Linear Differential Equations of the Type dx/dy + Px = Q

Not every linear-looking equation is linear in $y$. If, when solved for a derivative, the equation instead has $x^2$, $x\dfrac{dy}{dx}$, or another non-linear combination of $x$ and $\dfrac{dy}{dx}$, it may still be linear in $x$, if it can be arranged into the standard form

where now $P$ and $Q$ are given functions of $y$ only (or constants) -- $x$ and $\dfrac{dx}{dy}$ each appearing to the first power only. This is exactly Section 6's equation with the roles of $x$ and $y$ (and hence of $\dfrac{d}{dx}$ and $\dfrac{d}{dy}$) exchanged throughout.

When to prefer this form. If, on inspection, an equation is not linear in $y$ (e.g. it contains $y^2\dfrac{dy}{dx}$, or the "constants" $P,Q$ would need to depend on $y$ to fit Section 6's form), it is always worth checking whether the reciprocal relationship $\dfrac{dx}{dy}$ is linear in $x$ instead -- this happens often, since a curve's equation can equally be read as $x$ a function of $y$.

Deriving the integrating factor. Exactly as in Section 6 -- with $y \leftrightarrow x$ swapped -- multiplying $\dfrac{dx}{dy} + Px = Q$ by a function $\mu(y)$ chosen so that $\mu\!\left(\dfrac{dx}{dy}+Px\right) = \dfrac{d}{dy}(x\mu)$ requires $\dfrac{d\mu}{dy} = \mu P$, i.e. $\dfrac{d\mu}{\mu} = P\,dy$, giving

General solution. Multiplying through by $\mu = e^{\int P\,dy}$ turns the equation into $\dfrac{d}{dy}(x\mu) = \mu Q$; integrating both sides with respect to $y$:

$x\cdot\text{I.F.} = \int (Q\cdot\text{I.F.})\,dy + C.$ …