(aE1 + bE2 + c)z(k, l) = F(k,l), (5.183) e constants, and F(k, l) satisfies the condition (aE1 + bE2 + c)F(k,l) = 0. (5.184)
(aE1 + bE2 + c)z(k, l) = F(k,l), (5.183) e constants, and F(k, l) satisfies the condition (aE1 + bE2 + c)F(k,l) = 0. (5.184)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
Related questions
Question
Explain the determine blue and the eqaution is here

Transcribed Image Text:(aE1 + bE2 + c)z(k, l) = F(k,l),
(5.183)
where a, b, and c are constants, and F(k, l) satisfies the condition
(aE1 + bE2 + c)F(k,l) = 0.
(5.184)

Transcribed Image Text:5.5.5 Example E
O in equation (5.183) and continue to assume that F(k, l) is
a solution to the homogeneous equation. Under these conditions, we have
Assume that c=
(aE1 +bE2)z(k, l) = F(k,l)
(5.189)
and
(aE1 + bE2)F(k, l) = 0.
(5.190)
The solution to the last equation is
(F(k, e) = (-6/a)*f(e + k),
(5.191)
where f is an arbitrary function of l + k.
Examination of the left-hand side of equation (5.189) shows that it is of a
form such that Laplace's method can be used to obtain a solution. If we let
k +l = m = constant,
Vk = z(k, l) = z(k, m – k),
(5.192)
then vk satisfies the first-order inhomogeneous equation
avk+1 +bvk = (-b/a)*f(m),
(5.193)
where we have used the results of equations (5.191) and (5.192) to replace the
right-hand side of equation (5.189). Note that f(m) is a constant. Solving for
Vk gives
k
k
Uk = A
(5.194)
a
where A is an arbitrary constant. Replacing m by l+k and A by an arbitrary
function of l +k gives the complete solution to equation (5.189), under the
assumption of equation (5.190),
k
z(k, l) =
g(e + k) - (--) sce+ k).
f(l + k).
(5.195)
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