In the last section, we have studied how to solve nonhomogeneous constant coefficients linear differential equation. We now, consider nonhomogeneous constant coefficients, linear differential equation, say differential equation 1. It is a7 times and the derivative of y plus a7 minus 1 times n minus 1 derivative of y plus a sub 1 times y prime plus a sub 0 times y, which is a call to g of x, where the older coefficient as of I are real constants. And if we assume that the leading coefficient as event is not equal to 0. Then we know that the general solution of such a differential equation such a non homogeneous differential equation can be expressed as the sum of wise yc plus wiser p. Where the complementary solution wise of c is a general solution of corresponding homogeneous differential equation. Say, a7 times the ends that vertical y plus ace of n minus 1 times n minus 1 derivative of y plus ace a1 times y prime plus a sub 0 times y, which is equal to 0, okay? And y of p is there any particular solution of or isn't it nonhomogeneous differential equation 1, okay? In section 4.3, we have started how to find the complementary solution wise of c. So we now consider the problem of finding a particular solution wiser p of the given nonhomogeneous differential equation. The original differential equation we are considering is differential equation of 1 which is a7 times n still evocative ya7 minus 1 times n minus 1 study veritable y and plus ace a1 times y prime cross a sub 0 times y, and that is equal to g of x, right? So with the notation capital, this equal to deal with the ex and the family of a constant ace of I. We call any expression of the form p of capital d, which is equal to 87 times d to the n plus ace of n minus 1 times d to the n minus 1 plus ac1 times d plus ac1 of 0, a differential polynomial. Any such expression? And with the constant coefficients is the differential polynomial, and it's easy to see that any two differential polynomial pod, and the q of d commute, okay? In other words, p of d times q of d acting on y is equal to q of d times p of d acting on y, okay? We can switch the order of operation to differential pointers p of d and q of d, we can exchange them, okay? From pdqd into qdpd as a very simple example, let's consider the d minus 2 times d plus 1 over y, okay? What is it? D minus 2 times and the d plus 1 of y, right? First acting d plus 1 on y, then you will get d minus 2, dy means y prime plus 1 times y, that is y, okay? Apply d minus 2 again, then apply d on this expression, you will get y double prime, okay? Plus y prime minus 2 times of this is minus 2y prime minus 2y. So finally, we get y double prime minus y prime minus 2y, which is the same as this k minus d minus 2 times of y, okay? On the other hand, if you exchange the order of operation for these two differential polynomial is then you will get the plus 1 times d minus 2, right? On the other hand, d plus 1 times d minus 2y which is equal to d plus 1 and the acting on y is the y prime minus 2 times y and that is 2y. Now, apply deal again, then you will get y double prime minus 2y prime plus 1 times of this will be y prime minus 2y which is equal to y double prime minus y prime minus 2y, which is exactly equal to the this expression, right? So that we confirm that, okay? This too, right? D minus 2 times d plus 1 is the same as d plus 1 times d minus 2. In fact, they are equal to this care minus 1 minus 2, but be careful such a comment activity did not be true in general. For example, if the differential operator has a value of the coefficient, then you cannot exchange the order of those two differential operators. As a very simple example, consider the xd and the d, right? Acting on y, you will get x of d and the y prime that is equal to xy double prime, right? However, if we exchange the order of these two, then you will get d times x of d of y, so that is equal to d of xy prime. Then this is equal to buy the product rule xy double prime, and plus ny prime, okay? So that these two things are not the same thing, right? In other words, you cannot exchange the order of these two operations x of d times d it's not equal to d times x of d, right? It's because the one of the operator x of d has a variable coefficient, okay?
