How to Structure a Problem-Solving Attempt

Why an attempt needs structure

Most failed attempts do not fail at the hard step. They fail earlier, because the problem was misread, or later, because an answer was accepted without being checked. A fixed sequence of phases protects against both. The one used here extends George Pólya's classic four stages by splitting his final stage, looking back, into two separate jobs: checking the answer and reflecting on the method.

1. Understand: restate the problem, name its type, and predict features of the answer.
2. Plan: list candidate methods and choose one deliberately; keep a second one in reserve as a check.
3. Execute: carry out the plan, writing each step so that someone else could follow it.
4. Check: test the answer by means that do not repeat the execution.
5. Reflect: record what generalizes and what nearly went wrong.

The rest of this guide runs one problem through all five phases.

Phase 1: Understand

Name the problem as precisely as you can. This is a first-order, linear, constant-coefficient, nonhomogeneous ordinary differential equation with one initial condition. The unknown is a function $y(x)$; the data are the coefficient $3$, the forcing $\cos 2x$ and the value $y(0) = 0$. Linear first-order problems with continuous coefficients have exactly one solution through a given initial point, so you are looking for a single function, not a family.

Before solving, extract what the equation already tells you. At $x = 0$ it gives $y'(0) = 3y(0) + \cos 0 = 1$, so near the origin $y(x) \approx x$. Differentiating the equation once gives $y'' = 3y' - 2\sin 2x$, hence $y''(0) = 3$. The homogeneous equation $y' = 3y$ has the growing solution $e^{3x}$, so unless that mode is exactly cancelled you should expect exponential growth for large $x$. These predictions cost a few lines and become checks later.

Phase 2: Plan

Write down the candidate methods, not just the first one that comes to mind:

• Integrating factor. Multiply by $e^{-3x}$ to get $\left(e^{-3x}y\right)' = e^{-3x}\cos 2x$, then integrate. This needs the integral of $e^{-3x}\cos 2x$.
• Undetermined coefficients. The forcing $\cos 2x$ is not a solution of the homogeneous equation, so a trial $A\cos 2x + B\sin 2x$ will work. Only linear algebra is required.
• Laplace transform. The initial condition is at $0$ and the forcing has a standard transform, so the problem becomes algebra in $s$ followed by partial fractions.

Choose undetermined coefficients for the execution, because it involves the least integration. Keep the Laplace transform in reserve as an independent second method for the check phase. Deciding this now, rather than after an answer appears, keeps the check honest.

Phase 3: Execute

Each line of the execution states one operation and its result. That is not a matter of style: when an error appears later, a step-by-step record is the only way to find the line where it entered.

Phase 4: Check

Use checks that do not simply repeat the computation.

• Substitution. $y' = \tfrac{9}{13}e^{3x} + \tfrac{6}{13}\sin 2x + \tfrac{4}{13}\cos 2x$. Subtracting $3y = \tfrac{9}{13}e^{3x} - \tfrac{9}{13}\cos 2x + \tfrac{6}{13}\sin 2x$ leaves $\tfrac{13}{13}\cos 2x = \cos 2x$.
• Initial data and predictions. $y(0) = \tfrac{3}{13}(1 - 1) + 0 = 0$, and $y'(0) = \tfrac{9}{13} + \tfrac{4}{13} = 1$, matching the prediction from Phase 1. The coefficient of $e^{3x}$ is positive, so the solution grows, as expected.
• Second method. Transforming the equation with $y(0) = 0$ gives $(s - 3)Y(s) = s/(s^2+4)$. The partial-fraction decomposition below inverts term by term to $\tfrac{3}{13}e^{3x} - \tfrac{3}{13}\cos 2x + \tfrac{2}{13}\sin 2x$, the same function.
• Numerical spot check. The formula gives $y(1) = \tfrac{3}{13}(e^{3} - \cos 2) + \tfrac{2}{13}\sin 2 \approx 4.8710$. A fourth-order Runge–Kutta integration of the equation from $0$ to $1$ with step $10^{-4}$ produces the same value to better than $10^{-9}$.

Phase 5: Reflect

Reflection asks what this problem teaches beyond its own answer. Repeating the execution with a general rate $b$ and frequency $\omega$ (assuming $b$ and $\omega$ are not both zero) gives a reusable particular solution:

With $b = 3$ and $\omega = 2$ it reproduces the result above, which is itself a check. Two further lessons are worth recording. First, the trial function worked because $\cos 2x$ is not a homogeneous solution; if the forcing had been $e^{3x}$, the trial would have needed an extra factor of $x$. Second, the equation $y' - by = f(x)$ is exactly the form that Chapter 5 of the book treats with its Matrix Boundary Method, for forcing functions in a space closed under differentiation; the classical solution above is a useful baseline when reading that chapter.

What to write down when you are stuck

Being stuck is a normal phase of an attempt, not a sign to stop. What matters is that the time produces a record. Write down:

1. The problem in your own notationEvery given condition, the unknown, and the domain on which it lives.
2. The type and the methodName the problem class as precisely as you can and the method you tried.
3. The boundary of certaintyThe last line you are sure of and the first line you are not. Most stuck points sit between those two lines.
4. A smaller problem you can solveDrop the forcing, set the initial value to zero, or take a special parameter such as $\omega = 0$. Solving a special case often shows the structure of the general one.
5. What the answer must satisfyInitial values, signs, growth or decay, symmetry, units. In the example, $y(0) = 0$, $y'(0) = 1$ and eventual growth were known before any solving.
6. A specific questionTurn the stuck point into a question someone else can answer quickly. See How to Ask a Precise Mathematical Question.