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QUESTION

Obtain at least one solution of the form $$
y=x^{\sigma}\sum_{n=0}^{\infty}a_nx^n $$ for each of the following
differential equations. Where possible, obtain a second
independent solution of the same form, or comment on why it is not
possible to do so.

$4xy''+2y'+y=0$


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ANSWER

$4xy''+2y+y=0$, or $y''+\frac{1}{2x}y'+\frac{1}{4x}y=0$\\
 y=0 is a regular singular point, so we need a Frobenius
 series:\\ $\ds y=\ds \sum _{n=0}^\infty a_nx^{\sigma + n}$.

  Substitute:

 $\ds \sum_{n=0}^\infty a_n\left[\left(\sigma + n\right)\left(\sigma + n-1\right)x^{\sigma
 +n-2}+\frac{1}{2}\left(\sigma+n\right)x^{\sigma
 +n-2}+\frac{1}{4}x^{\sigma+n-1}\right]=0\\
 \ds \sum_{n=0}^\infty a_n\left[\left(\sigma+n\right)\left(\sigma
 +n-\frac{1}{2}\right)x^{\sigma+n-2}+\frac{1}{4}x^{\sigma+n-1}\right]=0$

Now reorder by powers of $x$.

Let $n=m+1$ in the first term and $n=m$ in the second.

 $\ds \sum_{-1}^\infty a_{m+1}\left(\sigma+m+1\right)\left(\sigma +m
+\frac{1}{2}\right)x^{\sigma +m-1}+\ds \sum_{m=0}^\infty
\frac{1}{4}a_mx^{\sigma +m-1}=0$

$m=-1$ gives us
$\sigma\left(\sigma-\frac{1}{2}\right)=0\Rightarrow \sigma=0$ or
$\sigma=\frac{1}{2}$.

$m=0$ gives
$a_{m+1}=-\frac{1}{4\left(\sigma+m+1\right)\left(\sigma+m+\frac{1}{2}\right)}a_m$

Case $\sigma=0:\
a_{m+1}=-\frac{1}{\left(2m+2\right)\left(2m+1\right)}a_m$ The
solution is $a_m=\frac{\left(-1\right)^m}{\left(2m\right)!}a_0$

Case $\sigma=\frac{1}{2}:\
a_{m+1}=-\frac{1}{\left(2m+3\right)\left(2m+2\right)}a_m,\
a_m=\frac{\left(-1\right)^m}{\left(2m+1\right)!}a_0$

The general solution is obtained by adding the two cases:
$$y=A\sum_{n=0}^\infty
\frac{\left(-1\right)^n}{\left(2n\right)!}x^n+B\sum_{n=0}^\infty
\frac{\left(-1\right)^n}{\left(2n+1\right)!}x^{n+\frac{1}{2}}$$

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