Solutions in terms of the hypergeometric functions
Hint:
For $\dfrac{d^2y}{dx^2}+\left(\dfrac{1}{x+8}+\dfrac{1}{3x}+\dfrac{1}{x-64}\right)\dfrac{dy}{dx}+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)y=0$ ,
Let $y=x^au$ ,
Then $\dfrac{dy}{dx}=x^a\dfrac{du}{dx}+ax^{a-1}u$
$\dfrac{d^2y}{dx^2}=x^a\dfrac{d^2u}{dx^2}+ax^{a-1}\dfrac{du}{dx}+ax^{a-1}\dfrac{du}{dx}+a(a-1)x^{a-2}u=x^a\dfrac{d^2u}{dx^2}+2ax^{a-1}\dfrac{du}{dx}+a(a-1)x^{a-2}u$
$\therefore x^a\dfrac{d^2u}{dx^2}+2ax^{a-1}\dfrac{du}{dx}+a(a-1)x^{a-2}u+\left(\dfrac{1}{x+8}+\dfrac{1}{3x}+\dfrac{1}{x-64}\right)\left(x^a\dfrac{du}{dx}+ax^{a-1}u\right)+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)x^au=0$
$\dfrac{d^2u}{dx^2}+\dfrac{2a}{x}\dfrac{du}{dx}+\dfrac{a(a-1)}{x^2}u+\left(\dfrac{1}{x+8}+\dfrac{1}{3x}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{a}{x(x+8)}+\dfrac{a}{3x^2}+\dfrac{a}{x(x-64)}\right)u+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)u=0$
$\dfrac{d^2u}{dx^2}+\left(\dfrac{6a+1}{3x}+\dfrac{1}{x+8}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{a(3a-2)}{3x^2}+\dfrac{a}{8x}-\dfrac{a}{8(x+8)}-\dfrac{a}{64x}+\dfrac{a}{64(x-64)}\right)u+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)u=0$
$\dfrac{d^2u}{dx^2}+\left(\dfrac{6a+1}{3x}+\dfrac{1}{x+8}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{48a(3a-2)+7}{144x^2}+\dfrac{7(48a-1)}{3072x}-\dfrac{a}{8(x+8)}+\dfrac{48a+7}{3072(x-64)}\right)u=0$
Choose $a=\dfrac{1}{12}$ , the ODE becomes
$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{2x}+\dfrac{1}{x+8}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{7}{1024x}-\dfrac{1}{96(x+8)}+\dfrac{11}{3072(x-64)}\right)u=0$
In fact according to http://www.wolframalpha.com/input/?i=y%27%27%2B(1%2F(x%2B8)%2B1%2F(3x)%2B1%2F(x-64))y%27%2B(7%2F(144x%5E2)-7%2F(3072x)%2B7%2F(3072(x-64)))y%3D0, it is possible to simplify to hypergeometric ODE luckily.
For $\dfrac{d^2y}{dx^2}+\left(\dfrac{1}{x+8}-\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x-4}\right)\dfrac{dy}{dx}+\left(\dfrac{1}{x^2}+\dfrac{3}{4x}-\dfrac{5}{6(x-1)}-\dfrac{1}{4(x-4)^2}\right)y=0$ ,
Let $y=x^a(x-4)^bu$ ,
Then $\dfrac{dy}{dx}=x^a(x-4)^b\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)u$
$\dfrac{d^2y}{dx^2}=x^a(x-4)^b\dfrac{d^2u}{dx^2}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u=x^a(x-4)^b\dfrac{d^2u}{dx^2}+2x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u$
$\therefore x^a(x-4)^b\dfrac{d^2u}{dx^2}+2x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u+\left(\dfrac{1}{x+8}-\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x-4}\right)\left(x^a(x-4)^b\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)u\right)+\left(\dfrac{1}{x^2}+\dfrac{3}{4x}-\dfrac{5}{6(x-1)}-\dfrac{1}{4(x-4)^2}\right)x^a(x-4)^bu=0$
$\dfrac{d^2u}{dx^2}+\left(\dfrac{2a}{x}+\dfrac{2b}{x-4}\right)\dfrac{du}{dx}+\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u+\left(\dfrac{1}{x+8}-\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x-4}\right)\dfrac{du}{dx}+\left(\dfrac{a}{x(x+8)}-\dfrac{a}{x^2}+\dfrac{a}{x(x-1)}+\dfrac{a}{x(x-4)}+\dfrac{b}{(x-4)(x+8)}-\dfrac{b}{x(x-4)}+\dfrac{b}{(x-1)(x-4)}+\dfrac{b}{(x-4)^2}\right)u+\left(\dfrac{1}{x^2}+\dfrac{3}{4x}-\dfrac{5}{6(x-1)}-\dfrac{1}{4(x-4)^2}\right)u=0$
$\dfrac{d^2u}{dx^2}+\left(\dfrac{2a-1}{x}+\dfrac{1}{x-1}+\dfrac{2b+1}{x-4}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}+\left(\dfrac{a(a-2)+1}{x^2}+\dfrac{3}{4x}+\dfrac{a}{x(x-1)}-\dfrac{5}{6(x-1)}+\dfrac{2ab+a-b}{x(x-4)}+\dfrac{a}{x(x+8)}+\dfrac{b}{(x-1)(x-4)}+\dfrac{b}{(x-4)(x+8)}+\dfrac{4b^2-1}{4(x-4)^2}\right)u=0$
Choose $a=1$ and $b=-\dfrac{1}{2}$ , the ODE becomes
$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}+\left(\dfrac{3}{4x}+\dfrac{1}{x(x-1)}-\dfrac{5}{6(x-1)}+\dfrac{1}{2x(x-4)}+\dfrac{1}{x(x+8)}-\dfrac{1}{2(x-1)(x-4)}-\dfrac{1}{2(x-4)(x+8)}\right)u=0$
