ODE Exam Study Guide

The key to this exam is to understand the methods and when to apply them, not just to memorize formulas. The notes repeatedly emphasize: "Don't memorise the formula – learn the approach."

Section 1: Fundamentals & Classification

First, identify the type of ODE you're dealing with. This determines your solution strategy.

Order: The order of the highest derivative in the equation.
- 1st Order: dy/dx = ...
- 2nd Order: d²y/dx² = ...
Linearity: An ODE is linear if it can be written in the form: \(\(a_n(x) \frac{d^n y}{dx^n} + a_{n-1}(x) \frac{d^{n-1} y}{dx^{n-1}} + \dots + a_1(x) \frac{dy}{dx} + a_0(x)y = f(x)\)\)
- Key check: The coefficients a_i(x) and the right-hand side f(x) can only depend on the independent variable (x or t). The dependent variable y and its derivatives cannot be multiplied together, raised to powers, or be inside functions like sin(y) or e^y.
Homogeneity (for linear ODEs):
- Homogeneous: The right-hand side is zero (f(x) = 0).
- Non-homogeneous: The right-hand side is not zero (f(x) ≠ 0).

Section 2: First-Order ODEs

A. Separable Equations

Form: dy/dx = g(x)f(y) or dy/dx = g(x)/h(y). Often nonlinear.
Method:
1. Separate the variables: \(\frac{1}{f(y)} dy = g(x) dx\).
2. Integrate both sides: \(\int \frac{1}{f(y)} dy = \int g(x) dx + C\).
3. Solve for y if possible.
Watch out for: The solution might only exist for a limited interval of x.

B. First-Order Linear Equations

Form: dy/dx + p(x)y = q(x).
General Solution Structure: The final solution is \(y(x) = y_p(x) + y_{cf}(x)\), where \(y_{cf}\) is the solution to the homogeneous part and \(y_p\) is a particular solution.
Method 1: Integrating Factors (Most Common)
1. Ensure the ODE is in standard form: \(y' + p(x)y = q(x)\).
2. Calculate the integrating factor: \(\mu(x) = e^{\int p(x) dx}\).
3. Multiply the entire standard-form ODE by \(\mu(x)\). The left side will become the derivative of a product: \(\mu(x)y' + \mu(x)p(x)y = \frac{d}{dx}(\mu(x)y)\).
4. The equation is now: \(\frac{d}{dx}(\mu(x)y) = \mu(x)q(x)\).
5. Integrate both sides and solve for y.
Method 2: Variation of Parameters
1. First, solve the homogeneous equation \(y' + p(x)y = 0\) to get the complementary function \(y_{cf}(x) = A y_1(x)\).
2. "Vary the parameter" by replacing the constant \(A\) with a function \(A(x)\). The ansatz for the full solution is \(y(x) = A(x)y_1(x)\).
3. Substitute this \(y(x)\) into the non-homogeneous ODE \(y' + p(x)y = q(x)\).
4. Many terms will cancel, leaving a simple equation to solve for \(A'(x)\).
5. Integrate to find \(A(x)\) and substitute back into \(y(x) = A(x)y_1(x)\).

Section 3: Second-Order Linear Homogeneous ODEs

A. Theory: Wronskian & Linear Independence

The general solution requires two linearly independent solutions, \(y_1\) and \(y_2\).
Wronskian: To test for linear independence, calculate the Wronskian: \(\(W(y_1, y_2) = \det \begin{pmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{pmatrix} = y_1y'_2 - y_2y'_1\)\)
If \(W ≠ 0\) on an interval, \(y_1\) and \(y_2\) are linearly independent and form a fundamental set of solutions.
The general solution is then \(y(x) = c_1y_1(x) + c_2y_2(x)\).

B. Constant Coefficient Equations

Form: \(ay'' + by' + cy = 0\).
Method:
1. Assume a solution of the form \(y = e^{rx}\).
2. This leads to the characteristic (or auxiliary) equation: \(ar^2 + br + c = 0\).
3. Find the roots \(r_1, r_2\) and write the solution based on three cases:
  - Case 1: Real, Distinct Roots (\(b^2 - 4ac > 0\)) \(\(y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)\)
  - Case 2: Complex Conjugate Roots (\(b^2 - 4ac < 0\)), roots \(r = \alpha \pm i\beta\) \(\(y(x) = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))\)\)
  - Case 3: Real, Repeated Roots (\(b^2 - 4ac = 0\)), root \(r\) \(\(y(x) = (c_1 + c_2 x) e^{rx}\)\)

C. Reduction of Order

Use this when you have a non-constant coefficient ODE and you are given (or can guess) one solution, \(y_1(x)\).
Method:
1. Assume the second solution is of the form \(y_2(x) = u(x)y_1(x)\).
2. Find \(y'_2\) and \(y''_2\) using the product rule.
3. Substitute them into the original homogeneous ODE.
4. After simplification, you will get a simpler ODE for \(u''\) and \(u'\). Let \(w = u'\), solve the resulting first-order ODE for \(w\), then integrate \(w\) to get \(u\).

