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

  • 1
    $\begingroup$ 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. $\endgroup$ – Ross Millikan Nov 27 '16 at 22:16
  • $\begingroup$ Finite solution basically means that you expect the particle (presumably an electron) a finite distance from the center (nucleus of the atom). If there were finite weight on the electron being infinitely far away then it would be possible for the electron to be unbound. Presumably that state isn't interesting/useful at the moment. The exponential is there to ensure that the weight goes to zero quickly as distance gets bigger. $\endgroup$ – Kitter Catter Nov 28 '16 at 3:43

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}


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