TITLE: Find solutions of the differential equation $3x^2y''+5xy'+3xy=0$.
QUESTION [7 upvotes]: Find all the solutions of the form $y(x)= x^m \sum_{n=0}^{\infty} a_nx^n, \ x>0 (m \in \mathbb{R})$ of the differential equation $3x^2y''+5xy'+3xy=0$.
That's what I have tried:
Since $x>0$ the differential equation can be written as follows.
$$y''+ \frac{5}{3x}y'+ \frac{1}{x}y=0$$
$$p(x)=\frac{5}{3x}, q(x)= \frac{1}{x}$$
The point $0$ is regular singular, i.e. the fuctions $xp(x), x^2q(x)$ can be written as power series at a region of $0$.
We are looking for solutions of the form $y(x)=x^m \sum_{n=0}^{\infty} a_n x^n$ for a suitable $m \in \mathbb{R}$ and for suitable $a_n \in \mathbb{R}$ and for $x \in (0,R)$ where $R$ is a suitable positive number.
Then we have:
$$y'(x)= \sum_{n=0}^{\infty} a_n (n+m) x^{n+m-1} \\ \Rightarrow xy'=\sum_{n=0}^{\infty} a_n (n+m) x^{n+m} \\ \Rightarrow 5xy'=\sum_{n=0}^{\infty} 5a_n (n+m) x^{n+m}$$
and
$$y''(x)= \sum_{n=0}^{\infty} a_n (n+m)(n+m-1) x^{n+m-2} \\ \Rightarrow x^2y''=\sum_{n=0}^{\infty} a_n (n+m)(n+m-1) x^{n+m} \\ \Rightarrow 3x^2y''=\sum_{n=0}^{\infty} 3a_n (n+m)(n+m-1) x^{n+m}$$
$$3xy= \sum_{n=0}^{\infty} 3a_n x^{n+m+1}$$
So it has to hold the following:
$$\sum_{n=0}^{\infty} \left[ 3a_n (n+m)(n+m-1) x^{n+m}+5a_n (n+m) x^{n+m}+ 3a_n x^{n+m+1}\right]=0 \Rightarrow \sum_{n=0}^{\infty} \left[ 3a_n (n+m)(n+m-1) +5a_n (n+m) + 3a_n x\right]x^{n+m}=0 $$
So it has to hold that:
$$a_n=\frac{-3a_{n-1}}{3m+3n+2}, \forall n=1,2,3, \dots$$
EDIT:  I am looking again at the exercise. For $m=0$ I got the following:
$$a_1=-\frac{3a_0}{5} \\ a_2=\frac{3^2 a_0}{5 \cdot 8} \\ a_3=-\frac{3^3 a_0}{5 \cdot 8 \cdot 11} \\ a-4= \frac{3^4 a_0}{5 \cdot 8 \cdot 11 \cdot 14}$$
So isn't for $m=0$ the general formula for $a_n$ the following?
$$$$
$$a_n=(-1)^n \frac{a_0}{\prod_{i=0}^{n-1} (3i+5)}$$
And for $m=-\frac{3}{2}$ isn't the formula for $a_n$ the following?
$$a_n=(-1)^n \frac{3^n a_0}{ \prod_{i=0}^{n-1} \frac{(6i+1)}{2}}$$
If so, then could we say the following?
$$y_1(x)= x^0 \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{\prod_{i=0}^{n-1}(3i+5)}$$
and
$$y_2(x)=x^{-\frac{3}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n 3^n}{ \prod_{i=0}^{n-1} \frac{(6i+1)}{2}}x^n$$
are solutions of the differential equation for $a_0=1$.
$$\left| \frac{\frac{(-1)^{n+1} x^{n+1}}{\prod_{i=0}^n (3i+5)}}{\frac{(-1)^{n} x^n}{\prod_{i=0}^{n-1} (3i+5)}}\right|=\left| \frac{x}{3n+5}\right| \to 0<1$$
Do we deduce from the latter that the radius of convergence is $+\infty$. If so, do we continue as follows?
Similarly we show that the radius of convergence of $y_2(x)$ is $+\infty$.
$$$$
$y_1, y_2$ are linearly independent in $(0,+\infty)$. Because if $c_1, c_2 \in \mathbb{R}$ with $c_1y_1(x)+c_2y_2(x)=0 \forall x \in (0,+\infty)$ then since $c_1 y_1(x)+ c_2y_2(x)$ is a power series with radius of convergence $+\infty$ we have $0= c_1 y_1(x)+c_2y_2(x)= \sum_{n=0}^{\infty} d_n x^n$ for some $d_n \in \mathbb{R}$ and thus $d_n=0 \forall n=0,1,2, \dots$
However $d_0=c_1=0$ and $d_1=-\frac{3}{5} c_2=0 \Rightarrow c_2=0$.
Thus, the general solution of the differntial equation is:
$$y(x)=c_1 \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{\prod_{i=0}^{n-1}(3i+5)}+ c_2 x^{-\frac{3}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n 3^n}{ \prod_{i=0}^{n-1} \frac{(6i+1)}{2}}x^n, c_1, c_2 \in \mathbb{R}$$
EDIT:
I remade the calculations for $m=0$ and now I got the following:
$$$$
For $n=1$: $a_1=-\frac{3a_0}{1 \cdot 5}$
$$$$
For $n=2$: $a_2=\frac{3^2 a_0}{2 \cdot 5 \cdot 8}$
$$$$
For $n=3$: $a_3=-\frac{3^3 a_0}{2 \cdot 3 \cdot 5 \cdot 8 \cdot 11}$
$$$$
For $n=4$: $a_4=\frac{3^4 a_0}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 8 \cdot 11 \cdot 14}$
$$$$
Are my current calculations right or were the previous one correct?
$$$$
If they are right how could we write the formula of $a_n$ without the use of the Gamma function?
EDIT:I retried it again. Couldn't we write the general formula for $a_n$ when $m=0$ as follows? $$$$
$$a_n=\frac{(-1)^n a_0}{n! \prod_{i=1}^n (3i+2)}$$
Or am I wrong?
Also it should be $m_2=-\frac{2}{3}$. Or am I wrong?
If it is like that isn't the general formula for $a_n$ in this case the following?
$$a_n=(-1)^n \frac{a_0}{n! \prod_{i=0}^{n-2} (2 \cdot 2+3 \cdot i)}$$
Or am I wrong?

