Let A, B and C represent three linear differential operators with constant coefficients; for example,

$$\begin{gathered} \mathbf{A}\mathbf{:}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{,}\mathbf{B}\mathbf{:}\mathbf{=}{\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{,} \\ \mathbf{C}\mathbf{:}\mathbf{=}{\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\mathbf{,} \end{gathered}$$

Where the a’s, b’s, and c’s are constants. Verify the following properties:

(a) Commutative laws:

$$\begin{gathered} \mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{B}\mathbf{}\mathbf{+}\mathbf{}\mathit{A}\mathbf{,}\mathbf{} \\ \mathit{A}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{B}\mathit{A} \end{gathered}$$

(b)Associative laws:

$$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,} \\ \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.} \end{gathered}$$

(c)Distributive law:$\mathbf{A}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{AB}\mathbf{+}\mathbf{AC}$

Commutative laws:

$$\begin{gathered} \mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}\mathbf{B}\mathbf{+}\mathbf{A}\mathbf{,} \\ \mathbf{AB}\mathbf{=}\mathbf{BA}\mathbf{.} \end{gathered}$$

Associative laws:

$$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,} \\ \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.} \end{gathered}$$

Distributive law:

$\mathbf{A}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{AB}\mathbf{+}\mathbf{AC}$.

Given that,

$\mathbf{A}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}$ …… (1)

$\mathbf{B}\mathbf{=}{\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}$…… (2)

$\mathbf{C}\mathbf{=}{\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}$…… (3)

Let us prove the commutative property.

Case (1):

Then, find the L.H.S.

$$\begin{gathered} \mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}} \\ \mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right) \end{gathered}$$

Now, R.H.S.

$\mathbf{B}\mathbf{+}\mathbf{A}\mathbf{=}\left({\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)$

.So, A + B = B + A.

Case (2):

$$\begin{gathered} \mathbf{AB}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right) \\ \mathbf{BA}\mathbf{=}\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right) \end{gathered}$$

Therefore, AB = BA

To prove:

$$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,} \\ \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.} \end{gathered}$$

Case (1):

$$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \\ \mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \end{gathered}$$

Henceforth, (A + B) + C = A + (B + C).

Case (2):

$$\begin{gathered} \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\left({\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \\ \mathbf{A}\left(\mathbf{BC}\right)\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\left({\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \end{gathered}$$

Hence, (AB) C = A (BC).

To prove:$\mathbf{A}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{AB}\mathbf{+}\mathbf{AC}$ .

Then,

$$\begin{gathered} \mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \\ \mathbf{AB}\mathbf{+}\mathbf{AC}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \end{gathered}$$

Consequently, A (B + C) = AB + AC.