In the world of differential equations, the function y and its derivatives are treated as variables, while expressions in x are often manipulated as though they were constants. With this in mind, you can probably guess what a linear equation is.
An nth order linear differential equation sets a linear combination of derivatives, from the nth derivative down to y itself, equal to a function r(x). The coefficients on the derivatives are functions of x. In other words, the equation has degree 1. Here is a third order linear equation.
xy′′′ + (sin(x)+x2)y′′ - log(x)y′ + 3y = x2-1
We usually assume functions are continuous in the interval of interest, and the lead function is nonzero throughout, so we can divide through by this function, whence the lead coefficient becomes 1.
When r(x), i.e. the expression on the right hand side, is 0, the set of solutions forms a vector space. Add two solutions together or scale a solution by a constant to find another solution.
When the right side is zero we have a homogeneous linear equation, which provides the "homogeneous solutions".
Depending on the equation, this may be a good way to proceed. If r(x) is nonzero, set it to 0 and solve the homogeneous equation first. Then find one solution, any solution, for r(x), and add it to the homogeneous solution set. This produces a shifted vector space.
Linear equations sound innocent enough, but they can be very complicated, and we're not going to analyze them all on this website. However, there is a general solution to the first order linear differential equation.
y′ + p(x)y = q(x)
Take the integral of -p, and raise E to that power. Call this function s(x). Note that s is everywhere positive.
Divide y by s, and call the result z. If we know y, we know z, and if we know z, we know y. As we shall see, it is easier to solve for z. Substitute y = zs and obtain the following.
(zs)′ + pzs = q
z′s - pzs + pzs = q
z′ = q/s
z′ = q×E∫p
This works because the derivative of the integral of p gives p back again. That is always the case when p is continuous.
There is always a constant when integrating z′, set by the initial conditions. Let's do a simple example.
y′ + y = 2
z′ = 2Ex
z = c + 2Ex
y = (c + 2Ex) × E-x
y = cE-x + 2
If y(0) = 7, then c = 5.
This procedure remains valid for complex functions, provided p is analytic. This is a stronger constraint than continuity, but we need it to assure complex integration and differentiation. Thus s(x) can be constructed, and it behaves as expected under differentiation, and it is everywhere nonzero, so z = y/s is well defined.