Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Let A, B and C represent three linear differential operators with constant coefficients; for example,
$A : = a_{2} D^{2} + a_{1} D + a_{0}, B : = b_{2} D^{2} + b_{1} D + b_{0}, C : = c_{2} D^{2} + c_{1} D + c_{0},$
Where the a’s, b’s, and c’s are constants. Verify the following properties:

(a) Commutative laws:
$A + B = B + A, A B = B A$
(b) Associative laws:
$(A + B) + C = A + (B + C), (AB) C = A (BC) .$
(c) Distributive law: $A (B + C) = AB + AC$

Therefore, the given equations satisfy the commutative law.
Hence, the given equations satisfy the associative law.
Consequently, the given equations satisfy the distributive law.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: General form

Commutative laws:

$A + B = B + A, AB = BA .$

Associative laws:

$(A + B) + C = A + (B + C), (AB) C = A (BC) .$

Distributive law:

$A (B + C) = AB + AC$ .

Step 2: Demonstrating the given equation

Given that,

$A = a_{2} D^{2} + a_{1} D + a_{0}$ …… (1)

$B = b_{2} D^{2} + b_{1} D + b_{0}$ …… (2)

$C = c_{2} D^{2} + c_{1} D + c_{0}$ …… (3)

Let us prove the commutative property.

Case (1):

Then, find the L.H.S.

$A + B = a_{2} D^{2} + a_{1} D + a_{0} + b_{2} D^{2} + b_{1} D + b_{0} = (a_{2} + b_{2}) D^{2} + (a_{1} + b_{1}) D + (a_{0} + b_{0})$

Now, R.H.S.

$B + A = (b_{2} + a_{2}) D^{2} + (b_{1} + a_{1}) D + (b_{0} + a_{0})$

.So, A + B = B + A.

Case (2):

$AB = (a_{2} D^{2} + a_{1} D + a_{0}) (b_{2} D^{2} + b_{1} D + b_{0}) BA = (b_{2} D^{2} + b_{1} D + b_{0}) (a_{2} D^{2} + a_{1} D + a_{0})$

Therefore, AB = BA

Step 3: Showing the given equation

To prove:

$(A + B) + C = A + (B + C), (AB) C = A (BC) .$

Case (1):

$(A + B) + C = (a_{2} + b_{2} + c_{2}) D^{2} + (a_{1} + b_{1} + c_{1}) D + (a_{0} + b_{0} + c_{0}) A + (B + C) = (a_{2} + b_{2} + c_{2}) D^{2} + (a_{1} + b_{1} + c_{1}) D + (a_{0} + b_{0} + c_{0})$

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

Case (2):

$(AB) C = (a_{2} D^{2} + a_{1} D + a_{0}) (b_{2} D^{2} + b_{1} D + b_{0}) (c_{2} D^{2} + c_{1} D + c_{0}) A (BC) = (a_{2} D^{2} + a_{1} D + a_{0}) (b_{2} D^{2} + b_{1} D + b_{0}) (c_{2} D^{2} + c_{1} D + c_{0})$

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

To prove: $A (B + C) = AB + AC$ .

Then,

$A + (B + C) = (a_{2} + b_{2} + c_{2}) D^{2} + (a_{1} + b_{1} + c_{1}) D + (a_{0} + b_{0} + c_{0}) AB + AC = (a_{2} + b_{2} + c_{2}) D^{2} + (a_{1} + b_{1} + c_{1}) D + (a_{0} + b_{0} + c_{0})$

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

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