Solution. Put \(\displaystyle {y = \sum _{n=0}^{\infty }a_nx^n}\). Differentiating twice, \[\frac {d^2y}{dx^2} = \sum _{n=2}^{\infty }n(n-1)a_nx^{n-2} = \sum _{n=0}^{\infty }(n+2)(n+1)a_{n+2}x^{n},\] where the last step replaces \(n\) by \(n+2\) so that both series run in powers of \(x^n\). Substituting, \[\sum _{n=0}^{\infty }\Big [(n+2)(n+1)a_{n+2} + a_n\Big ]x^n = 0 .\] A power series vanishes identically only if every coefficient vanishes, so \[\boxed {a_{n+2} = \frac {-a_n}{(n+2)(n+1)}}\qquad n = 0,1,2,\dots \]

The relation links \(a_{n+2}\) to \(a_n\), so the even and odd coefficients form two separate chains driven by \(a_0\) and \(a_1\).

Taking \(a_0 = 1\) and \(a_1 = 0\) kills every odd coefficient and gives \[a_2 = \frac {-1}{2\cdot 1} = -\frac {1}{2},\qquad a_4 = \frac {-a_2}{4\cdot 3} = \frac {1}{24},\qquad a_6 = \frac {-a_4}{6\cdot 5} = -\frac {1}{720},\] so that \[y = 1 - \frac {x^2}{2} + \frac {x^4}{24} - \frac {x^6}{720} + \cdots = \sum _{k=0}^{\infty }\frac {(-1)^kx^{2k}}{(2k)!} = \cos x .\]

Taking \(a_0 = 0\) and \(a_1 = 1\) gives the odd chain in the same way and yields \(\sin x\). The general solution is therefore \[y = A\cos x + B\sin x ,\] which agrees with the auxiliary equation method, as it must. The value of the series method is that it does not require the coefficients to be constant, and so it reaches equations the earlier methods cannot.