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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).

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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.