Case 8. Suppose the positive integers q is even and the positive integers p,r, s are odd. In this case Sn = Sn-q and Sn+1 = Sn-p = Sn-r = Sn-s· From Equation (1), we have arb + (b+ c)¢ dp + (e+ f)¢, ф аф + (b + c)) dø + (e + f)V. Thus, døp + (e+ f)² = a¢v + (b + c)¢², (10) %3D and døp + (e + f)u? = aørp + (b+ c)b?. (11) By subtracting (10) from (11), we deduce that (e + f)(² – v²) – (b+ c)(ø² – v?) = 0. Hence, we have [(e+ f) – (b+ c)](² – v²) = 0. Since b, c, e and f are nonzero positive real numbers, and o # p. This implies [(e + f) = (b+ c)], This contradicts the condition [(e+ f) # (b+c)]. %3D
Case 8. Suppose the positive integers q is even and the positive integers p,r, s are odd. In this case Sn = Sn-q and Sn+1 = Sn-p = Sn-r = Sn-s· From Equation (1), we have arb + (b+ c)¢ dp + (e+ f)¢, ф аф + (b + c)) dø + (e + f)V. Thus, døp + (e+ f)² = a¢v + (b + c)¢², (10) %3D and døp + (e + f)u? = aørp + (b+ c)b?. (11) By subtracting (10) from (11), we deduce that (e + f)(² – v²) – (b+ c)(ø² – v?) = 0. Hence, we have [(e+ f) – (b+ c)](² – v²) = 0. Since b, c, e and f are nonzero positive real numbers, and o # p. This implies [(e + f) = (b+ c)], This contradicts the condition [(e+ f) # (b+c)]. %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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