First-Order Linear Equations

A first-order linear equation $y' + P(x)y = Q(x)$ usually can't be separated — but multiply it by the integrating factor $\mu = e^{\int P\,dx}$ and the left side collapses into $(\mu y)'$, so both sides integrate directly.

By the end you'll be able to put a first-order linear equation into standard form, compute its integrating factor $\mu = e^{\int P\,dx}$, and integrate both sides of $(\mu y)' = \mu Q$ to solve for $y$.

Predict: the integrating factor turns the left side into the derivative of what product? Step k below and watch the middle line of the pipeline to check.

This is the family $y' + \frac{k}{x}y = 4x - 3$ (for $x > 0$). Stepping k changes $P(x) = k/x$, which changes μ = x^k — and the collapsed left side and the solution curve follow. Drag C to see how the constant of integration picks one curve from the family. Hover the chart (or focus the dot at x = 1) for exact values.

k (in P(x) = k/x) = 3

constant C = 1.00

standard form y′ + (3/x)·y = 4x − 3, so μ = e^∫(3/x)dx = e^{3 ln x} = x³

multiply by μ x³y′ + 3x²y = 4x⁴ − 3x³ ⇒ (x³·y)′ = 4x⁴ − 3x³

integrate ÷ μ y = 4x²/5 − 3x/4 + 1.00·x⁻³

Solution curves — the full solution and the C = 0 member of the family

full solution (your C) C = 0 curve

A first-order linear equation $y' + P(x)y = Q(x)$ is solved by multiplying through by the integrating factor $\mu = e^{\int P\,dx}$, which turns the left side into $(\mu y)'$ so both sides can be integrated.

Algebraic

Why does this particular $\mu$ work? Because $\mu = e^{\int P\,dx}$ is the exponential of an antiderivative of $P$, the chain rule gives $\mu' = \mu P$. Multiply the standard form equation by $\mu$: $$\mu y' + \mu P y = \mu Q.$$ Substituting $\mu P = \mu'$, the left side reads $\mu y' + \mu' y$ — exactly the product-rule expansion of $(\mu y)'$. So the equation becomes $(\mu y)' = \mu Q$, and integrating both sides gives $$\mu y = \int \mu Q\,dx \quad\Rightarrow\quad y = \frac{1}{\mu}\int \mu Q\,dx.$$ You saw this collapse live in the middle line of the pipeline above — that's what stepping k re-derives.

Procedural

The recipe is the same every time. (1) Put the equation in standard form $y' + P(x)y = Q(x)$ — divide through so the $y'$ coefficient is 1. (2) Compute $\mu = e^{\int P\,dx}$. (3) Multiply through by $\mu$; the left side becomes $(\mu y)'$ by construction — if it doesn't, recheck steps 1–2. (4) Integrate both sides, keeping the constant of integration $C$. (5) Divide by $\mu$ to isolate $y$, and use an initial condition (if given) to pin down $C$. Contrast this with separable equations from the earlier lesson: there you factor $y'$ as $f(x)g(y)$ and split the variables. When $y$ and $x$ terms appear as a sum — like $Q(x) - P(x)y$ — you can't factor, and the integrating factor is the tool instead.

Worked example

Solve $xy' + 3y = 4x^2 - 3x$ for $x > 0$.

Standard form — divide by $x$: $y' + \frac{3}{x}y = 4x - 3$, so $P(x) = \frac{3}{x}$.
Integrating factor: $\int \frac{3}{x}\,dx = 3\ln x$, so $\mu = e^{3\ln x} = x^3$.
Multiply through: $x^3y' + 3x^2y = 4x^4 - 3x^3$, i.e. $\frac{d}{dx}\!\left[x^3 y\right] = 4x^4 - 3x^3$.
Integrate: $x^3 y = \frac{4x^5}{5} - \frac{3x^4}{4} + C$.
Divide by $\mu = x^3$: $y = \frac{4x^2}{5} - \frac{3x}{4} + Cx^{-3}$.

Your turn

Same method, new numbers: solve $y' + \frac{4}{x}y = x$ for $x > 0$. Here $P(x) = \frac{4}{x}$, so $\int \frac{4}{x}\,dx = 4\ln x$ and $\mu = e^{4\ln x} = $ ____. Multiplying through gives $(\mu\,y)' = x\cdot\mu$; integrate and divide by $\mu$ to finish: $y = $ ____.

Reveal the answer

$\mu = e^{4\ln x} = x^4$. Multiplying through: $(x^4 y)' = x\cdot x^4 = x^5$. Integrate: $x^4 y = \frac{x^6}{6} + C$. Divide by $x^4$: $y = \frac{x^2}{6} + Cx^{-4}$. Set k = 4 on the slider above to watch the same collapse happen for $P(x) = 4/x$ (the demo's right-hand side differs, but μ depends only on P).

More info — the terminal-velocity view of the same method

The method isn't just for $P(x) = k/x$. A falling object with air resistance satisfies $v' + 0.196v = 9.8$ with $v(0) = 48$ — a first-order linear equation with constant $P(t) = 0.196$. Then $\mu = e^{\int 0.196\,dt} = e^{0.196t}$, the collapse gives $\left(e^{0.196t}v\right)' = 9.8\,e^{0.196t}$, and integrating yields $e^{0.196t}v = 50\,e^{0.196t} + C$, so $v(t) = 50 + Ce^{-0.196t}$. The initial condition forces $C = -2$, giving $v(t) = 50 - 2e^{-0.196t}$: the object approaches a terminal velocity of 50, exactly the equilibrium where $9.8 - 0.196v = 0$. Follow the full derivation in the Paul's Online Math Notes link under Dive deeper below.

Check your understanding

Question 1 of 4

Find the integrating factor for $y' + \frac{2}{x}y = x^3$ (for $x > 0$).

Question 2 of 4

Multiplying $y' + P(x)y = Q(x)$ through by $\mu = e^{\int P\,dx}$ turns the left-hand side into which single expression?

Question 3 of 4

Which of these first-order equations requires the integrating-factor method, because it cannot be solved by separation of variables?

Question 4 of 4

A falling object satisfies $v' + 0.196v = 9.8$. Multiplying by $\mu = e^{0.196t}$ and integrating gives $v(t) = 50 + Ce^{-0.196t}$, and $v(0) = 48$ forces $C = -2$. What velocity does the object approach as $t \to \infty$?

Recap

Standard form first: $y' + P(x)y = Q(x)$, with the $y'$ coefficient equal to 1.
The integrating factor is $\mu = e^{\int P\,dx}$; by construction $\mu' = \mu P$.
Multiplying through by $\mu$ collapses the left side into $(\mu y)'$, so $(\mu y)' = \mu Q$ and $y = \frac{1}{\mu}\int \mu Q\,dx$.
Keep the constant of integration; an initial condition pins down $C$.
Choose the technique by structure: a product $f(x)g(y)$ separates; a sum like $Q(x) - P(x)y$ is linear — use the integrating factor.

Dive deeper

Linear Differential Equations — Paul's Online Math Notes Follow the integrating-factor derivation and the terminal-velocity example.
OpenStax Calculus Volume 2, §4.5 First-order Linear Equations Work through the xy' + 3y = 4x^2 - 3x integrating-factor example.

Sources

First-order linear differential equations and the integrating factor