Addition: If L and M are differential operators, then (L + M)(f(x)) = L(f(x)) + M(f(x)). For example, (D+3)(x2 + 3x + 1) = (2x + 3) + (3x2 + 9x + 3) = 3x2 + 11x + 6.
Multiplication: If L and M are differential operators, then (LM)(f(x)) = L(M(f(x))). For example, (3D)(x2 + 3x + 1) = 3(2x + 3) = 6x + 9.
We can check that with these operations we have a non-commutative ring with identity. To see that the ring is non-commutative, observe that (xD)(f(x)) = xf '(x) while (Dx)(f(x)) = f(x) + xf '(x) from the product rule, so the operators D and x don't commute. On the other hand, (3D)(f(x)) = 3f '(x) = (D3)(f(x)) so differential operators will commute with multiplication by a constant.
In Math 240, Elementary Differential Equations, we work out that the solution of (D + p(x))(y(x)) = q(x) is y(x) = exp(-p(x)) ´ òexp(p(x))q(x)dx. We next want to solve more complicated equations like (D2 + 4D + 3)(y(x)) = 0. Here D2(f(x)) means D(D(f(x))) = f "(x). Equations of this form turn up in many different applications, including electrical circuits and spring-mass systems. In secondary school algebra, we solve higher-order equations like this by factoring them into linear equations which we can solve. We can do the same thing with these differential equations. We factor D2 + 4D + 3 = (D+3)(D+1), then use the formula given in the first sentence to work with (D+3)(y(x)) = 0 and (D+1)(y(x)) = 0. But this approach breaks down in the equation (D2 + 3xD + 2x2)(y(x)) = 0. If we try to factor D2 + 3xD + 2x2 according to the usual rules of algebra, we should look for two quantities whose sum is 3x and whose product is 2x2. It is fairly easy to guess that the quantities 2x and x have these properties. But when we try to plug them into the factoring, everything falls apart. (D + 2x)(D + x)(y(x)) = (D + 2x)(y'(x) + xy(x)) = y"(x) + 3xy'(x) + (2x2 + 1)y(x), where the extra +1 in the y(x) term comes from the product rule for D(xy(x)). The problem is that our usual rules of factoring depend on the commutative law. Since the commutative law fails for differential operators, we can't factor and that makes problems like the example above much more difficult. The first problem worked out fine because all the operators were either derivatives or constants, and they commute, so our factoring methods worked. In the second problem, we have multiplication by functions, which doesn't commute with the derivative operator, so factoring fails.
This is not an example that you can show to your secondary school classes. But it is an example you should be familiar with. It illustrates two key facts about algebra in non-commutative rings. First, our rules for factoring depend upon the commutative law. Without commutativity, factoring is enormously more difficult. In fact, good rules for factoring in the non-commutative case aren't known. Secondly, it is often helpful when working with non-commutative algebraic structures to consider what elements do commute with other elements. Once we know that differentiation and multiplication by constants commute, we are able to solve "constant-coefficient" problems because we do know how to factor in that case. In more theoretical algebra classes, there are a variety of other places where we find all the elements that commute with certain elements and use knowledge of these sets to study properties of non-commutative algebraic structures.