An nth-order linear differential equation has the form bn(x)y(n)+bn−1(x)y(n−1)+…+b2(x)y′′+b1(x)y′+b0(x)y=g(x), where g(x) and all coefficients bj(x),j=0..n depend solely on the variable x. In other words, they do not depend on y or on any derivative of y.
If g(x)≡0, the differential equation is called homogeneous; otherwise, it is called non-homogeneous.
For example, x3y′′′+x1y′′+2y′=0 is homogeneous, and y′′+xy′=cos(x) is non-homogeneous.
A linear differential equation has constant coefficients, if all the coefficients bj(x) are constants; if one or more of these coefficients are not constant, the differential equation has variable coefficients.
For example, y(4)+x2y′+xy=6 has variable coefficients, and 3y′′+y′=ex has constant coefficients.
Theorem. Consider an initial value problem given by a linear differential equation and n initial conditions y(x0)=y0, y′(x0)=y0′, y′′(x0)=y0′′, ..., y(n−1)(x0)=y0n−1. If g(x) and bj(x),j=0..n are continuous in some interval I containing x0 and if bn(x)=0 in I, the given differential equation together with the initial conditions has a unique (only one) solution defined throughout I.
When the conditions on bn(x) in the theorem hold, we can divide the differential equation by bn(x) to obtain y(n)+an−1(x)y(n−1)+…+a2(x)y′′+a1(x)y′+a0(x)y=ϕ(x), where aj(x)=bn(x)aj(x),j=0..n−1 and ϕ(x)=ϕ(x)g(x).
Now, let's define the differential operator L(y) by L(y)=y(n)+an−1(x)y(n−1)+…+a2(x)y′′+a1(x)y′+a0(x)y; then, the differential equation can be rewritten as L(y)=ϕ(x), and, in particular, the linear homogeneous differential equation can be expressed as L(y)=0.
The differential operator has two properties:
If L(y)=0, we have that L(cy)=0 for any constant c.
If L(y1)=0 and L(y2)=0, we have that L(y1+y2)=0.
These two properties can be combined into one property: if L(y1)=0, L(y2)=0, ..., L(yn)=0, we have that L(c1y1+c2y2+…+cnyn)=0 for any constants ci,i=1..n.
What does it give us? It gives us the following fact: if we have n solutions y1, y2, ..., yn that satisfy the given homogeneous differential equation, their linear combination will also satisfy this homogeneous differential equation. The only question is what form y1, y2, ..., yn should have for their linear combination to provide the general solution of the homogeneous differential equation.
Example. y1=cos(t) and y2=sin(t) are solutions of the differential equation y′′+y=0. So, yg=c1y1+c2y2=c1cos(t)+c2sin(t) is also the solution for any constants c1 and c2.