Differential Equations – Second Order (FP2.11)

Hello future Mathematician! Welcome to the world of second-order differential equations (DEs). Don't worry if the name sounds intense; these are simply mathematical tools used to describe how systems change based on their acceleration, which is incredibly useful for modelling physical phenomena like vibrations, electrical circuits, and even economics.

In this chapter, we will learn how to find the full, exact formula for \(y\) given an equation involving \(y\), \(\frac{dy}{dx}\), and \(\frac{d^2y}{dx^2}\).

Quick Context: Why are they "Second Order"?

A DE's "order" is determined by the highest derivative present. If it has \(\frac{d^2y}{dx^2}\) (the acceleration), it is second-order. This means we expect two arbitrary constants (usually \(A\) and \(B\)) in our general solution, just like in mechanics where you need initial displacement and initial velocity to fully describe a projectile's path!

1. Structure of Linear Second-Order Differential Equations

We focus exclusively on Linear Differential Equations with Constant Coefficients. This means the derivatives and \(y\) itself appear only to the power of one, and the numbers multiplying them (\(a, b, c\)) are constants (integers, specifically, according to the syllabus).

The Two Main Types

Second-order linear DEs fall into two categories, depending on the function on the right-hand side, \(f(x)\):

  • 1. Homogeneous Equations (or Unforced Equations)
    This is the "base case" where \(f(x)=0\).

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

    Analogy: This describes the natural oscillation of a system, like a spring bouncing in a vacuum (no external forces).
  • 2. Non-Homogeneous Equations (or Forced Equations)
    This is when \(f(x)\) is a non-zero function of \(x\).

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

    Analogy: This describes the system when an external force (like an engine running or continuous pushing) is applied.

Key Takeaway: Solving non-homogeneous equations requires two parts: the solution to the homogeneous part (called the Complementary Function) and an extra solution for the forcing term (called the Particular Integral).

2. Solving Homogeneous Equations: The Complementary Function (CF)

The solution to the homogeneous equation \( a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 \) is known as the Complementary Function (\(y_{CF}\)).

The secret to solving these is the Auxiliary Equation (AE). We assume the solution has the form \(y = e^{\lambda x}\), and when we substitute this in, the DE simplifies into a quadratic equation in terms of \(\lambda\):

The Auxiliary Equation (AE)

If \( a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 \), the AE is:
$$ a\lambda^2 + b\lambda + c = 0 $$

The form of the CF depends entirely on the nature of the roots of this quadratic equation.

Case 1: Two Distinct Real Roots (\(\lambda_1 \neq \lambda_2\))

This happens when the discriminant \((b^2 - 4ac)\) is positive.

The general solution (CF) is a linear combination of the two exponential solutions: $$ y_{CF} = Ae^{\lambda_1 x} + Be^{\lambda_2 x} $$

Analogy: This often represents "overdamped" motion, where the system returns smoothly to equilibrium without oscillating.

Case 2: One Repeated Real Root (\(\lambda_1 = \lambda_2 = \lambda\))

This happens when the discriminant \((b^2 - 4ac)\) is zero.

The general solution (CF) must include an extra factor of \(x\) to ensure the two terms are independent: $$ y_{CF} = (A + Bx)e^{\lambda x} $$

Common Mistake Alert: Don't forget the factor of \(x\) on the second term! If you write \(Ae^{\lambda x} + Be^{\lambda x}\), it just simplifies to \((A+B)e^{\lambda x}\), which is only one arbitrary constant, not two.

Case 3: Complex Conjugate Roots (\(\lambda = \alpha \pm i\beta\))

This happens when the discriminant \((b^2 - 4ac)\) is negative. Since \(a, b, c\) are real integers, the roots must come in conjugate pairs.

While the raw exponential solution is \(y = A'e^{(\alpha + i\beta)x} + B'e^{(\alpha - i\beta)x}\), this is usually converted into the much cleaner trigonometric form using Euler's identity (\(e^{i\theta} = \cos\theta + i\sin\theta\)): $$ y_{CF} = e^{\alpha x} (A \cos(\beta x) + B \sin(\beta x)) $$

Analogy: This represents oscillating motion. \(\alpha\) dictates whether the oscillation grows (\(\alpha > 0\)), decays (\(\alpha < 0\), damping), or stays constant (\(\alpha = 0\), Simple Harmonic Motion).

✅ Quick Review: CF Formulas

Let \(\lambda_1, \lambda_2\) be the roots of \( a\lambda^2 + b\lambda + c = 0 \).

  • Distinct Real: \( y_{CF} = Ae^{\lambda_1 x} + Be^{\lambda_2 x} \)
  • Repeated Real (\(\lambda\)): \( y_{CF} = (A + Bx)e^{\lambda x} \)
  • Complex (\(\alpha \pm i\beta\)): \( y_{CF} = e^{\alpha x} (A \cos(\beta x) + B \sin(\beta x)) \)

3. Solving Non-Homogeneous Equations: The Particular Integral (PI)

When solving \( a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x) \), the General Solution (\(y_{GS}\)) is the sum of the Complementary Function (CF) and the Particular Integral (PI):

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

The CF handles the natural behaviour (as derived in Section 2). The PI is any function that satisfies the original DE when substituted in. Since \(f(x)\) is restricted to certain forms in this syllabus, finding \(y_{PI}\) involves an educated guess.

