Let
[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots[/tex]
Differentiating twice gives
[tex]\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots[/tex]
[tex]\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n[/tex]
When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.
Substitute these into the given differential equation:
[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0[/tex]
[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0[/tex]
Then the coefficients in the power series solution are governed by the recurrence relation,
[tex]\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}[/tex]
Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.
• If n is even, then n = 2k for some integer k ≥ 0. Then
[tex]k=0 \implies n=0 \implies a_0 = a_0[/tex]
[tex]k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}[/tex]
[tex]k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}[/tex]
[tex]k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}[/tex]
It should be easy enough to see that
[tex]a_{n=2k} = \dfrac{a_0}{(2k)!}[/tex]
• If n is odd, then n = 2k + 1 for some k ≥ 0. Then
[tex]k = 0 \implies n=1 \implies a_1 = a_1[/tex]
[tex]k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}[/tex]
[tex]k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}[/tex]
[tex]k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}[/tex]
so that
[tex]a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}[/tex]
So, the overall series solution is
[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)[/tex]
[tex]\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}[/tex]
