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...

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Extracted 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