First-order linear equation

To recognize it, write it in standard form: the coefficient of \(y'\) must be 1. So \(x\,y'+2y=x^2\) is divided by \(x\) and becomes \(y'+(2/x)\,y=x\), with \(p=2/x\) and \(q=x\). Moving terms across changes the sign: \(y'=3y+x^2\) is equivalent to \(y'-3y=x^2\), with \(p=-3\).

If \(q=0\) the equation is called homogeneous, \(y'+p(x)\,y=0\), and it is separable: its solution is \(y=Ce^{-\int p\,dx}\). For example, \(y'+2xy=0\) gives \(y=Ce^{-x^2}\). If \(q\ne0\), the equation is non-homogeneous and its general solution has the form "one particular solution plus the homogeneous solution". For \(y'+y=x\) the particular solution is \(x-1\), and the general one is \(x-1+Ce^{-x}\).

Linearity has a key consequence: solutions of the homogeneous equation can be added and multiplied by constants and remain solutions. In the non-homogeneous case this fails: if \(y_p\) solves \(y'+y=1\), then \(2y_p\) solves \(y'+y=2\), not the same equation. The difference of two solutions of the non-homogeneous equation is a solution of the homogeneous one.

When \(p\) and \(q\) are continuous on an interval, the existence and uniqueness theorem guarantees a single solution for each initial condition, defined on the whole interval. That sets linear equations apart from nonlinear ones such as \(y'=y^2\). A nonzero \(q\) is not handled by separating variables: the general method for that case is the integrating factor.