Why the integrating factor method uses dx in first-order linear differential equations

Key concept: This is one of those spots where notation looks more mysterious than it really is. The key idea is that $y$ is still a function of $x$, so once you divide by $y(x)$, the left side becomes a derivative with respect to $x$ that can be integrated in the usual way.

Step by step

For the homogeneous equation

$y'(x)=p(x)y(x),$

you can divide by $y(x)$ only where $y(x)\neq 0$:

$\frac{y'(x)}{y(x)}=p(x).$

Now notice what $\frac{y'(x)}{y(x)}$ really is. Since $y$ depends on $x$,

$\frac{d}{dx}\big(\ln|y(x)|\big)=\frac{y'(x)}{y(x)}.$

So integrating with respect to $x$ gives

$\int \frac{y'(x)}{y(x)}\,dx=\int p(x)\,dx,$

which becomes

$\ln|y(x)|=Q(x)+C.$

Exponentiating:

$|y(x)|=e^{Q(x)+C}=e^C e^{Q(x)}.$

Since $e^C>0$, the arbitrary constant can absorb the sign, so we write

$y(x)=A e^{Q(x)}, \qquad A\in\mathbb R.$

Why this is allowed

The integral is with respect to $x$ because the whole expression is already a function of $x$. You are not pretending $y$ is independent of $x$; you are using the chain rule in reverse.

Why separable equations often use $dy$

For a separable equation like

$y'=f(x)g(y),$

we usually rewrite it as

$\frac{1}{g(y)}\,dy=f(x)\,dx$

because that is the cleanest way to separate the variables.

If you instead wrote

$\frac{y'}{g(y)}=f(x),$

and integrated both sides with respect to $x$, that is still valid if you treat the left side as

$\int \frac{y'(x)}{g(y(x))}\,dx.$

Then you can use the substitution $u=y(x)$, $du=y'(x)dx$, which gives

$\int \frac{1}{g(u)}\,du.$

So the $dy$ notation is not a different rule; it is just a shortcut for substitution.

The real rule

You may integrate with respect to $x$ or $y$, but the differential has to match the variable you are integrating over. The common separable-equation form

$\frac{1}{g(y)}\,dy=f(x)\,dx$

is just the shortest way to write that substitution step.

Pitfall alert

A common mistake is to treat $\frac{y'}{g(y)}$ as if it were already a function of $x$ only and then integrate without substitution. That can hide the chain rule and lead to confusion. Another trap is dividing by $y$ or $g(y)$ without checking whether they can be zero; you may lose constant solutions if you skip that check.

Try different conditions

If the equation is written as

$y'=f(x)g(y),$

you can still integrate with respect to $x$:

$\int \frac{y'(x)}{g(y(x))}\,dx=\int f(x)\,dx.$

Using $u=y(x)$, the left side becomes

$\int \frac{1}{g(u)}\,du.$

So both approaches are equivalent. The difference is mostly notation: $dy$ is the substitution result, while $dx$ keeps everything in one variable and makes the chain rule visible.

Further reading

integrating factor, separable differential equation, chain rule