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**Unformatted text preview: **we find that the only non-zero coefficients u n are when n = 3 k +1. The recurrence relation u 3 k +1 = u 3 k − 2 (3 k + 1)(3 k ) yields u 3 k +1 = 1 (3 k + 1)(3 k )(3 k − 2)(3 k − 3) ··· 7 · 6 · 4 · 3 . The resulting solution is tildewide u ( x ) = x + 1 12 x 4 + 1 504 x 7 + ··· = x + ∞ summationdisplay k = 1 x 3 k +1 (3 k + 1)(3 k )(3 k − 2)(3 k − 3) ··· 7 · 6 · 4 · 3 . (12 . 80) Again, the denominator skips every third term in the product. Every solution to the Airy equation can be written as a linear combination of these two basis power series solutions: u ( x ) = a hatwide u ( x ) + b tildewide u ( x ) , where a = u (0) , b = u ′ (0) . Both power series (12.79, 80), converge quite rapidly, and so the first few terms will provide a reasonable approximation to the solutions for moderate values of x . We have, in fact, already encountered another solution to the Airy equation. Accord- ing to formula (9.93), the integral Ai( x ) = 1 π integraldisplay ∞ cos ( sx + 1 3 s 3 ) ds (12 . 81) defines the Airy function of the first kind . Let us prove that it satisfies the Airy differential equation (12.78): d 2 dx 2 Ai( x ) = x Ai( x ) . Before differentiating, we recall the integration by parts argument in (9.92) to re-express the Airy integral in absolutely convergent form: Ai( x ) = 2 π integraldisplay ∞ s sin ( sx + 1 3 s 3 ) ( x + s 2 ) 2 ds. We are now permitted to differentiate under the integral sign, producing (after some alge- bra) Ai ′′ ( x ) − x Ai( x ) = 2 π integraldisplay ∞ d ds bracketleftBigg s ( x + s 2 ) cos ( sx + 1 3 s 3 ) − sin ( sx + 1 3 s 3 ) ( x + s 2 ) 3 bracketrightBigg ds = 0 . Thus, the Airy function must be a certain linear combination of the two basic series solutions: Ai( x ) = Ai(0) hatwide u ( x ) + Ai ′ (0) tildewide u ( x ) ....

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