# $2^{nd}$ order ODE $4x(1-x)y''-y=0$ with $y'(0)=1$ at $x=0$

it has two singular point $x_0 =1$ and $x_0=1$
I rearrange the equation to
$$y” – \frac{1}{4x(1-x)}y =0$$
and by $v(x)=\frac{(x-x_0)^2}{4x(1-x)}$ are analytic at $x_0$ for both $0$ and $1$;

that is, $x_0=1,0$ are regular singular points,

so that, I can find a solution by series… (Here, what kind of method should I apply..?)

I just tried the Method of Frobenius:

where the indicial equation is $$r(r-1)-1=0$$ and $r_1=\frac{1+\sqrt5}{2}$and $r_2=\frac{1-\sqrt5}{2}$ where $r_1 – r_2 =\sqrt5$ is not ingeter.

I require to use $$y_1 = x^\frac{1+\sqrt5}{2} \sum_{n=0}^{\infty}a_n x^n$$
$$y_2 = x^\frac{1-\sqrt5}{2} \sum_{n=0}^{\infty}b_n x^n$$
it seems not sensible to solve this with $x^\frac{1-\sqrt5}{2}$ trem,

should I keep working with Method of Frobenius…?… or in any other way..?

#### Solutions Collecting From Web of "$2^{nd}$ order ODE $4x(1-x)y''-y=0$ with $y'(0)=1$ at $x=0$"

Hint

If you use Frobenius method, you write $$y=\sum_{i=0}^{\infty}a_ix^i$$ $$y’=\sum_{i=0}^{\infty}ia_ix^{i-1}$$ $$y”=\sum_{i=0}^{\infty}i(i-1)a_ix^{i-2}$$ Now, for $$4x(1-x)y”-y=4xy”-4x^2y”-y$$ and replacing, you have $$\sum_{i=0}^{\infty}4i(i-1)a_i x^{i-1}=\sum_{i=0}^{\infty}(4i(i-1)+1)a_i x^{i}$$ So, by identification of the coefficients for the same power in both lhs and rhs, you arrive to $$a_{i+1}=\frac{4i(i-1)+1}{4i(i+1)}a_i$$ The initial condition gives $a_1=1$. So, you have all coefficients (I suppose that you noticed $a_0=0$).

Added after OP’s work and comment

Less lazy than I am, the OP reported in a comment to this answer that

$$a_n=\frac{(4*2*1+1)(4*3*2+1)…(4(n-1)(n-2)+1)}{4^{n-1}n!(n-1)!}$$ (in the comment, I identified a small typo for th power of $4$ in the denominator). So the problem was to simplify the numerator and the overall expression. The result is $$a_n=\frac{\Gamma \left(n-\frac{1}{2}\right)^2}{\pi (n-1)! n!}$$ and then $$\sum_{i=1}^{\infty} a_n x^n=x \, _2F_1\left(\frac{1}{2},\frac{1}{2};2;x\right)$$

Frobenius method reads, after proving that $x=0$ and $x=1$ are regular singular points:

$$y(x) = \sum^\infty_{n=0} b_n x^{n+s}, \quad s \in \mathbb{C}, \quad b_0 \neq 0.$$

Substitute this in the ODE, assuming uniform convergence to have:

$$4\sum^\infty_{n=0} b_n (n+s)(n+s-1)x^{n+s-1}-4\sum^\infty_{n=0} b_n (n+s)(n+s-1) x^{n+s} – \sum^\infty_{n=0} b_n x^{n+s} = 0,$$

group the second and third terms and write the first one conveniently to come up with:

$$4 \sum^{\infty}_{n=-1}b_{n+1} (n+s+1)(n+s)x^{n+s} -\sum^\infty_{n=0} b_n\left( 4 (n+s)(n+s-1)- 1\right) x^{n+s} = 0,$$ for which we can say that for:

\begin{align} n = -1: & \quad 4 b_0 s(s-1) = 0 \implies s_1 = 1 \text{ or } s_2 = 0 \\ n \geq 0: & \quad b_{n+1} = \frac{1+4(n+s)(n+s-1)}{4(n+s+1)(n+s)} \, b_n \end{align}

Since $|s_1-s_2| \in \mathbb{N}$ we have pathology (of course, note that $s=0$ is not a valid value of $s$ in the recurrence relation) and we must therefore choose $s = s_1 = 0$ for which we will be able to write one part of the solutions, say $y_1(x)$, while the other one remains unkown until variation of parameters is applied, i.e., setting $y(x) = A(x) y_1(x)$.

Hope this helps.

Cheers!

If you write $y_1(x) = b_0 x \, (1+c_1 x + c_2 x^2 + \ldots + c_n x^n) = b_0 x \, (1+P(x))$, where $c_i = b_i/b_0$ and $P(x)$ is a polynomial such that $P(0) = 0$, then after applying VOP you seem to get the following:

$$y(x) = A_1 y_1(x) + A_2 \underbrace{ y_1(x) \int \frac{\mathrm{d}x}{x^2 (1+P)^2} }_{y_2(x)},$$

where it is convenient to recall that:

$$(1+P)^{-2} = 1-2P+3P^2 – 4P^3+ \mathcal{O}(P^4)$$

so you can ultimately express $y_2(x)$ as a power sieres as well (see @ClaudeLeibovici’s comment below).

$$a_n=\frac{(4*2*1+1)(4*3*2+1)…(4(n-1)(n-2)+1)}{4^{n-1}n!(n-1)!}=\frac{4^{n-2}(2*1+\frac{1}{4})(3*2+\frac{1}{4})…((n-1)(n-2)+\frac{1}{4})}{4^{n-1}n!(n-1)!}=\frac{(2*1+\frac{1}{4})(3*2+\frac{1}{4})…((n-1)(n-2)+\frac{1}{4})}{4*n!(n-1)!}$$

Thanks, I just tried it again..for this part

$$I=(2*1+\frac{1}{4})(3*2+\frac{1}{4})…((n-1)(n-2)+\frac{1}{4})$$

$$I=\prod_{n=3}^{+\infty}[(n-1)(n-2)+\frac{1}{4}]$$$$=\prod_{n=3}^{+\infty}[n^2 – 3n +\frac{9}{4}]$$$$=\prod_{n=3}^{+\infty}[(n-\frac{3}{2})^2]$$

$$I=\Gamma^2(n-\frac{3}{2}+1)=\Gamma^2(n-\frac{1}{2})$$

substitute this back, there is $$a_n=\frac{\Gamma^2(n-\frac{1}{2})}{4*\Gamma(n+1)\Gamma(n)}$$

where has a difference of $\pi$ and $4$ of denominator..

b.tw..the computer also gives $\pi$… how does this comes?