# How can someone explain finite solutions

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

• That is not an equation. Presumably the expression you typed is supposed to equal zero or maybe some energy level. Is that so? It appears $R(x)$ has some meaning, like a wave function or energy. Please give the context. – Ross Millikan Nov 27 '16 at 22:16
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.
$$R(x)=e^{-\frac{x}{2}}y\\R'(x)=e^{-\frac{x}{2}}\left(y'-{y\over2}\right)\\R^{"}=e^{-\frac{x}{2}}\left(y^{"}-y'+{y\over4}\right)$$ Second step
Now we are going to get the new equation just substituting: $$R'' + \frac {2}{x}R' + \left(\frac{A}{x} - \frac{1}{4}\right)R=0\\e^{-\frac{x}{2}}\left(y^{"}-y'+{y\over4}\right) + \frac {2}{x}e^{-\frac{x}{2}}\left(y'-{y\over2}\right) + \left(\frac{A}{x} - \frac{1}{4}\right)e^{-\frac{x}{2}}y=0\\\left(y^{"}-y'+{y\over4}\right) + \frac {2}{x}\left(y'-{y\over2}\right) + \left(\frac{A}{x} - \frac{1}{4}\right)y=0\\y^{"}+y'\left({2\over x}-1\right)+y\left({1\over4}-{1\over4}+{A\over x}-{1\over x}\right)=0\\y^{"}+y'\left({2\over x}-1\right)+y\left({A\over x}-{1\over x}\right)=0\\y^{"}+y'{\left(2-x\right)\over x}+y{\left(A-1\right)\over x}=0$$