$4x^2y′′-8x^2y′+(4x^2+1)y=0$ solve by Frobenius Method. I would like to ask if someone can explain to me how can we solve the following DE using this method.
$4x^2y′′-8x^2y′+(4x^2+1)y=0$
 A: Given differential equation $4x^2y^{\prime\prime}-8x^2y^{\prime}+(4x^2+1)y=0$. $P(x)=-2$ 
and $Q(x)=\frac{1+4x^2}{4x^2}$, since $Q(x)$ is not analytic at $x=0$, we say $x=0$ is a 
singular point of this differential equation . Singular points are futher classified into regular singular 
and irregular singular points. Since $xP(x)$ and $x^2Q(x)$ are analytic at $x=0$, we say $x=0$ is a regular singular point. So we can assume that $y=\displaystyle\sum_{n=0}^{\infty} a_n x^{n+m}$ for real number $m$ which we will find out.
Then $y^{\prime}=\displaystyle\sum_{n=0}^{\infty} a_n (n+m)x^{n+m-1}$ and $y^{\prime\prime}=\displaystyle\sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-2}$. We will substitute this in the given differential equation. We get,
$$(4a_0m(m-1)+a_0)x^m+(4a_1m(m+1)-8a_0m+a_1)x^{m+1}+\sum_{n=0}^{\infty}\left[ 4a_n+4a_{n+2}(n+m+1)(n+m+2)-8a_{n+1}(n+m+1)+a_n+2\right]x^{n+m+2}=0$$
$m=\frac{1}{2}$, $a_1=a_0$ and $a_{n+2}(4(n+m+2)(n+m+1)+1)-8a_{n+1}(n+m+1)=4a_n=0$
Put $m=\frac{1}{2}$ , the recussion relation is $a_{n+2} = \dfrac{1}{(n+2)^2}\left((2n+3)a_{n+1}-a_n\right)$. Now by induction you can prove that $a_n = \dfrac{a_0}{n!}$. Therefore $y_1 = \sqrt{x}\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!} = \sqrt{x}e^x$. Since the roots of the Indicial equation equation are equal, we cannot find the second Inependent solution using this method.  
It can be solved as $y_2 = y_1\displaystyle\int \dfrac{1}{y_1^2}e^{-\int p(x)dx}dx$. So $y_2 = \sqrt{x}e^x\displaystyle\int \dfrac{1}{xe^{2x}}e^{2x}dx = \sqrt{x}e^x\ln{|x|}$. Thefore our final answer wil be $y=c_1y_1+c_2y_2 = c_1\sqrt{x}e^x+c_2\sqrt{x}e^x\ln{|x|}$. Where $c_1$ and $c_2$ are arbitrary constants. 
Hope this helps :)     
