8.7.4 Example D Let ur(x) satisfy the following equation Aur(x) = aDuk(x), (8.288) where dur (x) Aur (x) = uk+1(x) – Uk(x), Du:(x) = (8.289) dx Therefore, equation (8.288) becomes Uk+1(x) – Uk(x) = abuz(x), (8.290) %3D where, for the moment, D is replaced by the "constant" b. The solution to equation (8.290) is (uk (2) = (1+ ab)*C(x), (8.291) where C(x) is an arbitrary function of x. Putting b= D gives %3D Cue) = (1+) cle) k d. 1+ a dx Uk (2æ) = C(x) k 1 + a d -) C(2) a' dx k d et/a C(x). dx = ak -x/a e (8.292) If we define Co(x) = e-=/"C(x), (8.293) then the general solution to equation (8.288) is Uk (x) = ake-r/a dx (금) d $(x). (8.294)

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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8.7.4 Example D
Let uk (x) satisfy the following equation
Auk (x) = aDux(x),
(8.288)
where
dur (x)
Див (х) %3 ив+1(х) — и; (), Dи, (х) %3D
dx
(8.289)
Therefore, equation (8.288) becomes
Ик+1(2) — и (х) — abuk (х),
(8.290)
where, for the moment, D is replac
equation (8.290) is
by the "constant" b. The solution to
(Uk (a) = (1+ ab)*C(2),
(8.291)
where C(x) is an arbitrary function of x. Putting b = D gives
Uk (x).
k
d
1+ a
dx
C(x)
k
d
1
+
dx
-) C(x)
k
k
d
= ak
eT/a C(x).
dx
(8.292)
If we define
Co(a) = e=/"C{x),
(8.293)
then the general solution to equation (8.288) is
k
d
U16(x) = ake¬#/a
$(x).
(8.294)
dx
Transcribed Image Text:8.7.4 Example D Let uk (x) satisfy the following equation Auk (x) = aDux(x), (8.288) where dur (x) Див (х) %3 ив+1(х) — и; (), Dи, (х) %3D dx (8.289) Therefore, equation (8.288) becomes Ик+1(2) — и (х) — abuk (х), (8.290) where, for the moment, D is replac equation (8.290) is by the "constant" b. The solution to (Uk (a) = (1+ ab)*C(2), (8.291) where C(x) is an arbitrary function of x. Putting b = D gives Uk (x). k d 1+ a dx C(x) k d 1 + dx -) C(x) k k d = ak eT/a C(x). dx (8.292) If we define Co(a) = e=/"C{x), (8.293) then the general solution to equation (8.288) is k d U16(x) = ake¬#/a $(x). (8.294) dx
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