Elementary Differential Equations
Elementary Differential Equations
10th Edition
ISBN: 9780470458327
Author: William E. Boyce, Richard C. DiPrima
Publisher: Wiley, John & Sons, Incorporated
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Chapter 6.6, Problem 1P

(a)

To determine

To establish: The commutative law of the convolution integral.

(a)

Expert Solution
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Explanation of Solution

Theorem used:

If F(s)=L{f(t)} and G(s)=L{g(t)} both exist for s>a0, then

H(s)=F(s)G(s)=L{h(t)},     s>a                                           (1)

Where h(t)=0tf(tτ)g(τ)dτ=0tf(τ)g(tτ)dτ                     (2)

The function h is known as the convolution of f and g; the integrals in equation (2) are called convolution integrals.

Calculation:

Consider left hand side (fg)(t)=0tf(tτ)g(τ)dτ.

Let u=tτ then du=dτ.

(fg)(t)=t0f(u)g(tu)(du)=0tf(u)g(tu)du=(gf)(t)

Hence, the commutative law is proved.

(b)

To determine

To establish: The distributive law of the convolution integral.

(b)

Expert Solution
Check Mark

Explanation of Solution

Consider left hand side (f(g1+g2))(t)=0tf(tτ)(g1+g2)(τ)dτ.

(f(g1+g2))(t)=0tf(tτ)(g1(τ)+g2(τ))dτ=0t[f(tτ)g1(τ)+f(tτ)g2(τ)]dτ=0tf(tτ)g1(τ)dτ+0tf(tτ)g2(τ)dτ=(fg1)(t)+(fg2)(t)

Hence, the distributive law is proved.

(c)

To determine

To establish: The associative law of the convolution integral.

(c)

Expert Solution
Check Mark

Explanation of Solution

Consider left hand side (f(gh))(t)=0tf(tτ)(gh)(τ)dτ.

(f(gh))(t)=0t0τ[f(tτ)g(τη)h(η)]dτdη=0tηt[f(tτ)g(τη)h(η)]dτdη

Let u=τη then dτ=du.

(f(gh))(t)=0t0tη[f(tηu)g(u)du]dηhη=0t(fg)(tη)h(η)d(η)=((fg)h)(t)

Hence, the associative law is proved.

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Chapter 6 Solutions

Elementary Differential Equations

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