Introduction to Differential Equations (DEs)
Hello! Welcome to the final topic in Further Pure Mathematics 2: Differential Equations. Don't worry if this chapter seems tricky at first; DEs are arguably the most powerful tool in advanced mathematics, linking pure calculus directly to real-world physics, engineering, and population models.
In A-Level Mathematics (9709), you learned how to solve simple separable DEs. Here in Further Maths, we equip you with techniques to tackle two main types of complex DEs: First-Order Linear equations and Second-Order Linear equations, using specialized tools like the Integrating Factor (IF), Complementary Functions (CF), and Particular Integrals (PI).
Let's dive in!
1. First-Order Linear Differential Equations
A first-order DE involves only \(\frac{dy}{dx}\), \(y\), and functions of \(x\). The key technique for solving the linear form is using the Integrating Factor (IF).
1.1 Standard Form and the Integrating Factor
A first-order linear differential equation must be arranged into the standard form:
$$ \frac{dy}{dx} + P(x)y = Q(x) $$
Where \(P(x)\) and \(Q(x)\) are functions of \(x\) (or constants).
Key Term: The Integrating Factor (\(\mu\))
The integrating factor is a function, \(\mu\), that we multiply the entire equation by, making the Left-Hand Side (LHS) the exact derivative of a product: \(\frac{d}{dx} (\mu y)\).
$$ \mu = e^{\int P(x) dx} $$
Analogy: Think of the Integrating Factor as the magic glue. It ensures that when you multiply it by the left side of the DE, everything sticks together neatly into a single, easy-to-integrate derivative.
1.2 Step-by-Step Solution using the IF
- Standardize: Ensure the equation is in the form \(\frac{dy}{dx} + P(x)y = Q(x)\).
- Find \(P(x)\): Identify the function multiplying \(y\).
- Calculate IF: Find \(\mu = e^{\int P(x) dx}\). (Remember: you do not need the arbitrary constant \(C\) here.)
- Multiply: Multiply the standardized DE by the IF, \(\mu\).
- Simplify the LHS: The left side must simplify to \(\frac{d}{dx} (\mu y)\). If it doesn't, check your IF calculation!
- Integrate: Integrate both sides with respect to \(x\): $$ \mu y = \int Q(x) \mu dx + C $$
- Solve for \(y\): Isolate \(y\) to find the General Solution.
Quick Review: IF Method
| Form | IF (\(\mu\)) | Integrated Form | | :--- | :--- | :--- | | \(\frac{dy}{dx} + P(x)y = Q(x)\) | \(e^{\int P(x) dx}\) | \(\mu y = \int Q(x) \mu dx + C\) |
2. Second-Order Linear Differential Equations (Constant Coefficients)
Second-order linear DEs have the form: $$ a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x) $$
The General Solution (GS) is always the sum of two parts:
$$ y_{GS} = y_c + y_p $$
$$ y_{GS} = \text{Complementary Function (CF)} + \text{Particular Integral (PI)} $$
2.1 Part 1: The Complementary Function (\(y_c\))
The CF is the solution to the Homogeneous Equation, where \(f(x)=0\): $$ a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 $$
We solve this by setting up the Auxiliary Equation (AE), a quadratic equation found by replacing the derivatives with a variable, usually \(m\):
$$ am^2 + bm + c = 0 $$
The nature of the roots of the AE determines the form of the CF.
Case 1: Two Distinct Real Roots (\(m_1 \neq m_2\))
If the discriminant (\(b^2 - 4ac\)) is positive, the roots are real and different.
$$ y_c = Ae^{m_1 x} + Be^{m_2 x} $$
Example: If roots are \(m=3\) and \(m=-1\), then \(y_c = Ae^{3x} + Be^{-x}\).
Case 2: One Repeated Real Root (\(m_1 = m_2 = m\))
If the discriminant is zero, there is one repeated root \(m\).
$$ y_c = (A + Bx)e^{mx} $$
Memory Tip: When a root repeats, you introduce the factor \(x\) to the second term to ensure the solutions are linearly independent.