$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}-\left(\dfrac{1}{4x}-\dfrac{1}{3(x-1)}+\dfrac{1}{12(x-4)}+\dfrac{1}{12(x+8)}\right)u=0$
Let us focus on the second ODE since at the first glance it has three regular singular points and as such it should be possible to map it onto the hyper-geometric equation which too has three singular points. For 2nd order ODEs we always reduce it to the normal form, i.e. such that has no coefficient at the first derivative. This is done by writing $y(x)=m(x) \cdot v(x)$ where $m(x):=\exp(-1/2 \int a_1(x) dx)$and $a_1(x)$ is the coeffcient at the 1st derivative. In our case: \begin{equation} a_1(x)=\frac{1}{x+8} + \frac{1}{3 x} + \frac{1}{x-64} \end{equation} therefore $m(x)=(x+8)^{-1/2} x^{-1/6} (x-64)^{-1/2}$ and the function $v(x)$ satisfies the following ODE: \begin{equation} v^{''}(x) + \frac{48(1024+112 x+25 x^2)}{(-64+x)^2 x^2 (8+x)^2} v(x)=0 \quad (I) \end{equation}
Now what we want to do is to relate the ODE above to the appropriately transformed hypergeometric ODE. \begin{equation} x(1-x) Y^{''}(x) + (c-(a+b+1)x)Y^{'}(x)-a b Y(x)=0 \end{equation} We will start from that ODE and carry out two transformations. Let us take the Moebius function $f(x):=(A x+B)/(C x+D)$ and firstly change the abscissa $x \rightarrow f(x)$ and $d/dx \rightarrow 1/f^{'}(x) d/d x$ and then reduce the equation to the normal form by writing \begin{equation} Y(x):= (A x+B)^{-c/2} (B-D+(A-C) x)^{-(1+a+b-c)/2} (D+C x)^{(-1+a+b)/2}\cdot V(x) \end{equation} Having done this we obtain the following ODE for the function $V(x)$. We have: \begin{eqnarray} V^{''}(x)- \frac{(B C-A D)^2}{4} \cdot \frac{{\mathfrak A_0}+{\mathfrak A_1} x+{\mathfrak A_2} x^2}{(B+A x)^2 (B-D+(A-C) x)^2 (D+C x)^2} \cdot V(x)=0 \quad(II) \end{eqnarray} where: \begin{eqnarray} {\mathfrak A_0}&:=&B^2 \left(a^2-2 a b+b^2-1\right)+2 B D (2 a b-a c-b c+c)+(c-2) c D^2\\ {\mathfrak A_1}&:=&2 \left(A \left(B \left(a^2-2 a b+b^2-1\right)+D (2 a b-a c-b c+c)\right)+C (a B (2 b-c)+c (-b B+B+(c-2)D))\right)\\ {\mathfrak A}_2&:=&A^2 \left(a^2-2 a b+b^2-1\right)+2 A C (2 a b-a c-b c+c)+(c-2) c C^2 \end{eqnarray}
Note that ODE $(I)$ has exactly the same form as $(II)$. All we need to do is to adjust the parameters accordingly to map the later to the former. Firstly we find the upper case letters by matching the zeros of the denominator. Choosing \begin{eqnarray} B&=&-64 A\\ C&=&-8 A\\ D&=&-64 A \end{eqnarray} does the job. Now comes the toughest part meaning choosing the lower case parameters to match the numerators in ODEs $(I)$ and $(II)$. We have: \begin{eqnarray} {\mathfrak A_0}=-16 \left(4096 a^2+8192 a b-8192 a c+4096 b^2-8192 b c+4096 c^2-4096\right)=48 \times 1024\\ {\mathfrak A_1}=-16 \left(-128 a^2+2048 a b-896 a c-128 b^2-896 b c+1024 c^2-1152 c+128\right)=48 \times 112\\ {\mathfrak A_2}=-16 \left(a^2-34 a b+16 a c+b^2+16 b c+64 c^2-144 c-1\right)=48 \times 25 \end{eqnarray} These are just quadratic equations so they can be solved. In fact there are two sets of solutions: \begin{eqnarray} (a,b,c)&=&(\frac{1}{4},\frac{1}{4},1)\\ (a,b,c)&=&(\frac{3}{4},\frac{3}{4},1) \end{eqnarray} Now we have completed the job. We made sure that $v(x)=V(x)$. All we have to do is to bring everything together and simplify. We have: \begin{eqnarray} y(x)=C_1 x^{\frac{1}{12}} (8+x)^{-\frac{1}{4}} F_{2,1}\left[1/4,1/4,1;\frac{x-64}{-8 x-64}\right] + C_2 x^{\frac{7}{12}} (8+x)^{-\frac{3}{4}} F_{2,1}\left[3/4,3/4,1;\frac{x-64}{-8 x-64}\right] \end{eqnarray} The final step is to verify the result using a Computer Algebra System. We have:
In[295]:=
FullSimplify[(D[#, {x, 2}] + (1/(x + 8) + 1/(3 x) + 1/(x - 64)) D[#,
x] + (7/(144 x^2) - 7/(3072 x) +
7/(3072 (x - 64))) #) & /@ { (x)^(1/(12)) ( (8 + x))^(-1/4)
Hypergeometric2F1[1/4, 1/4, 1, (x - 64)/(-8 x - 64)], ((x)^(
7/(12)))/((8 + x))^(3/4)
Hypergeometric2F1[3/4, 3/4, 1, (x - 64)/(-8 x - 64)]}]
Out[295]= {0, 0}