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Given that y=

[infinity]
Σ
n=0
aₙxⁿ, compute y' and y'' and write out the first four terms of each series, as well as the coefficient of xⁿ in the general term. Show that if y''=y, then the coefficients a₀ and a₁ are arbitrary, and determine a₂ and a₃ in terms of a₀ and a₁. Show that aₙ₊₂=aₙ/(n+2)(n+1), n = 0,1,2,3,....

1 Answer

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If y'' = y, then a₀ and a₁ are arbitrary, a₂ = -a₀/2, a₃ = -a₁/6, and aₙ₊₂ = aₙ/((n+2)(n+1)) for n = 0,1,2,3,...

In the scenario of a power series denoted by
y= a_nx^n, if the second derivative y" is equated to \y (i.e., y ′′ =y), certain coefficients become arbitrary. Specifically, a0 and a1 can be selected freely, and this condition yields explicit expressions for subsequent coefficients: a2​=− a0/2 and a3=-a1/6 .

The overarching recurrence relation governing these coefficients is a_n+2​ = a_n/(n+2)(n+1) for n=0,1,2,3,…. This finding highlights the freedom in choosing initial coefficients while providing a systematic formula for their subsequent values.

User Nehz
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