Step-by-Step Guide to Finding \(y_{PI}\)

The PI must be of the same general form as \(f(x)\).

Step 1: Guess the form of \(y_{PI}\) based on \(f(x)\).

\( f(x) \) (The Forcing Term) Guess for \( y_{PI} \) Maximum Degree Restriction
Polynomial (e.g., \(x^3\)) General polynomial of the same degree (e.g., \(Ax^3 + Bx^2 + Cx + D\)) Degree 4
Exponential (e.g., \(e^{kx}\)) \( Pe^{kx} \) N/A
Trigonometric (e.g., \(\sin(kx)\) or \(\cos(kx)\)) \( P \cos(kx) + Q \sin(kx) \) (Must include both sin and cos!) N/A

Step 2: Differentiate and Substitute.
Differentiate your guess (\(y_{PI}\)) twice to find \(\frac{dy_{PI}}{dx}\) and \(\frac{d^2y_{PI}}{dx^2}\). Substitute all three terms into the original non-homogeneous DE.

Step 3: Equate Coefficients.
Compare the coefficients of \(e^{kx}\), \(\cos(kx)\), \(\sin(kx)\), or powers of \(x\) on both sides of the equation. This will give you simultaneous equations to solve for the unknown constants (\(A, B, P, Q\), etc.) in your PI guess.

The Crucial Overlap Rule (The Modification Rule)

This is the most common place to make a mistake!

If your chosen guess for \(y_{PI}\) (from Step 1) is already a term in the Complementary Function (\(y_{CF}\)), the standard guess will not work (it will simply differentiate to zero when substituted into the homogeneous part).

If there is an overlap, you must multiply your standard guess by the lowest possible power of \(x\) (usually \(x\) or \(x^2\)) until it is linearly independent of the CF terms.

Example: If \(f(x) = 5e^{3x}\), your standard guess is \(y_{PI} = Pe^{3x}\). But if your CF was \(y_{CF} = Ae^{3x} + Be^{-x}\), the PI guess overlaps with \(Ae^{3x}\).
The modified PI guess must then be: \( y_{PI} = Px e^{3x} \).

Did you know? This multiplication by \(x\) is necessary because the forced input frequency (from \(f(x)\)) matches the natural frequency (from the CF), causing a phenomenon called resonance. The system's response grows linearly with time, hence the \(x\) term!

Key Takeaway: Always find \(y_{CF}\) first, then compare it to your guessed \(y_{PI}\) form. If they match, multiply \(y_{PI}\) by \(x\).

4. Finding the Particular Solution (Using Boundary Conditions)

Once you have the General Solution \(y_{GS} = y_{CF} + y_{PI}\), it will still contain two arbitrary constants, \(A\) and \(B\), from the CF. To find the unique Particular Solution, you need two additional pieces of information, known as Boundary Conditions or Initial Conditions.

These conditions usually specify the value of \(y\) or \(\frac{dy}{dx}\) at a specific point \(x\) (often at \(x=0\)).

Step-by-Step for Particular Solutions

  1. Find \(y_{GS}\): Calculate \(y_{CF}\) and \(y_{PI}\) and combine them: \(y = y_{CF} + y_{PI}\).
  2. Differentiate \(y_{GS}\): Calculate \(\frac{dy}{dx}\). (You will need this if one of your conditions involves the derivative).
  3. Apply Condition 1: Substitute the first boundary condition (e.g., \(y(0)=5\)) into \(y_{GS}\). This creates an equation linking \(A\) and \(B\).
  4. Apply Condition 2: Substitute the second boundary condition (e.g., \(y'(0)=1\)) into the derivative \(\frac{dy}{dx}\). This creates a second equation linking \(A\) and \(B\).
  5. Solve: Solve the two simultaneous equations for \(A\) and \(B\).
  6. Final Answer: Substitute the values of \(A\) and \(B\) back into \(y_{GS}\) to write the final Particular Solution.

Tip for struggling students: Always check your algebra when finding the derivatives! Errors here will inevitably lead to wrong values for A and B. Take your time, especially with the product rule if you have complex CFs or PI terms.

📝 FINAL KEY TAKEAWAYS

1. Identify the DE type: Homogeneous (\(=0\)) or Non-Homogeneous (\(=f(x)\)).

2. The General Solution structure is: \( y_{GS} = y_{CF} + y_{PI} \).

3. CF (Complementary Function): Found using the Auxiliary Equation \( a\lambda^2 + b\lambda + c = 0 \). The form depends on whether roots are distinct real, repeated real, or complex.

4. PI (Particular Integral): Found by guessing a form similar to \(f(x)\). Crucially, check for overlap with the CF and multiply by \(x\) if necessary.

5. Particular Solution: Use the two given boundary conditions to find the specific values of the constants \(A\) and \(B\).