Integrating factor

An integrating factor is a function $\mu (x)$ that you multiply through a first-order linear ODE to make the left-hand side become an exact derivative. Once it’s an exact derivative, you can integrate directly.

For the standard-form first-order linear ODE:

$\frac{dy}{dx}+P(x)y=Q(x)$

the integrating factor is

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

Multiplying both sides by $\mu (x)$:

$\mu \frac{dy}{dx}+\mu Py=\mu Q$

The left side is exactly $\frac{d}{dx}[\mu (x)y(x)]$ by the product rule (since ${\mu }^{\prime }=\mu P$). So:

$\frac{d}{dx}[\mu (x)y(x)]=\mu (x)Q(x)$

Integrate both sides:

$\mu (x)y(x)=\int \mu (x)Q(x)dx+C$

Solve for $y$:

$y(x)=\frac{1}{\mu (x)}\left[\int \mu (x)Q(x)dx+C\right]$

Or, expanded:

$y(x)=e^{-\int P(x)dx}\left[\int Q(x)e^{\int P(x)dx}dx+C\right]$

Why this works

The key step: choosing $\mu =e^{\int Pdx}$ makes ${\mu }^{\prime }=P\mu$. Then the left-hand side $\mu y^{\prime }+P\mu y=\mu y^{\prime }+{\mu }^{\prime }y$, which is the product rule expansion of $(\mu y{)}^{\prime }$. Once the LHS is a single derivative, the equation is trivially integrable.

The trick of “find a multiplier that makes the LHS an exact derivative” is more general — see Exact equation for the related concept where you find a function $F(x,y)$ that’s constant along solutions.

Why we need an integrating factor in the first place

Plain integration doesn’t work on $y^{\prime }+Py=Q$ because of the $Py$ term — it mixes $x$-dependence (via $P$) with $y$-dependence. The ODE isn’t separable.

The integrating factor is a clever multiplier that re-bundles the $y^{\prime }+Py$ chunk into $(\mu y{)}^{\prime }$, removing that mixing.

Worked example 1

$y^{\prime }+y=1$, with $P=1$, $Q=1$.

Integrating factor: $\mu =e^{\int 1dx}=e^x$.

Multiply through:

$e^x{y}^{\prime }+e^{x}y=e^x$

Left side is $\frac{d}{dx}(e^{x}y)$. So:

$\frac{d}{dx}(e^{x}y)=e^x$

Integrate: $e^{x}y=e^x+C$, hence $y=1+C{e}^{-x}$.

Worked example 2

$y^{\prime }+\frac{3}{x}y=3x-2$, $y(1)=1$.

$P(x)=3/x$, $Q(x)=3x-2$.

Integrating factor: $\mu =e^{\int 3/xdx}=e^{3\ln x}=x^3$.

Multiply through:

$x^3{y}^{\prime }+3x^{2}y=x^3(3x-2)=3x^4-2x^3$

Left side is $\frac{d}{dx}(x^{3}y)$:

$\frac{d}{dx}(x^{3}y)=3x^4-2x^3$

Integrate:

$x^{3}y=\frac{3}{5}{x}^5-\frac{1}{2}{x}^4+C$

Solve for $y$:

$y(x)=\frac{3}{5}{x}^2-\frac{1}{2}x+\frac{C}{x^3}$

Apply initial condition $y(1)=1$:

$1=\frac{3}{5}-\frac{1}{2}+C\Rightarrow C=\frac{9}{10}$

Final:

$y(x)=\frac{3}{5}{x}^2-\frac{1}{2}x+\frac{9}{10x^3}$

When to use it

The integrating factor method works for any first-order linear ODE — that’s its strength. It’s the universal hammer for first-order linear.

It does not work for nonlinear first-order ODEs (no useful $\mu$ exists), and it’s overkill for separable equations (where direct separation is faster).

For first-order ODEs that aren’t linear but are exact, see Exact equation. For higher-order linear ODEs, see Characteristic equation (constant coefficients) and Method of variation of parameters (variable coefficients).