First Order Linear Differential Equations

Definition. A first-order differential equation of the form

— where $P$ and $Q$ are functions of $x$ only — is called linear: no product of $y$ with $\dfrac{dy}{dx}$ occurs, and $y$ together with its derivative occur only to the first degree.

Deriving the integrating factor. Consider first the associated homogeneous equation $\dfrac{dy}{dx}+Py=0$. Separating variables, $\dfrac{dy}{y}=-P\,dx$; integrating gives $y\,e^{\int P\,dx}=C$. Now examine what happens on differentiating the quantity $y\,e^{\int P\,dx}$ by the product rule:

using the original (non-homogeneous) equation in the last step. So the left side is an exact derivative whenever $y$ solves the linear equation. Integrating both sides with respect to $x$ gives the general solution

The factor $e^{\int P\,dx}$ is called the integrating factor (I.F.) of the equation.

Remarks.

1. The solution is often written compactly as $y\times(I.F.)=\displaystyle\int Q\times(I.F.)\,dx+C$.
2. In the integrating factor $e^{\int P\,dx}$, $P$ is the coefficient of $y$ provided the coefficient of $\dfrac{dy}{dx}$ has first been made unity — if the equation is not already in this normalised form, divide through by the coefficient of $\dfrac{dy}{dx}$ first.
3. The mirror equation $\dfrac{dx}{dy}+Px=Q$ (with $P,Q$ functions of $y$ only, no product of $x$ with $\dfrac{dx}{dy}$, both to the first degree) is solved the same way, with the roles of $x$ and $y$ exchanged: $x\,e^{\int P\,dy}=\displaystyle\int Q\,e^{\int P\,dy}\,dy+C$.

Worked pattern (Example 10.22). $\dfrac{dy}{dx}+2y=e^{-x}$: here $P=2,\ Q=e^{-x}$, so $I.F.=e^{2x}$; the solution is $ye^{2x}=\displaystyle\int e^{-x}e^{2x}dx+C=e^{x}+C$, i.e. $y=e^{-x}+Ce^{-2x}$.

Worked pattern (Example 10.23, a trigonometric IF). After dividing through by $x$ to normalise, $P=\dfrac{\tan x}{x}-\dfrac{1}{x}$-type expressions arise; $\int P\,dx$ collapses via $\log(\cos x)$ and $\log x$ identities to give $I.F.=\dfrac{1}{x\cos x}$, and the solution follows.

Worked pattern (Example 10.24, a power-of-trig IF). $\dfrac{dy}{dx}+2y\cot x=3x^2\operatorname{cosec}^2x$: $P=2\cot x$, $\int P\,dx=2\log(\sin x)=\log(\sin^2x)$, so $I.F.=\sin^2x$; the solution is $y\sin^2x=x^3+C$. …