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91Ó°ÊÓ

Let Dstand for d/dx, that is, Dy=dy/dx; then

D2y=D(Dy)=ddx(dydx)=d2ydx2, D3y=d3ydx3,etc.

D(or an expression involving D) is called a differential operator. Two operators are equal if they give the same results when they operate on yFor example,

D(D+x)y=ddx(dydx+xy)=d2ydx2+xdydx+y=(D2+xD+1)y

So, we say that

D(D+x)=D2+xD+1

In a similar way show that:

(a) (D−a)(D−b)=(D−b)(D−a)=D2−(b+a)D+abFor constantand.

(b).D3+1=(D+1)(D2−D+1)

(c)Dx=xD+1. (Note thatDand xdo not commute, that is,Dx≠xD.)

(d),(D−x)(D+x)=D2−x2+1but.(D+x)(D−x)=D2−x2−1

Comment: The operator equations in (c) and (d) are useful in quantum mechanics; see Chapter 12, Section 22.

(D+x)(D−x)=D2−x2−1

Short Answer

Expert verified

a) It isproved that.(D−a)(D−b)=(D−b)(D−a)=D2−(b+a)D+ab

b)It isproved that.(D3+1)=(D+1)(D2−D+1)

c)It is proved that.Dx=xdydx+1

d)It is proved that.(D−x)(D+x)=D2−x2+1

Step by step solution

01

Given information from question 

Given information are

D=ddxDy=dydxD2y=ddx(dydx)

It is given that

(D−a)(D−b)=(D−b)(D−a)=D2−(b+a)D+ab,here.D=ddx

02

Algebraic equation 

Use the algebraic equation:

(a3+b3)=(a+b)(a2−ab+b2)

03

Prove(D−a)(D−b)=(D−b)(D−a)=D2−(b+a)D+ab

a)

Let us consider:

(D−a)(D−b)=D(D−b)−a(D−b)=ddx(dydx−by)−a(ddx−by)=d2ydx2−bdydx−adydx−aby

Further solve the equation

=d2ydx2−(b+a)dydx−aby=D2−(b+a)D+ab

Similarly,

(D−b)(D−a)=D(D−a)−b(D−a)=ddx(dydx−ay)−b(ddx−ay)=d2ydx2−adydx−bdydx−aby

Solve further

=D2−(b+a)D+ab

Hence, it is proved

04

Prove D3+1=(D+1)(D2−D+1)

(b)

Let us consider:

(D3+1)=(D+1)(D2−D+1)=D(D2−D+1)+1(D2−D+1)=ddx(d2ydx2−dydx+y)+1(d2ydx2−dydx+y)=d3ydx3−d2ydx2+dydx+d2ydx2−dydx+y

Further solve the equation

=d3ydx3+0+0+y=d3ydx3+y=D3+1

05

ProveDx=xD+1

(c)

Let us consider:

Dx=ddx(xy)=xdydx+dxdxy=xdydx+(1)y

Solve further,

=xdydx+y=xdydx+1

06

Prove (D−x)(D+x)=D2−x2+1 

(d)

Let us consider:

(D−x)(D+x)=D(D+x)−x(D+x)=ddx(dydx+xy)+x(dydx+xy)=d2ydx2+xdydx+y−xdydx−x2y

Solving further,

=d2ydx2+y−x2y=d2ydx2+1−x2=D2+1−x2=D2−x2+1

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