Case 3: Conjugate Complex Roots (\(m = \alpha \pm i\beta\))
If the discriminant is negative, the roots are complex.
$$ y_c = e^{\alpha x}(A \cos \beta x + B \sin \beta x) $$
Here, \(\alpha\) is the real part and \(\beta\) is the imaginary part (we use the positive value of \(\beta\)).
Example: If roots are \(m = 2 \pm 3i\), then \(\alpha=2\) and \(\beta=3\), so \(y_c = e^{2x}(A \cos 3x + B \sin 3x)\).
Key Takeaway (CF): The CF uses two arbitrary constants (\(A\) and \(B\)) because it comes from a second-order equation. Its form depends entirely on the roots of the Auxiliary Equation.
2.2 Part 2: The Particular Integral (\(y_p\))
The PI (\(y_p\)) is any solution that satisfies the full non-homogeneous equation \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)\). It does not contain arbitrary constants.
We find the PI by proposing a Trial Function based on the form of \(f(x)\).
| If \(f(x)\) is... | Try \(y_p\) (Trial Function) | Example |
|---|---|---|
| A Polynomial of degree \(n\) | A general Polynomial of degree \(n\) | If \(f(x) = x^2 - 1\), try \(y_p = Px^2 + Qx + R\) |
| An Exponential \(ke^{bx}\) | \(Pe^{bx}\) | If \(f(x) = 5e^{3x}\), try \(y_p = Pe^{3x}\) |
| A Trigonometric function \(a \cos px + b \sin px\) | \(P \cos px + Q \sin px\) | If \(f(x) = \sin 2x\), try \(y_p = P \cos 2x + Q \sin 2x\) |
To find the constants \(P, Q, R\)...:
- Propose the trial function \(y_p\).
- Differentiate \(y_p\) once (\(\frac{dy_p}{dx}\)) and twice (\(\frac{d^2y_p}{dx^2}\)).
- Substitute these back into the original DE.
- Compare coefficients (or terms) on both sides of the equation to solve for the constants.
2.3 The Critical Case: Duplication (Resonance)
This is the most common pitfall!
Common Mistake to Avoid: If your proposed Particular Integral \(y_p\) is already part of the Complementary Function \(y_c\) (i.e., it duplicates a term in \(y_c\)), it will fail when substituted.
The Fix: If duplication occurs, multiply your trial function \(y_p\) by \(x\). If the new trial function still duplicates a term in \(y_c\), multiply it by \(x^2\).
Example Scenario:
- CF: \(y_c = Ae^{2x} + Be^{-x}\).
- \(f(x)\) is \(4e^{2x}\).
- Initial trial PI: \(P e^{2x}\). This duplicates \(Ae^{2x}\).
- New Trial PI: \(y_p = Px e^{2x}\).
Syllabus Link: The syllabus explicitly mentions finding coefficients (like \(k\)) when a special form like \(kx \cos 2x\) is given as the PI. This happens precisely when the standard PI (\(P \cos 2x + Q \sin 2x\)) fails because it duplicates terms in the CF (where the auxiliary roots were \(\pm 2i\)).
Key Takeaway (PI): Choose the trial function based on \(f(x)\). ALWAYS check if it duplicates any term in the CF. If it does, multiply by \(x\).
3. Using Substitutions to Simplify DEs
Sometimes, a differential equation isn't linear or separable in its given form. The syllabus requires using specific substitutions to transform the difficult equation into one you know how to solve (like a linear first-order DE or a linear second-order DE with constant coefficients).
3.1 Substitution to Reduce to Constant Coefficients (\(x = e^t\))
This substitution is used for equations containing terms like \(x^2\frac{d^2y}{dx^2}\) and \(x\frac{dy}{dx}\). This is often called the Euler-Cauchy equation.
