Further Mathematics | Second order differential equations

📚 FP2 Study Notes: Second Order Differential Equations

Hello future mathematician! This chapter, Second Order Differential Equations (S.O.D.E.s), is where Further Pure Mathematics really starts to shine. Don't worry if the name sounds intense—we're essentially learning the language of motion, oscillations, and electrical circuits.

Understanding S.O.D.E.s allows us to model systems where the rate of change depends on both the current value and the rate of change itself. Think springs, pendulums, or even how quickly a rumour spreads!

What You Will Master in This Chapter:

• Solving S.O.D.E.s with constant coefficients.
• Understanding the Complementary Function (C.F.) for homogeneous equations.
• Finding the Particular Integral (P.I.) for non-homogeneous equations.
• Using initial and boundary conditions to find the unique particular solution.

1. Introduction to Second Order Differential Equations

1.1. The Basic Form and Terminology

A second-order differential equation involves a second derivative, \(\frac{\text{d}^2y}{\text{d}x^2}\). In FP2, we focus exclusively on linear S.O.D.E.s with constant coefficients.

The general form we study is:

$$ a \frac{\text{d}^2y}{\text{d}x^2} + b \frac{\text{d}y}{\text{d}x} + cy = f(x) $$

• \(a, b, c\): These are fixed constants (numbers), which makes our life much easier!
• \(\frac{\text{d}^2y}{\text{d}x^2}\): The second derivative (acceleration, curvature).
• \(f(x)\): This is the driving function or forcing term.

We classify these equations into two types:

• Homogeneous Equation: When \(f(x) = 0\). This describes the natural, unforced behaviour of the system.
• Non-Homogeneous Equation: When \(f(x) \neq 0\). This describes the system under an external influence.

The Goal: The General Solution (G.S.)

The final solution, \(y\), is always the sum of two parts:

$$ \mathbf{y} = \mathbf{y_{CF}} + \mathbf{y_{PI}} $$
$$ \text{General Solution} = \text{Complementary Function} + \text{Particular Integral} $$

Analogy: Think of a car's motion. The Complementary Function (C.F.) is how the car moves naturally (e.g., coasting to a stop). The Particular Integral (P.I.) is the movement caused by the external force (e.g., constantly pressing the accelerator).

2. Step 1: Solving the Homogeneous Equation (The Complementary Function, C.F.)

To find the C.F. (\(y_{CF}\)), we must solve the equation:

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

2.1. The Auxiliary Equation

We guess that the solution is in the form \(y = e^{mx}\). Substituting this into the homogeneous equation leads directly to a simple quadratic equation in \(m\), called the Auxiliary Equation (or Characteristic Equation):

$$ \mathbf{am^2 + bm + c = 0} $$

Solving this quadratic gives us the roots (\(m\)), which determine the form of the C.F. Since it's a quadratic, there are three possibilities for the roots, which lead to three different solution forms.

2.2. Case 1: Two Distinct Real Roots (\(m_1 \neq m_2\))

If the discriminant \((b^2 - 4ac)\) is positive, you get two different real numbers, \(m_1\) and \(m_2\).

Example: \(m^2 + 5m + 6 = 0 \implies (m+2)(m+3) = 0\). Roots are \(m_1 = -2\), \(m_2 = -3\).

The C.F. Form: $$ \mathbf{y_{CF} = Ae^{m_1x} + Be^{m_2x}} $$

Note: \(A\) and \(B\) are arbitrary constants that we find later using initial conditions.

2.3. Case 2: One Repeated Real Root (\(m_1 = m_2 = m\))

If the discriminant \((b^2 - 4ac)\) is zero, you get only one real root, \(m\). This means the auxiliary equation is a perfect square.

Example: \(m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0\). Root is \(m = 2\).

The C.F. Form: $$ \mathbf{y_{CF} = (Ax + B)e^{mx}} $$

💡 Memory Aid: When the root repeats, you need the extra term \(Ax\) to ensure the solution is truly "second order" (i.e., has two independent constants, A and B).

2.4. Case 3: Complex Conjugate Roots (\(\alpha \pm i\beta\))

If the discriminant \((b^2 - 4ac)\) is negative, you get two complex roots, which always appear as a conjugate pair, \(\alpha + i\beta\) and \(\alpha - i\beta\).

Example: \(m^2 + 2m + 5 = 0\). Using the quadratic formula gives \(m = -1 \pm 2i\). Here, \(\alpha = -1\) and \(\beta = 2\).

The C.F. Form: $$ \mathbf{y_{CF} = e^{\alpha x}(A \cos(\beta x) + B \sin(\beta x))} $$

Did you know? This is the form that describes oscillations (like a pendulum swinging) that are either damped (\(\alpha < 0\)) or growing (\(\alpha > 0\)).

