Solutions in terms of the hypergeometric functions

Question

Is it possible to somehow express the solutions to the differential equations: $$\frac{d^2y}{dx^2} + \bigg(\frac{1}{x + 8} - \frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 4}\bigg) \frac{dy}{dx} + \bigg(\frac{1}{x^2} + \frac{3}{4x} - \frac{5}{6(x - 1)} - \frac{1}{4(x - 4)^2}\bigg) y = 0$$ and $$\frac{d^2y}{dx^2} + \bigg(\frac{1}{x + 8} + \frac{1}{3x} + \frac{1}{x - 64}\bigg) \frac{dy}{dx} + \bigg(\frac{7}{144x^2} - \frac{7}{3072x} + \frac{7}{3072(x - 64)}\bigg) y = 0$$ in terms of the hypergeometric functions?

Answer 1

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

$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}+\left(\dfrac{3}{4x}+\dfrac{1}{x(x-1)}-\dfrac{5}{6(x-1)}+\dfrac{1}{2x(x-4)}+\dfrac{1}{x(x+8)}-\dfrac{1}{2(x-1)(x-4)}-\dfrac{1}{2(x-4)(x+8)}\right)u=0$

$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}-\left(\dfrac{1}{4x}-\dfrac{1}{3(x-1)}+\dfrac{1}{12(x-4)}+\dfrac{1}{12(x+8)}\right)u=0$