i'm dealing with series solutions to differential equations and i'm slightly confused about the term "finite solutions" and how it corresponds to the problem below:
The question refers to this equation: $R'' + \dfrac {2}{x}R' + (\dfrac{A}{x} - \dfrac{1}{4})R$
and then asks to find this result:
enter image description here I'm confused why we get the Y(x), and how that appears because of the bit about a finite solution
thanks in advance
I don't know all the details about your problem but it seems that you have to solve a differential equation with a boundary condition for the solution $R(x)$:$$\lim_{x\to +\infty}R(x)=0$$ so the idea is to write $R(x)$ as something going to zero when $x\to 0$ (in this case $e^{-\frac{x}{2}}$), multiplying a "good" function ($y(x)$) that it will be got solving a new (easier) equation.
First step
\begin{equation} R(x)=e^{-\frac{x}{2}}y\\R'(x)=e^{-\frac{x}{2}}\left(y'-{y\over2}\right)\\R^{"}=e^{-\frac{x}{2}}\left(y^{"}-y'+{y\over4}\right) \end{equation} Second step
Now we are going to get the new equation just substituting: \begin{equation} R'' + \frac {2}{x}R' + \left(\frac{A}{x} - \frac{1}{4}\right)R=0\\e^{-\frac{x}{2}}\left(y^{"}-y'+{y\over4}\right) + \frac {2}{x}e^{-\frac{x}{2}}\left(y'-{y\over2}\right) + \left(\frac{A}{x} - \frac{1}{4}\right)e^{-\frac{x}{2}}y=0\\\left(y^{"}-y'+{y\over4}\right) + \frac {2}{x}\left(y'-{y\over2}\right) + \left(\frac{A}{x} - \frac{1}{4}\right)y=0\\y^{"}+y'\left({2\over x}-1\right)+y\left({1\over4}-{1\over4}+{A\over x}-{1\over x}\right)=0\\y^{"}+y'\left({2\over x}-1\right)+y\left({A\over x}-{1\over x}\right)=0\\y^{"}+y'{\left(2-x\right)\over x}+y{\left(A-1\right)\over x}=0 \end{equation}