Section 4: Second-Order Linear Non-homogeneous ODEs

Form: \(y'' + p(x)y' + q(x)y = f(x)\).
General Solution Structure: \(y(x) = y_p(x) + y_{cf}(x)\).
1. First, always find the complementary function \(y_{cf}(x)\) by solving the homogeneous version.
2. Then, find one particular solution \(y_p(x)\) using one of the methods below.

A. Method of Undetermined Coefficients

When to use: Only for constant coefficient ODEs, and only when \(f(x)\) is a polynomial, exponential, sine, cosine, or a product/sum of these.
Method: "Judiciously guess" the form of \(y_p(x)\) based on \(f(x)\).
If \(f(x) = \text{poly of degree n}\) → \(y_p(x) = A_n x^n + \dots + A_0\)
If \(f(x) = C e^{kx}\) → \(y_p(x) = A e^{kx}\)
If \(f(x) = C \cos(kx) \text{ or } C \sin(kx)\) → \(y_p(x) = A \cos(kx) + B \sin(kx)\)
The Crucial Rule: If your guess for \(y_p\) contains any term that is already in your \(y_{cf}\), you must multiply your entire guess by \(x\). If it's still in \(y_{cf}\), multiply by \(x\) again.

B. Variation of Parameters (The General Method)

When to use: Works for non-constant coefficients and any function \(f(x)\). It is more powerful but often more work.
Method:
1. Find the complementary solution \(y_{cf}(x) = c_1y_1(x) + c_2y_2(x)\).
2. The ansatz for the particular solution is \(y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)\).
3. The functions \(u_1(x)\) and \(u_2(x)\) are found by solving the following system for \(u'_1\) and \(u'_2\): \(\(\begin{cases} u'_1 y_1 + u'_2 y_2 = 0 \\ u'_1 y'_1 + u'_2 y'_2 = f(x) \end{cases}\)\) (Note: \(f(x)\) is the RHS when the ODE is in standard form \(y'' + ... = f(x)\))
4. Solve this system (e.g., using Cramer's rule, which involves the Wronskian).
5. Integrate \(u'_1\) and \(u'_2\) to find \(u_1\) and \(u_2\).
6. Substitute back into the \(y_p\) ansatz.

Section 5: Key Applications

Population Dynamics:
Malthusian Model: \(\frac{dP}{dt} = rP\). Exponential growth/decay.
Logistic Model: \(\frac{dP}{dt} = rP(1 - P/K)\). Growth limited by a carrying capacity \(K\).
Mechanical Vibrations: \(my'' + cy' + ky = F(t)\).
The homogeneous solution (\(y_{cf}\)) describes the natural behavior (damped oscillations).
The particular solution (\(y_p\)) describes the long-term response to the external force \(F(t)\).
Resonance: Occurs in undamped systems (\(c=0\)) when the forcing frequency equals the natural frequency. The amplitude grows without bound.
Newton's Law of Cooling: \(\frac{dT}{dt} = -k(T - T_{ambient})\). First-order linear.
Mixing Problems: Set up a balance equation for the rate of change of a substance. Often results in a first-order linear ODE.

Final Exam Strategy

Classify First: For any problem, first determine its order, linearity, and homogeneity.
Choose the Right Tool:
- 1st Order Linear -> Integrating Factor.
- 1st Order Separable -> Separate and Integrate.
- 2nd Order Homogeneous, Constant Coeff. -> Characteristic Equation.
- 2nd Order Non-homo, Constant Coeff., nice f(x) -> Undetermined Coefficients.
- 2nd Order, Non-constant Coeff. or nasty f(x) -> Variation of Parameters.
- Given one solution to 2nd Order -> Reduction of Order.
Show Your Work: Derive results like the integrating factor or the solution to the characteristic equation. Don't just write them down.
Check the Complementary Function: The most common mistake in non-homogeneous problems is failing to check if the \(y_p\) guess is part of \(y_{cf}\).
Apply Initial Conditions Last: Find the full general solution \(y(x) = y_p + y_{cf}\) before using the initial conditions to find the constants \(c_1\) and \(c_2\).