Pure Mathematics | Second order differential equations

👋 Welcome to Second Order Differential Equations!

Hey there, future mathematician! Welcome to one of the most powerful chapters in Further Pure Mathematics: Second Order Differential Equations (S.O.D.E.s).

Don't worry if the name sounds intense. In simple terms, these equations help us model things in the real world that involve acceleration, like springs oscillating, electrical circuits stabilizing, or even the movement of a pendulum.

The goal of this chapter is to find a function \(y\) that satisfies an equation involving \(y\), \(\frac{dy}{dx}\), and \(\frac{d^2y}{dx^2}\). By the end of these notes, you'll know exactly how to solve them, step-by-step!

Section 1: The Structure of Second Order Linear Equations

1.1 Standard Form and Constant Coefficients

In FP2, we only deal with linear second order differential equations that have constant coefficients. This means the numbers multiplying the derivatives are just constants, not functions of \(x\).

The general form looks like this:

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

• \(a, b, c\) are constant numbers (coefficients).
• \(f(x)\) is the function of \(x\) that forces the system (or 0, if it's unforced).

🔑 Key Concept: The General Solution

The final solution to an S.O.D.E. is always composed of two parts:

\[ y = y_{CF} + y_{PI} \]

• \(y_{CF}\): The Complementary Function (CF). This describes the natural behavior of the system (ignoring the forcing function \(f(x)\)).
• \(y_{PI}\): The Particular Integral (PI). This describes the specific response due to the forcing function \(f(x)\).

Let's start by finding the CF.

Section 2: Finding the Complementary Function (\(y_{CF}\))

To find the CF, we must first solve the Homogeneous Equation, which means setting the right-hand side to zero:

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

2.1 The Auxiliary Equation (AE)

We assume that the solution has an exponential form, \(y = e^{mx}\). When we substitute this into the homogeneous equation, we get a simple quadratic equation called the Auxiliary Equation (AE):

\[ am^2 + bm + c = 0 \]

The nature of the roots \(m\) of this quadratic equation determines the form of the CF. There are three cases based on the discriminant, \(b^2 - 4ac\).

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

If the AE has two different real roots, \(m_1\) and \(m_2\).

The CF Solution:

\[ y_{CF} = A e^{m_1 x} + B e^{m_2 x} \]

(A and B are arbitrary constants determined by initial conditions.)

Example: Solve \(\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = 0\).
AE: \(m^2 + 5m + 6 = 0 \Rightarrow (m+2)(m+3) = 0\). Roots are \(m_1 = -2\) and \(m_2 = -3\).
CF: \(y_{CF} = A e^{-2x} + B e^{-3x}\).

2.3 Case 2: Real and Repeated Roots (\(m_1 = m_2 = m\))

If the AE has only one real root (the quadratic is a perfect square).

The CF Solution:

\[ y_{CF} = (A x + B) e^{m x} \]

💡 Why the \(x\)? If we just wrote \(Ae^{mx} + Be^{mx}\), we could combine the constants and only have one term, which is not a general solution for a second-order equation (we need two arbitrary constants). Multiplying one term by \(x\) ensures we have two truly independent solutions.

Example: Solve \(\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0\).
AE: \(m^2 - 4m + 4 = 0 \Rightarrow (m-2)^2 = 0\). Root is \(m = 2\).
CF: \(y_{CF} = (A x + B) e^{2x}\).

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

If the discriminant is negative, the roots are complex conjugates, where \(\alpha\) is the real part and \(\beta\) is the imaginary part.

The CF Solution:

\[ y_{CF} = e^{\alpha x} (A \cos(\beta x) + B \sin(\beta x)) \]

Did you know? This solution structure comes from Euler's Identity, linking exponentials to sines and cosines. This form perfectly models systems that oscillate (like a swinging pendulum or a spring bouncing).

Example: Solve \(\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0\).
AE: \(m^2 + 2m + 5 = 0\). Using the quadratic formula:
\(m = \frac{-2 \pm \sqrt{4 - 4(5)}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i\).
Here, \(\alpha = -1\) and \(\beta = 2\).
CF: \(y_{CF} = e^{-x} (A \cos(2x) + B \sin(2x))\).

Section 3: Finding the Particular Integral (\(y_{PI}\))

The Particular Integral (PI) handles the right-hand side \(f(x)\) of the non-homogeneous equation:

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

The method involves "guessing" the form of the PI based on the function \(f(x)\) and then substituting this guess back into the differential equation to find the exact coefficients.

3.1 Guessing the Form of \(y_{PI}\)

The guess must have the same form as \(f(x)\) but include undetermined coefficients (like \(P, Q, R\)).

Case A: Polynomial \(f(x)\)

If \(f(x)\) is a polynomial of degree \(n\), guess a general polynomial of degree \(n\).

If \(f(x)\) is:	Guess \(y_{PI}\) is:
\(5\) (Constant)	\(P\)
\(2x - 1\) (Degree 1)	\(Px + Q\)
\(x^2 + 3\) (Degree 2)	\(P x^2 + Q x + R\)

Case B: Exponential \(f(x)\)

If \(f(x)\) is \(k e^{kx}\).

If \(f(x)\) is:	Guess \(y_{PI}\) is:
\(4 e^{5x}\)	\(P e^{5x}\)

Case C: Trigonometric \(f(x)\)

If \(f(x)\) involves sine or cosine, the guess must include both sine and cosine terms (because differentiating a sine gives cosine, and vice versa).

