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 hereI'm confused why we get the Y(x), and how that appears because of the bit about a finite solution
thanks in advance
$\endgroup$ 21 Answer
$\begingroup$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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