Question

Let Dstand for $\mathit{d}\mathbf{/}\mathit{d}\mathit{x}$, that is, $\mathit{D}\mathit{y}\mathbf{=}\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}$; then

${\mathit{D}}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathit{D}\mathbf{\left(}\mathit{D}\mathit{y}\mathbf{\right)}\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{,}{\mathit{D}}^{\mathbf{3}}\mathit{y}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{3}}}\mathbf{,}\mathbf{\text{etc.}}$

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,

$\mathit{D}\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathit{y}\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathbf{x}\mathbf{y}\mathbf{)}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathit{x}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{y}$

So, we say that

$\mathit{D}\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathit{D}\mathbf{+}\mathbf{1}$

In a similar way show that:

(a) $\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{=}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}\mathbf{(}\mathit{b}\mathbf{+}\mathit{a}\mathbf{)}\mathit{D}\mathbf{+}\mathit{a}\mathit{b}$For constantand.

(b).${\mathit{D}}^{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{(}\mathit{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\mathbf{+}\mathbf{1}\mathbf{)}$

(c)$\mathit{D}\mathit{x}\mathbf{=}\mathit{x}\mathit{D}\mathbf{+}\mathbf{1}$. (Note thatDand xdo not commute, that is,$\mathit{D}\mathit{x}\mathbf{\ne }\mathit{x}\mathit{D}$.)

(d),$\mathbf{(}\mathit{D}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$but.$\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$

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

$\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$

Short answer

a) It isproved that.$(D-a)(D-b)=(D-b)(D-a)=D^2-(b+a)D+ab$

b)It isproved that.$(D^3+1)=(D+1)(D^2-D+1)$

c)It is proved that.$Dx=x\frac{dy}{dx}+1$

d)It is proved that.$(D-x)(D+x)=D^2-x^2+1$

Step-by-step solution

Given information from question 

Given information are

$\begin{array}{l}D=\frac{d}{dx}\\ Dy=\frac{dy}{dx}\\ D^{2}y=\frac{d}{dx}\left(\frac{dy}{dx}\right)\end{array}$

It is given that

$\begin{array}{c}(D-a)(D-b)=(D-b)(D-a)\\ =D^2-(b+a)D+ab\end{array}$,here.$D=\frac{d}{dx}$

Algebraic equation 

Use the algebraic equation:

$\mathbf{(}{\mathit{a}}^{\mathbf{3}}\mathbf{+}{\mathit{b}}^{\mathbf{3}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}\mathbf{a}\mathbf{b}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{)}$

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

a)

Let us consider:

$\begin{array}{c}(D-a)(D-b)\\ =D(D-b)-a(D-b)\\ =\frac{d}{dx}(\frac{dy}{dx}-by)-a(\frac{d}{dx}-by)\\ =\frac{d^{2}y}{d{x}^2}-b\frac{dy}{dx}-a\frac{dy}{dx}-aby\end{array}$

Further solve the equation

$\begin{array}{c}=\frac{d^{2}y}{d{x}^2}-(b+a)\frac{dy}{dx}-aby\\ =D^2-(b+a)D+ab\end{array}$

Similarly,

$\begin{array}{c}(D-b)(D-a)\\ =D(D-a)-b(D-a)\\ =\frac{d}{dx}(\frac{dy}{dx}-ay)-b(\frac{d}{dx}-ay)\\ =\frac{d^{2}y}{d{x}^2}-a\frac{dy}{dx}-b\frac{dy}{dx}-aby\end{array}$

Solve further

=$D^2-(b+a)D+ab$

Hence, it is proved

Prove D3+1=(D+1)(D2−D+1)

(b)

Let us consider:

$\begin{array}{c}(D^3+1)=(D+1)(D^2-D+1)\\ =D(D^2-D+1)+1(D^2-D+1)\\ =\frac{d}{dx}(\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}+y)+1(\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}+y)\\ =\frac{d^{3}y}{d{x}^3}-\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}+y\end{array}$

Further solve the equation

$\begin{array}{c}=\frac{d^{3}y}{d{x}^3}+0+0+y\\ =\frac{d^{3}y}{d{x}^3}+y\\ =D^3+1\end{array}$

ProveDx=xD+1

(c)

Let us consider:

$\begin{array}{c}Dx=\frac{d}{dx}\left(xy\right)\\ =x\frac{dy}{dx}+\frac{dx}{dx}y\\ =x\frac{dy}{dx}+\left(1\right)y\end{array}$

Solve further,

$\begin{array}{c}=x\frac{dy}{dx}+y\\ =x\frac{dy}{dx}+1\end{array}$

Prove (D−x)(D+x)=D2−x2+1 

(d)

Let us consider:

$\begin{array}{c}(D-x)(D+x)=D(D+x)-x(D+x)\\ =\frac{d}{dx}(\frac{dy}{dx}+xy)+x(\frac{dy}{dx}+xy)\\ =\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y-x\frac{dy}{dx}-x^{2}y\end{array}$

Solving further,

$\begin{array}{c}=\frac{d^{2}y}{d{x}^2}+y-x^{2}y\\ =\frac{d^{2}y}{d{x}^2}+1-x^2\\ =D^2+1-x^2\\ =D^2-x^2+1\end{array}$