If \(f(x)\) is:	Guess \(y_{PI}\) is:
\(7 \sin(3x)\)	\(P \cos(3x) + Q \sin(3x)\)
\(\cos(x)\)	\(P \cos(x) + Q \sin(x)\)

Step-by-Step Process for PI:

1. Formulate the guess \(y_{PI}\).
2. Find \(\frac{dy_{PI}}{dx}\) and \(\frac{d^2y_{PI}}{dx^2}\).
3. Substitute \(y_{PI}\), \(\frac{dy_{PI}}{dx}\), and \(\frac{d^2y_{PI}}{dx^2}\) back into the original S.O.D.E.
4. Equate coefficients on the left side with the coefficients on the right side \(f(x)\).
5. Solve the resulting simultaneous equations for \(P, Q, R, ...\).

Don't worry if this seems tricky at first; substitution and comparing coefficients are the hardest parts, requiring careful algebra!

3.2 The Critical Rule: The Resonance Condition (Overlap)

This is the most common pitfall for students! If your initial guess for \(y_{PI}\) is already part of the \(y_{CF}\), it means that input naturally makes the homogeneous equation zero. Substituting it won't balance the \(f(x)\) part.

Analogy: Imagine pushing a swing. If you push at the natural frequency of the swing (the CF), the swing gets huge, but the forcing function \(f(x)\) doesn't work as expected. To fix it, you need to change your solution structure.

The Fix: If the proposed \(y_{PI}\) term is already in the \(y_{CF}\), you must multiply the entire guess by \(x\).

Example of Overlap (Case B - Exponential): Suppose the AE roots were \(m=3\) and \(m=-1\). \(y_{CF} = A e^{3x} + B e^{-x}\).
If \(f(x) = 5e^{3x}\), your initial guess would be \(y_{PI} = P e^{3x}\).
Since \(e^{3x}\) is already in the CF, the guess is invalid.
Corrected Guess: \(y_{PI} = P x e^{3x}\).

What if multiplying by \(x\) still causes overlap?
This only happens in the Repeated Root case (Case 2). If the AE had the root \(m\), the CF is \((Ax+B)e^{mx}\). If \(f(x) = k e^{mx}\), then both \(e^{mx}\) and \(x e^{mx}\) are solutions to the homogeneous equation. In this very rare scenario, you must multiply by \(x^2\).

Corrected Guess (Double Overlap): \(y_{PI} = P x^2 e^{mx}\).

Section 4: The General Solution and Boundary Conditions

4.1 Forming the General Solution

Once you have successfully found the Complementary Function \(y_{CF}\) (with arbitrary constants \(A\) and \(B\)) and the Particular Integral \(y_{PI}\) (with specific coefficients \(P, Q, ...\)), you combine them to get the General Solution:

\[ y = y_{CF} + y_{PI} \]

This general solution contains the two arbitrary constants \(A\) and \(B\).

4.2 Using Initial and Boundary Conditions

To find the unique solution for a specific physical problem, we must find the values of \(A\) and \(B\) using boundary conditions or initial conditions.

Since we have a second-order equation, we always need two conditions. These usually involve the value of \(y\) at a specific point (e.g., \(y(0)=5\)) and/or the value of the derivative \(\frac{dy}{dx}\) at a specific point (e.g., \(y'(0)=1\)).

Step-by-Step Process for Final Solution:

1. Find the General Solution \(y = y_{CF} + y_{PI}\).
2. Differentiate the General Solution to find \(\frac{dy}{dx}\).
3. Apply the first condition (e.g., substitute \(x=0\) and \(y=5\) into the \(y\) equation) to get an equation in \(A\) and \(B\).
4. Apply the second condition (e.g., substitute \(x=0\) and \(y'=1\) into the \(\frac{dy}{dx}\) equation) to get a second equation in \(A\) and \(B\).
5. Solve the simultaneous equations for \(A\) and \(B\).
6. Substitute \(A\) and \(B\) back into the General Solution to get the unique solution.

Common Mistake Alert!

When applying conditions involving \(\frac{dy}{dx}\), students often forget to include the derivative of the \(y_{PI}\) part. Remember: Differentiate the entire general solution before substituting the initial conditions for the gradient!

Section 5: Summary and Review

Solving second order differential equations is a procedural skill. If you follow the steps, you will succeed!

Flowchart to Solving S.O.D.E.s

1. Identify Homogeneous Part: \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0\)
2. Solve Auxiliary Equation: Find roots \(m\) from \(am^2 + bm + c = 0\).
3. Determine \(y_{CF}\): Use the correct formula based on the nature of the roots (Distinct, Repeated, Complex).
4. Determine \(y_{PI}\): Look at \(f(x)\) and formulate the appropriate guess (Polynomial, Trig, Exponential).
5. Check for Overlap: If the guess for \(y_{PI}\) matches a term in \(y_{CF}\), multiply the guess by \(x\).
6. Find Coefficients: Substitute \(y_{PI}\) and its derivatives into the original non-homogeneous equation and solve for the coefficients (P, Q, ...).
7. General Solution: Write \(y = y_{CF} + y_{PI}\).
8. Apply Conditions: Use the two boundary/initial conditions on \(y\) and \(\frac{dy}{dx}\) to find the unique constants \(A\) and \(B\).

You’ve got this! Second order DEs combine quadratics, differentiation, and algebra into one satisfying process. Practice makes perfect!