I've done the following:

Consider the differential equation given by $\displaystyle y^{\prime\prime} - 2y^\prime + \frac{4x^2 + 1}{4x^2} y = 0$. As one can see immediately, $x=0$ is a singular. Is it regular? Yes, as the limits of $xp(x) = -2x$ and $x^2 q(x) = x^2 + 1/4$ both exist as $x$ approaches $0$.

The indicial equation is of the form $m(m-1) + p_0 m + q_0 = m(m-1) + 1/4 = 0$. Solving this yields the repeated root $m = 1/2$.

We have that $m_1 = m_2 = 1/2$, so our Frobenius solution is of the form $$y = x^{1/2} \sum_{n=0}^\infty a_nx^n.$$ Note also the following system of equations: \begin{align} & a_0[m(m-1) + \frac{1}{4}] = 0\\ & a_1[(m+1)m +\frac{1}{4}] + a_0m = 0\\ & a_2[(m+2)(m+1) + \frac{1}{4}] + a_1 (m+1) = 0\\ & a_3[(m+3)(m+2) + \frac{1}{4}] + a_2 (m+2) = 0 \end{align} If we fix $\displaystyle m=\frac{1}{2}$, we solve for all of the above to yield $\displaystyle a_1 = -\frac{a_0}{2}$ and $\displaystyle a_2 = -\frac{3}{8}a_1 = \frac{3a_0}{16}$, etc. Thus, the first of our solutions is $$y_1(x) = x^{1/2}\sum_{n=0}^\infty a_nx^n = x^{1/2}(1-\frac{x}{2} + \frac{3}{16}x^2-\frac{5}{96}x^3 + \frac{35}{3072}x^4 -...)a_0.$$

However, I am told the first solution has a closed form, which I am unable to find. Have I made a mistake somewhere?


Yes, you made a mistake somewhere.

To make the solution more simple, use the form you already found to simplify the original equation:

$$y(x)=\sqrt{x} f(x)$$

The original equation now reads:

$$f''-\left( 2-\frac{1}{x} \right) f'+ \left( 1-\frac{1}{x} \right) f=0 $$

If we search for the solution in the form:

$$f(x)=\sum_{n=0}^\infty a_n x^n$$

We can multiply the equation by $x$:

$$xf''-\left( 2x-1\right) f'+ \left( x-1 \right) f=0 $$

Subsituting the series into the last equation we find:


$$a_{n \geq 2}=\frac{(2n-1) a_{n-1}-a_{n-2}}{n^2}$$

Computing the first few coefficients, we see that:




Without proof, we can guess that:

$$f(x)=a_0 e^x$$

Simply substituting this into the ODE is enough to prove the guess.

So our first solution will be:

$$y_1=a_0 \sqrt{x} e^x$$

The second linearly independent solution for the original equation will be:

$$y_2=c \sqrt{x} e^x \ln x$$

Which can't be expanded as a series around $x=0$, but can around $x=1$.


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