Integrating Factor Method

The idea in one sentence

The integrating factor $\mu(x)$ is a clever multiplier that turns the left side of a first-order linear ODE $\frac{dy}{dx} + P(x) \, y = Q(x)$ into an exact derivative — a single $\dfrac{d}{dx}[\mu \, y]$ — so you can integrate once and be done.

This is the same machinery as problem #174, but here we'll derive $\mu$ from scratch rather than just applying it.

Why $\mu(x) = e^{\int P\, dx}$ works

We want $\mu(x)$ to make the left side collapse: $\mu \, y' + \mu P \, y \;=\; \frac{d}{dx}\bigl[\mu \, y\bigr] \;=\; \mu \, y' + \mu' \, y$

Matching the coefficient of $y$ on both sides forces $\mu' = \mu \, P \quad\Longrightarrow\quad \frac{d\mu}{\mu} = P \, dx \quad\Longrightarrow\quad \ln \mu = \int P \, dx$ $\boxed{\, \mu(x) = \exp\!\left( \int P(x) \, dx \right) \,}$

No integration constant is needed — any non-zero $\mu$ that satisfies $\mu' = \mu P$ does the job, and the simplest choice is the one with constant of integration zero.

The given equation

$\frac{dy}{dx} + \frac{1}{x} \, y = x$

Here $P(x) = 1/x$ and $Q(x) = x$.

Step-by-step solution

Step 1 — Compute the integrating factor. $\mu(x) = e^{\int \frac{1}{x}\, dx} = e^{\ln|x|} = |x|$

Working on an interval where $x > 0$ (e.g. $x > 0$), take $\mu(x) = x$.

Step 2 — Multiply through by $\mu$. $x \cdot y' + x \cdot \frac{1}{x} \, y = x \cdot x$ $x \, y' + y = x^{2}$

Step 3 — Recognize the collapse. The left side is exactly $\dfrac{d}{dx}[x \, y]$ by the product rule: $\frac{d}{dx}[x \, y] = x \, y' + 1 \cdot y = x \, y' + y \quad\checkmark$

So the equation becomes $\frac{d}{dx}[x \, y] = x^{2}$

This is the whole point of the method — the left side is now a single derivative.

Step 4 — Integrate both sides with respect to $x$. $x \, y = \int x^{2} \, dx = \frac{x^{3}}{3} + C$

Step 5 — Solve for $y$. $\boxed{\, y(x) = \frac{x^{2}}{3} + \frac{C}{x} \,}$

Structure of the solution

Two pieces again, like every first-order linear ODE:

• Particular piece: $y_p = x^{2}/3$. One specific solution of the full equation (no free constant).
• Homogeneous piece: $y_h = C/x$. The general solution of $y' + y/x = 0$ — these are the solutions you can add without breaking the equality, because the equation is linear.

General solution = particular + homogeneous: $y = y_p + y_h$.

Verification

$y = \frac{x^{2}}{3} + \frac{C}{x} \;\Longrightarrow\; y' = \frac{2x}{3} - \frac{C}{x^{2}}$

Plug into the left side $y' + y/x$: $\left(\frac{2x}{3} - \frac{C}{x^{2}}\right) + \frac{1}{x}\left(\frac{x^{2}}{3} + \frac{C}{x}\right) = \frac{2x}{3} - \frac{C}{x^{2}} + \frac{x}{3} + \frac{C}{x^{2}} = x \quad\checkmark$

Initial value problem

Say $y(1) = 2$. Plug in: $2 = \frac{1}{3} + C \implies C = \frac{5}{3}$ $y(x) = \frac{x^{2}}{3} + \frac{5}{3 x}$

The full recipe (memorize this)

For $\dfrac{dy}{dx} + P(x) \, y = Q(x)$:

1. $\mu(x) = \exp\!\left(\displaystyle\int P(x) \, dx\right)$
2. Multiply: $\dfrac{d}{dx}[\mu \, y] = \mu(x) \, Q(x)$
3. Integrate: $\mu \, y = \displaystyle\int \mu(x) \, Q(x) \, dx + C$
4. Divide: $y(x) = \dfrac{1}{\mu(x)}\!\left[\displaystyle\int \mu(x) \, Q(x) \, dx + C\right]$

Three common choices of $P$:

• $P(x) = k$ (constant) → $\mu = e^{kx}$.
• $P(x) = 1/x$ → $\mu = x$ (or $\mu = 1/x$ for $-P$).
• $P(x) = -\tan x$ → $\mu = \cos x$ (integral of $-\tan x$ is $\ln|\cos x|$).

Knowing these by heart will speed you up on exams.

Common mistakes

• Not putting the equation in standard form first. If it's $3 y' + 6 y = 12 x$, divide by $3$ before reading $P$: $y' + 2y = 4x$, so $P = 2$, not $6$.
• Adding a constant when computing $\mu$. You don't need one — any non-zero $\mu$ works, and the clean choice has $C = 0$.
• Forgetting to divide by $\mu$ at the end. Easy to leave the answer as $\mu y = \ldots$ and call it done.
• Domain slip with $\mu = |x|$. Choose a side of $0$ and stick with it; solutions rarely cross $x = 0$ anyway because $P$ is singular there.

Try it in the visualization

Animate the multiplication $y' + \dfrac{1}{x} y \xrightarrow{\times\, x} (x y)'$: watch the left side of the equation physically collapse into a single derivative panel. Sweep the initial condition to see the $C/x$ tail change sign and magnitude.