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Found in: Page 251

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Let A, B and C represent three linear differential operators with constant coefficients; for example,${\mathbf{A}}{\mathbf{:}}{\mathbf{=}}{{\mathbf{a}}}_{{\mathbf{2}}}{{\mathbf{D}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{a}}}_{{\mathbf{1}}}{\mathbf{D}}{\mathbf{+}}{{\mathbf{a}}}_{{\mathbf{0}}}{\mathbf{,}}{\mathbf{B}}{\mathbf{:}}{\mathbf{=}}{{\mathbf{b}}}_{{\mathbf{2}}}{{\mathbf{D}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{b}}}_{{\mathbf{1}}}{\mathbf{D}}{\mathbf{+}}{{\mathbf{b}}}_{{\mathbf{0}}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{\mathbf{C}}{\mathbf{:}}{\mathbf{=}}{{\mathbf{c}}}_{{\mathbf{2}}}{{\mathbf{D}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{c}}}_{{\mathbf{1}}}{\mathbf{D}}{\mathbf{+}}{{\mathbf{c}}}_{{\mathbf{0}}}{\mathbf{,}}$ Where the a’s, b’s, and c’s are constants. Verify the following properties:(a) Commutative laws: ${\mathbit{A}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbit{B}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbit{B}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbit{A}}{\mathbf{,}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbit{A}}{\mathbit{B}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbit{B}}{\mathbit{A}}$ (b) Associative laws: $\left(\mathbf{A}\mathbf{+}\mathbf{B}\right){\mathbf{+}}{\mathbf{C}}{\mathbf{=}}{\mathbf{A}}{\mathbf{+}}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right){\mathbf{,}}\phantom{\rule{0ex}{0ex}}\left(\mathbf{AB}\right){\mathbf{C}}{\mathbf{=}}{\mathbf{A}}\left(\mathbf{BC}\right){\mathbf{.}}$ (c) Distributive law: ${\mathbf{A}}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right){\mathbf{=}}{\mathbf{AB}}{\mathbf{+}}{\mathbf{AC}}$

1. Therefore, the given equations satisfy the commutative law.
2. Hence, the given equations satisfy the associative law.
3. Consequently, the given equations satisfy the distributive law.
See the step by step solution

## Step 1: General form

Commutative laws:

$\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}\mathbf{B}\mathbf{+}\mathbf{A}\mathbf{,}\phantom{\rule{0ex}{0ex}}\mathbf{AB}\mathbf{=}\mathbf{BA}\mathbf{.}$

Associative laws:

$\left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,}\phantom{\rule{0ex}{0ex}}\left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.}$

Distributive law:

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

## Step 2: Demonstrating the given equation

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.

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

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):

$\mathbf{AB}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\phantom{\rule{0ex}{0ex}}\mathbf{BA}\mathbf{=}\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)$

Therefore, AB = BA

## Step 3: Showing the given equation

To prove:

$\left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,}\phantom{\rule{0ex}{0ex}}\left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.}$

Case (1):

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

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

Case (2):

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

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

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

Then,

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

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