In the last section, we have studied how to solve homogeneous constant coefficients linear differential equation. We now consider a nonhomogeneous constant coefficients linear differential equation. Say, differential equation one, which is a_sub_n times nth derivative of y plus a_sub_(n-1) times (n-1) derivative of y, plus a_sub_1 times y prime plus a_sub_0 times y, which is equal to g(x), where the all the coefficient a_sub_i are real constants, and we assume that the leading coefficient a_sub_n is not equal to zero. Then we know that the general solution of such differential equation, such a nonhomogeneous differential equation, can be expressed as a sum of y_sub_c plus y_sub_p, where the complementary solution y_sub_c is a general solution of corresponding homogeneous differential equation. Say a_sub_n times the nth derivative of y, plus a_sub_(n-1) times (n-1) derivative of y, plus a_sub_1 times y prime, plus a_sub_0 times y which is equal to zero. And y_sub_p is any particular solution of original nonhomogeneous differential equation one. In section 4.3, we have studied how to find the complementary solution y_sub_c. So, we now consider the problem of finding a particular solution y_sub_p of the given nonhomogeneous differential equation. The original differential equation we are considering is differential equation one which is a_sub_n times nth derivative of y, a_sub_(n-1) times (n-1) derivative of y, and plus a_sub_1 times y prime, plus a_sub_0 times y, and that is equal to g(x). So with a notation capital D is equal to d over dx and the family of constant a_sub_i, we call any expression of the form P of capital D which is equal to a_sub_n times D to the n, plus a_sub_(n-1) times D to the n-1, plus a_sub_1 times D plus a_sub_0 are differential polynomial. Any such expression with the constant coefficient is a differential polynomial. Then it's easy to see that any two differential polynomials P(D) and the Q(D), they commute. In other words, P(D) times Q(D) acting on y is equal to Q(D) times P(D) acting on y. We can switch the order of operation to differential operators P(D) and Q(D). We can exchange them. From P(D)Q(D) into Q(D)P(D). As a very simple example, let's consider the (D-2) times (D+1) of y. What is it? (D-2) times and the (D+1) of y, forced acting (D+1) on y, then you will get (D-2), D y means y prime and plus one times y that is y. Apply (D-2) again, then apply D on this expression. You will get y double prime plus y prime minus 2 times of this is minus 2 y prime, minus 2 y. So, finally, we'll get y double prime minus y prime minus 2 y, which is the same as this D squared minus D minus 2 times of y. On the other hand, if you exchange the order of operations for these two differential polynomials, then you will get d plus one times d minus two. On the other hand, (D+1) times (D-2) of y which is equal to (D+1) and D acting on y is the y prime and minus two times of y, that is 2 y. Now apply D again, then you will get y double prime minus 2 y prime plus one times of this will be y prime minus 2 y, which is equal to y double prime minus y prime minus 2 y, which is exactly equal to this expression. So that we confirm that these two. (D-2) times (D+1) is the same as (D+1) times (D-2). In fact, they are equal to this D squared - D - 2. But be careful, such a commutativity then not be true in general. For example, if the differential operator has a variable coefficient then you cannot exchange the order of those two differential operators. As a very simple example, consider the x D and D acting on y, you will get x of D and y prime that is equal to x y double prime. However, if you exchanged the order of these two, then you will get D times x D of y. So that is equal to D of x y prime. Then, this is equal to the by the product rule, x y double prime and plus y prime. So that these two things are not the same thing. In other words, you cannot exchange the order of these two operations, x D times D is not equal to D times x of D. It's because the one of the operator x D has variable coefficient.
