8.7.5 Example E The mixed differential-difference equation duk+1(x) duk Ик+2(г) — Зк- dx + 2k(k – 1) dx? = 0, (8.295) has, in the x variable, the structure of a Euler or Cauchy-Euler equation....

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Extracted text: 8.7.5 Example E The mixed differential-difference equation duk+1(x) duk Ик+2(г) — Зк- dx + 2k(k – 1) dx? = 0, (8.295) has, in the x variable, the structure of a Euler or Cauchy-Euler equation. Consequently, we assume the solution to have the form Ux (x) = (k – 2)!vx(x). (8.296) The substitution of this u (x) into equation (8.295) gives for vk (x) Uk+2(x) – 3dek+1(2) + 2&vx(x) , dx? Vk+2(x) – 3- dx dvr(2) = 0. (8.297) Using Evr (x) = Vk+1(x), we may rewrite this last equation and obtain C(E? - 3E + 2) d? Vr (x) = 0, d. (8.298) dx² or )( d d - 2) vr (a) = 0. (8.299) E E dx dx Therefore, (2) Va (x) = v" (x) + v (x), (1) (8.300) where d d E) »P (2) = 0, (E - 2 (=) = 0. (1) v" (x) dx (8.301) E = 0. dx These solutions are given by the expressions k d k d. v (2) = - (유) A(2), 맛이(2) -안 () B(2), В(), (8.302) dx dx and [(4)° k Uk (x) = (k – 2)! A(x) + 2k k d B(x) (8.303) dx dx where A(x) and B(x) are arbitrary functions of x.