Let \(y = \displaystyle\sum^ \infty_{i=0}a_ix^i\)
Then
\(\frac{dy}{dx}= \displaystyle\sum^\infty_{i = 1} i \,a_i \,x ^{i -1}\)
\(\frac{d^2y}{dx^2}= \displaystyle\sum^\infty_{i = 2} i(i -1) \,a_i \,x ^{i -2}\)
Now,
\(\frac{d^2y}{dx^2}+ y = 0\)
⇒ \(\displaystyle \sum ^\infty_{i = 2}i(i -1) \,a_i \,x ^{i -2} + \displaystyle\sum^ \infty_{i=0}a_ix^i = 0\)
⇒ \(\displaystyle\sum^\infty_{i = 0} (i + 2) (i + 1)a_{i + 2}x^i + \displaystyle\sum^\infty_{i = 0}a_i x^i = 0\)
⇒ \(\displaystyle\sum^\infty_{i = 0}((i + 2)(i + 1)a_{i + 2} + a_i)x^i = 0\)
⇒ \((i + 2)(i + 1)a_{i + 2} + a_i = 0\)
⇒ \((i+2 (i + 1)a_{i + 2}) = -a_i\)
⇒ \(a_{i+2} = \frac{-a_i}{(i + 2)(i + 1)}\)
∴ \(a_2 = \frac{-a_0}{2 \times 1}\)
\(a_3 = \frac{-a_1}{3\times2}\)
\(a_4 = \frac{-a_2}{4\times3}= \frac{a_2}{4\times3\times2\times1}\)
\(a_5 = \frac{-a_3}{5\times 4} = \frac{a_1}{5\times4\times 3 \times 2}\)
∴ \(y = a_0 + a_1x + a_2x^2 + a_3 x^3 + a_4x^4+ a_5x^5+...\)
\(= a_0+ a_1x- \frac{x^2}{2!}a_0 -\frac{x^3}{3!}a_1+ \frac{x^4}{4!}a_0 + \frac{x^5}{5!}a_1 + ..\)
\(= a_0\left(1- \frac{x^2}{2!}+ \frac{x^4}{4!}-...\right)+ a_1\left(x - \frac{x^3}{3!} + \frac{x^5}{5!}\right)\)
\(= a_0 \,cos\,x + a_1sin\,x\).
Hence, \(y= a_0 \,cos\,x + a_1sin\,x\) is solution of given differential equation.
