.owlsprep.com/study/edexcel-ial-further-math-fp2-first-order-differential-equations/); [Pure Maths differentiation and integration](https://www.owlsprep.com/study/edexcel-ial-maths-pure-differentiation-integration/)

## Learning objectives

- Solve linear constant-coefficient second order ODEs using the complementary function + particular integral method
- Identify auxiliary equation root types (distinct, equal, complex) and write corresponding complementary functions
- Select correct trial particular integrals for exponential, polynomial, and sinusoidal f(x), including resonance cases
- Reduce non-standard ODEs to solvable form using given substitutions
- Apply initial/boundary conditions to find particular solutions from general solutions

## 1. Complementary Functions from the Auxiliary Equation

The standard form of a linear constant-coefficient second order ODE is $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$ 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 $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$, and a particular integral (PI), which solves the full non-homogeneous equation.

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

**Derivation:** Derive the auxiliary equation for homogeneous ODEs

*Starting from:* $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$

1. Assume a trial solution of the form $y = e^{mx}$ 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 $e^{mx}$ (which is never zero):
6. $$a m^2 + b m + c = 0$$

*Conclusion:* 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 $m_1, m_2$ | $b^2 - 4ac > 0$ | $y = A e^{m_1 x} + B e^{m_2 x}$ |
| Equal real roots $m$ | $b^2 - 4ac = 0$ | $y = (A + Bx) e^{m x}$ |
| Complex conjugate roots $\alpha \pm \beta i$ | $b^2 - 4ac < 0$ | $y = e^{\alpha x} (A \cos \beta x + B \sin \beta x)$ |

**Worked example:** Find the complementary function for the ODE $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 3\cos x$

1. First write the auxiliary equation for the homogeneous ODE $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0$:
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 $\alpha = -1$ and $\beta = 2$
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.

*Calculator:* allowed

## 2. Particular Integrals for Non-Homogeneous ODEs

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.

**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 |
| --- | --- |
| $k e^{px}$ | $\lambda e^{px}$ |
| Linear polynomial $A + Bx$ | $\alpha + \beta x$ |
| Quadratic polynomial $p + qx + rx^2$ | $\alpha + \beta x + \gamma x^2$ |
| $m \cos \omega x + n \sin \omega x$ | $\alpha \cos \omega x + \beta \sin \omega x$ |

> **warning**
>
> If your standard trial PI duplicates any term in the complementary function (this is called resonance), multiply the trial PI by $x$ before substituting. For example, if $f(x) = \sin 2x$ and the CF has terms with $\cos 2x$ and $\sin 2x$, use trial PI $x(\alpha \cos 2x + \beta \sin 2x)$.

**Worked example:** Find the general solution to $\frac{d^2y}{dx^2} + 4y = \sin 2x$

1. First find the complementary function by solving the homogeneous ODE:
2. $$m^2 +4 =0 \implies m = \pm 2i$$
3. So CF is $y_{CF} = A \cos 2x + B \sin 2x$
4. Notice that $f(x) = \sin 2x$ matches a term in the CF, so we have a resonance case. Multiply the standard trial PI by $x$:
5. Trial PI: $y_{PI} = x(\alpha \cos 2x + \beta \sin 2x)$
6. Compute first and second derivatives of $y_{PI}$ 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 $y_{PI}$ 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 $\cos 2x$ and $\sin 2x$:
14. Coefficient of $\cos 2x$: $4\beta = 0 \implies \beta = 0$
15. Coefficient of $\sin 2x$: $-4\alpha = 1 \implies \alpha = -1/4$
16. So $y_{PI} = -\frac{1}{4}x \cos 2x$
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.

*Calculator:* allowed

## 3. Finding Particular Solutions from Conditions

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 $\frac{dy}{dx}=0$.

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 $\frac{dy}{dx}=0$ 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.

*Calculator:* allowed

## 4. Solving Reducible ODEs with Given Substitutions

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

1. Start with the given substitution: $z = y e^{2x}$, so rearrange to get $y = z e^{-2x}$
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 $e^{-2x}$ (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 $z = y e^{2x}$ 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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using wrong sign in auxiliary equation coefficients
  - Why it fails: Mismatching coefficients between the ODE and auxiliary equation leads to incorrect roots and wrong CF
  - Correct: Copy coefficients of d²y/dx², dy/dx, y directly into the auxiliary equation, preserving all signs
- **Wrong:** Forgetting to multiply trial PI by x in resonance cases
  - Why it fails: 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: Always check if your standard trial PI matches any term in the CF before substitution; if yes, multiply the trial PI by x
- **Wrong:** Substituting initial conditions into the CF before finding the PI
  - Why it fails: The constants in the CF apply to the full general solution, not just the homogeneous part
  - Correct: Form the full general solution (CF + PI) first, then substitute initial/boundary conditions to find A and B
- **Wrong:** Misapplying product rule for resonance trial PIs or substitution problems
  - Why it fails: Trial PIs for resonance and substitution variables have multiplied functions of x, requiring careful product rule use
  - Correct: Write out each derivative step explicitly, separating terms before combining like terms
- **Wrong:** Mixing up CF forms for different root types
  - Why it fails: Using the wrong CF form leads to a solution that does not satisfy the homogeneous ODE
  - Correct: Memorize the three CF forms for distinct real, equal real, and complex conjugate roots, and match them to your auxiliary equation roots

## Cheatsheet

| Component | Rule/Form |
| --- | --- |
| Auxiliary Equation | $a m^2 + b m + c = 0$ for $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$ |
| CF: Distinct real roots $m_1,m_2$ | $A e^{m_1 x} + B e^{m_2 x}$ |
| CF: Equal real roots $m$ | $(A + Bx)e^{mx}$ |
| CF: Complex roots $\alpha \pm \beta i$ | $e^{\alpha x}(A \cos \beta x + B \sin \beta x)$ |
| PI Trial: $k e^{px}$ | $\lambda e^{px}$ (multiply by x if matches CF) |
| PI Trial: $p + qx + rx^2$ | $\alpha + \beta x + \gamma x^2$ (multiply by x if matches CF) |
| PI Trial: $m\cos\omega x + n\sin\omega x$ | $\alpha\cos\omega x + \beta\sin\omega x$ (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 |

## 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.

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