Foundations — First-order linear ODEs — integrating factor method (derivation)

Before you can follow the parent derivation, every symbol on that page must mean something concrete to you. Below we build each one from absolute zero, in the order they depend on each other — no symbol is used before its own section.

1. The symbols $x$ and $y$, and what a "function of $x$" is

Picture a curve drawn on a page. The horizontal axis is $x$, the vertical axis is $y$. As you slide $x$, your finger traces up and down along the curve — that trace is the function.

Why the topic needs it: an ODE is an equation whose unknown is a whole curve $y(x)$, not a single number. Solving it means finding which curve fits.

2. The derivative $\frac{dy}{dx}$ — steepness of the curve

Here is the first piece of genuinely new notation. Read $\frac{dy}{dx}$ out loud as "dee-y dee-x", and understand it as a single object meaning "how fast $y$ climbs as $x$ moves."

Why "derivative" and not just "slope of a line"? A curve's steepness changes from point to point. The derivative is the tool that gives you the steepness at every $x$ separately — it is itself a function of $x$. This is exactly why an ODE can hold at every point along the curve.

3. The coefficients $P(x)$ and $Q(x)$

Picture two extra dials on a machine, each labelled with a number that changes as you slide $x$. At $x=2$, maybe $P=3$ and $Q=e^2$; at $x=5$, different numbers. They are given to you — your job is to find $y$.

Now the parent equation reads, in full, $\frac{dy}{dx}+P(x)y=Q(x),$ and every symbol in it — $\frac{dy}{dx}$ (section 2), $y$ (section 1), $P,Q$ (here) — has now been built.

Why the topic needs the "$x$ only" restriction: if $P$ contained $y$ (say $P=y^2$), the equation would be nonlinear and the product-rule trick collapses. The whole method leans on $P$ and $Q$ being blind to $y$.

4. What "linear" means — the exact power rule

Left side of the figure: allowed shapes — $y$, $\frac{dy}{dx}$, times a function of $x$, added up. Right side: forbidden shapes — $y^2$, $y\frac{dy}{dx}$, $\sin y$.

Why the topic needs it: "linear" is precisely the class of equations for which the whole left side can later be absorbed into a single derivative. It is the entry ticket.

5. The integral sign $\int$ — undoing the derivative

The figure shows the round trip: a curve $\to$ (derivative) $\to$ its slope $\to$ (integral) $\to$ back to the curve, plus a possible vertical shift (the accent-red arrow) — because sliding a curve up or down doesn't change its slope.

Why the topic needs it: the final move of the whole method is "integrate both sides." That $+C$ is exactly the freedom that lets one ODE describe a whole family of curves, pinned down only by an initial condition. See Separable ODEs for where this backwards-machine gets used inside the derivation that follows.

6. The exponential $e^{(\cdots )}$ and its inverse $\ln$

The figure: $e^x$ and $\ln x$ are mirror images across the line $y=x$ (accent red). Feeding one into the other lands you back where you started — that mirror is what "inverse" means.

Why the topic needs it: the special function we are about to build turns out to be an exponential, and simplifying it always means cancelling an $e$ against a $\ln$. Without this you get stuck at $e^{\ln |x|}$ and can't finish.

7. The product rule — the engine of the whole trick

This is the single most important prerequisite. Study Product Rule until it is automatic.

8. Building the special multiplier $\mu (x)$

Everything above now lets us define the one new object the parent note calls the integrating factor.

Multiply the standard equation $\frac{dy}{dx}+Py=Q$ by an as-yet-unknown $\mu$: $\mu \frac{dy}{dx}+\mu Py=\mu Q.$ Now set $u=\mu$ and $v=y$ in the product rule from section 7: $\frac{d}{dx}(\mu y)=\mu \frac{dy}{dx}+\frac{d\mu }{dx}y.$

Solving $\frac{d\mu }{dx}=\mu P$ by separation of variables

This little equation is itself solvable — it is a separable ODE for $\mu$. Here "separate the variables" means: get everything involving $\mu$ on one side and everything involving $x$ on the other.

Step 1 — divide by $\mu$ (WHAT & WHY): to peel $\mu$ off the right, divide both sides by $\mu$, moving all $\mu$-stuff left: $\frac{1}{\mu }\frac{d\mu }{dx}=P(x).$

Step 2 — multiply by $dx$ (WHAT & WHY): to line each side up under its own integral, gather the $\mu$-differential on the left and the $x$-differential on the right: $\frac{d\mu }{\mu }=P(x)dx.$

Step 3 — integrate both sides (WHAT & WHY): now each side is a pure "reverse-slope" problem (section 5). The left is the classic integral whose answer is a logarithm, $\int \frac{d\mu }{\mu }=\ln |\mu |$; the right is $\int Pdx$: $\ln |\mu |=\int P(x)dx+C_0.$ This is why integrating $P$ produces the exponent — the $\int Pdx$ is literally what comes out of the right-hand side, and it lands inside a logarithm.

Step 4 — undo the logarithm (WHAT & WHY): to free $\mu$ from inside $\ln$, apply its inverse $e^{(\cdots )}$ (section 6) to both sides: $|\mu |=e^{\int Pdx+C_0}=e^{C_0}{e}^{\int Pdx}.$

Why the topic needs it: without recognising the product rule running in reverse, the collapse $\mu \frac{dy}{dx}+\mu Py=\frac{d}{dx}(\mu y)$ looks like magic. With it, it is inevitable.

How the foundations feed the topic

Read it top-down: the notions of variable and derivative feed both the meaning of "linear" and the product rule; the product rule plus the exponential plus integration build the integrating factor $\mu$; and $\mu$ together with integration deliver the method itself.

Equipment checklist

Cover the right side and test yourself. If any answer is fuzzy, revisit its section above before opening the parent note.

Connections

• Product Rule — the engine that turns $\mu \frac{dy}{dx}+\mu Py$ into $\frac{d}{dx}(\mu y)$.
• Separable ODEs — the method used inside the derivation to solve $\frac{d\mu }{dx}=\mu P$.
• Exact ODEs — integrating factors reappear there to make equations exact.
• Bernoulli Equations — become linear (this topic) after a substitution.
• Linear Constant-Coefficient ODEs — the special case where $P,Q$ are constants.