3. Step 2: Finding the Particular Integral (P.I.)

The P.I. (\(y_{PI}\)) is a specific solution that satisfies the full non-homogeneous equation:

$$ a \frac{\text{d}^2y}{\text{d}x^2} + b \frac{\text{d}y}{\text{d}x} + cy = f(x) $$

The key here is guessing the form of \(y_{PI}\) based on the function \(f(x)\).

3.1. The Guessing Strategy

We choose a trial solution for \(y_{PI}\) that looks like \(f(x)\) and includes unknown constant coefficients (e.g., \(P, Q, R\)).

Table of Trial Solutions (P.I. Forms)

Let's say \(f(x)\) is the right-hand side:

3.2. Step-by-Step P.I. Process

1. Guess: Choose the correct trial form for \(y_{PI}\) based on \(f(x)\).
2. Differentiate: Find the first derivative (\(y'_{PI}\)) and second derivative (\(y''_{PI}\)) of your guess.
3. Substitute: Plug \(y_{PI}\), \(y'_{PI}\), and \(y''_{PI}\) into the original S.O.D.E.
4. Compare Coefficients: Equate the coefficients of matching terms on both sides of the equation (e.g., equate coefficients of \(x^2\), then \(x\), then the constant terms) to solve for \(P, Q, R, \dots\).

Don't worry if this generates long equations; take your time and organise your terms carefully!

3.3. The Special Case: Duplication (The Resonance Rule)

This is the most common pitfall! It occurs when your normal trial solution for the P.I. contains a term that is already present in the Complementary Function (C.F.).

If you don't adjust the P.I. guess in this scenario, the substitution step will result in \(0 = f(x)\), which is impossible!

The Rule for Duplication:
If the standard P.I. guess duplicates a term in the C.F., you must multiply the entire trial solution by \(\mathbf{x}\).

Example of Duplication: Suppose the C.F. is \(y_{CF} = Ae^{2x} + Be^{-3x}\).
If \(f(x) = 5e^{2x}\), your initial guess would be \(y_{PI} = Pe^{2x}\).
Since \(e^{2x}\) is already in the C.F., this guess fails.
The correct adjusted guess must be: \(y_{PI} = Px e^{2x}\).

Advanced Duplication (Repeated Roots): If the C.F. had a repeated root, \(y_{CF} = (Ax + B)e^{2x}\), and \(f(x) = 5e^{2x}\), multiplying by \(x\) once (\(Pxe^{2x}\)) would still duplicate a term (\(Axe^{2x}\) is in the C.F.). In this very specific case, you must multiply by \(\mathbf{x^2}\).
Correct adjusted guess: \(y_{PI} = Px^2 e^{2x}\).

4. Step 3 & 4: The General Solution and Particular Solution

4.1. Forming the General Solution (G.S.)

Once you have successfully found the C.F. (with constants \(A\) and \(B\)) and the P.I. (with constants \(P, Q, \dots\) determined), you combine them:

$$ y = y_{CF} + y_{PI} $$

This G.S. contains two arbitrary constants \(A\) and \(B\). This represents a whole family of solutions.

4.2. Finding the Particular Solution

In applied problems, we usually need a specific, unique solution. We use boundary or initial conditions to find the exact values of \(A\) and \(B\).

Since the equation is second order, you need two independent conditions:

1. Initial Conditions: Both conditions are given at the same starting point (e.g., \(x=0\)).
   Example: \(y(0) = 5\) (starting position) AND \(y'(0) = 1\) (starting velocity).
2. Boundary Conditions: Conditions are given at two different points.
   Example: \(y(0) = 5\) AND \(y(1) = 10\).

Step-by-Step for Finding A and B

1. Write the G.S.: \(y = y_{CF} + y_{PI}\).
2. Differentiate the G.S.: Find \(y' = y'_{CF} + y'_{PI}\). (You need this if initial conditions involve \(y'\)).
3. Substitute Conditions: Plug the given values (e.g., \(x=0\) and \(y=5\)) into the equations for \(y\) and \(y'\).
4. Solve Simultaneous Equations: You will now have a pair of simultaneous linear equations involving \(A\) and \(B\). Solve them to find the specific values.
5. State the Particular Solution: Substitute the values of \(A\) and \(B\) back into the General Solution formula.

Tip: Always try to use initial conditions at \(x=0\), as \(e^0 = 1\), \(\sin(0)=0\), and \(\cos(0)=1\), which greatly simplifies the resulting simultaneous equations!