If we let \(x = e^t\), then \(t = \ln x\). We use the chain rule to transform the derivatives:
First Derivative: $$ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt} \cdot \frac{1}{x} $$ $$ x\frac{dy}{dx} = \frac{dy}{dt} $$
Second Derivative: $$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{x}\frac{dy}{dt} \right) $$ (Using the product rule on the right side) $$ x^2\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt} $$
Substituting these expressions back changes the original variable equation into a new DE in terms of \(t\) with constant coefficients, which you can solve using the CF/PI method from Section 2.
3.2 Substitution to Reduce to Separable Form (\(y = ux\))
This substitution is typically used for first-order homogeneous DEs of the form \(\frac{dy}{dx} = F\left(\frac{y}{x}\right)\).
Example Syllabus Form: Reduce \(\frac{dy}{dx} = \frac{x+y}{x-y}\) to separable form.
The substitution is \(y = ux\). We need to find \(\frac{dy}{dx}\) in terms of \(u, x,\) and \(\frac{du}{dx}\) using the product rule:
$$ \frac{dy}{dx} = u(1) + x\frac{du}{dx} $$ $$ \frac{dy}{dx} = u + x\frac{du}{dx} $$
Substitute \(y=ux\) and the new derivative into the original equation:
$$ u + x\frac{du}{dx} = \frac{x + ux}{x - ux} = \frac{x(1 + u)}{x(1 - u)} = \frac{1 + u}{1 - u} $$
Now, rearrange to separate the variables \(u\) and \(x\):
$$ x\frac{du}{dx} = \frac{1 + u}{1 - u} - u = \frac{1 + u - u(1 - u)}{1 - u} = \frac{1 + u^2}{1 - u} $$
$$ \frac{1 - u}{1 + u^2} du = \frac{1}{x} dx $$
This new equation is now separable and can be solved by direct integration, finally replacing \(u\) with \(\frac{y}{x}\).
Key Takeaway (Substitution): Substitutions allow us to tackle non-standard DEs by converting them into standard linear forms we already know how to solve (either IF or CF/PI methods).
4. Initial Conditions and Interpretation
The General Solution (GS) you find contains arbitrary constants (\(C\) for first order, \(A\) and \(B\) for second order). This represents an infinite family of possible solutions.
4.1 Finding the Particular Solution
To find a unique solution—the Particular Solution—you must use the Initial Conditions (ICs) provided in the problem.
For a first-order DE, you need one condition (e.g., \(y(0)=5\)).
For a second-order DE, you need two conditions (e.g., \(y(0)=2\) and \(y'(0)=1\)).
Step-by-Step Application:
- Find the General Solution \(y\).
- If necessary (for second-order equations), differentiate \(y\) to find \(\frac{dy}{dx}\) or \(y'\).
- Substitute the first IC into \(y\) to find one constant (or a relationship between constants).
- If required, substitute the second IC into \(y'\) to find the remaining constant.
- Substitute these values back into the General Solution to obtain the Particular Solution.
4.2 Interpretation in Context
Differential equations are often used to model dynamic systems (motion, population, heat transfer). The syllabus requires you to interpret the solution in terms of the problem modelled.
- CF (Complementary Function): Represents the natural response or the transient behavior of the system. If the roots are negative, these terms decay over time (e.g., oscillations dying down).
- PI (Particular Integral): Represents the forced response or the steady-state behavior. This is the long-term behavior of the system, often driven by the external input \(f(x)\).
- Particular Solution: Once constants are found, this describes the specific behavior of the system starting from the given initial state.
Did you know?
If you are solving a DE modelling a mass attached to a spring, complex roots in the Auxiliary Equation mean the system will oscillate (like a sine wave), while real roots mean it returns to equilibrium without oscillating (it's "overdamped"). Mathematics shows you exactly what the physical system will do!
Key Takeaway (ICs and Interpretation): Initial conditions fix the constants to define a unique path. Always relate the mathematical components (CF, PI) back to the physical meaning of the quantities in the problem (e.g., time, displacement, charge).