$$ \begin{align} (x^2-1)y”+10xy’+8y&=0 \\ y(0)&=1 \\ y’(0)&=0 \end{align} $$
I have attempted this problem about 10 times with no avail. I know that the part of the solution that is $x^{2n+1}=0$. I have gotten an expression for the left side that is $C_{n+2}=C_n(n+8)/(n+2)$. After writing terms out to $C_{10}$, I obtained $(16*14*12*10*8)/(10*8*6*4*2)$
I can easily identify the denominator as $2^n*n!$. My problem lies with the numerator. I attempted to pull out a 2 & enter the solution as $2*(n+4)!/(2^n*n!)$ or variations of that like with no two in the numerator and $2^n-1$ in the denominator but that is not correct. Any advice?
You should cancel to see what's really going on.
Let $a_m=c_{2m}$. Then $a_{m+1}=\dfrac{2m+8}{2m+2}a_m=\dfrac{m+4}{m+1}a_m$.
Now we compute $$\require{cancel} \begin{align*} c_0=a_0&=1\\ c_2=a_1&=\frac41\\ c_4=a_2&=\frac52\times\frac41\\ c_6=a_3&=\frac63\times\frac52\times\frac41\\ c_8=a_4&=\frac7{\cancel4}\times\frac63\times\frac52\times\frac{\cancel4}1\\ c_{10}=a_5&=\frac{8}{\cancel 5}\times\frac7{\cancel 4}\times\frac63\times\frac{\cancel 5}2\times\frac{\cancel 4}1\\ &\vdots \end{align*} $$ which is more neatly expressed as $$ c_{2m}=\frac{(m+3)(m+2)(m+1)}{3!}=\binom{m+3}{3}=\binom{-4}{m}(-1)^m. $$ So $y=(1-x^2)^{-4}$. The appearence of $(x^2-1)^{-4}$ shouldn't come as a surprise because the original differential equation can be written as $$ \frac{\mathrm{d}}{\mathrm{d}x}\left[(x^2-1)^5y'\right]+8(x^2-1)^4y=0. $$