REPLY [4 votes]: I think  your elaboration provided in your question   together with some of your comments is  quite near to a correct solution.

Let's calculate $y_1(x)$.
We start with $a_n$ which was correctly specified in one of your comments:
\begin{align*}
a_n=\frac{-3a_{n-1}}{(m+n)(3m+3n+2)}\qquad\forall  n=1,2,3,\ldots
\end{align*}
  Setting $m=0$ we obtain
  \begin{align*}
a_n&=\frac{-3a_{n-1}}{n(3n+2)}\\
&=\frac{-a_{n-1}}{n\left(n+\frac{2}{3}\right)}\\
&=\frac{a_{n-2}}{n(n-1)\left(n+\frac{2}{3}\right)\left(n-\frac{1}{3}\right)}\\
&=\ldots\\
&=\frac{(-1)^na_0}{n!\left(n+\frac{2}{3}\right)\left(n+\frac{2}{3}-1\right)\cdots\left(n+\frac{2}{3}-(n-1)\right)}\\
&=\frac{(-1)^na_0}{n!{\left(n+\frac{2}{3}\right)}_n}\tag{1}\\
&=(-1)^n\frac{a_0}{n!}\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(n+\frac{5}{3}\right)}\tag{2}
\end{align*}

In (1) we   apply the  Pochhammer symbol $(x)_n=x(x-1)\cdots (x-n+1)$ and in (2) we use the identity 
$$(x)_n=\frac{\Gamma(x+1)}{\Gamma(x-n+1)}$$

So, we find the solution
$$y_1(x)=a_0\Gamma\left(\frac{5}{3}\right)\sum_{n=0}^{\infty}\frac{(-1)^n}{\Gamma\left(n+\frac{5}{3}\right)}\frac{x^n}{n!}$$

According to a comment of @JackDAurizio let's have a look at the Bessel function of the  first kind

\begin{align*}
J_\alpha(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\alpha+1)}\left(\frac{x}{2}\right)^{2n+\alpha}
\end{align*}
We observe
\begin{align*}
x^{-\frac{1}{3}}J_{\frac{2}{3}}(2\sqrt{x})=x^{-\frac{1}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma\left(n+\frac{5}{3}\right)}x^{n+\frac{1}{3}}
=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma\left(n+\frac{5}{3}\right)}x^{n}
\end{align*}
We conclude
\begin{align*}
y_1(x)=a_0\Gamma\left(\frac{5}{3}\right)x^{-\frac{1}{3}}J_{\frac{2}{3}}(2\sqrt{x})
\end{align*}

Similarly we can find
$$y_2(x)=a_0\Gamma\left(\frac{1}{3}\right)x^{-\frac{1}{3}}J_{-\frac{2}{3}}(2\sqrt{x})$$

and the general solution is therefore
  $$y(x)=c_1\Gamma\left(\frac{5}{3}\right)x^{-\frac{1}{3}}J_{\frac{2}{3}}(2\sqrt{x})+c_2\Gamma\left(\frac{1}{3}\right)x^{-\frac{1}{3}}J_{-\frac{2}{3}}(2\sqrt{x})$$


[2015-06-04]: Note according to OPs comment

Avoiding the Gamma function $\Gamma(x)$ a representation via the Pochhammer symbol (1) is convenient:
$$a_n=\frac{(-1)^na_0}{n!{\left(n+\frac{2}{3}\right)}_n}$$

We observe for $n=3$:
\begin{align*}
a_3&=-\frac{3^3a_0}{2\cdot3\cdot5\cdot8\cdot11}\\
&=-\frac{3^3a_0}{3!\cdot5\cdot8\cdot11}\\
&=-\frac{a_0}{3!\cdot\frac{5}{3}\cdot\frac{8}{3}\cdot\frac{11}{3}}\\
&=-\frac{a_0}{3!\cdot\left(3-\frac{4}{3}\right)\left(3-\frac{1}{3}\right)\left(3+\frac{2}{3}\right)}\\
&=-\frac{a_0}{3!\left(3+\frac{2}{3}\right)_3}
\end{align*}
Similarly the case $n=4$:
\begin{align*}
a_4&=\frac{3^4a_0}{2\cdot3\cdot4\cdot5\cdot8\cdot11\cdot14}\\
&=\frac{3^4a_0}{4!\cdot5\cdot8\cdot11\cdot14}\\
&=\frac{a_0}{4!\cdot\frac{5}{3}\cdot\frac{8}{3}\cdot\frac{11}{3}\cdot\frac{14}{3}}\\
&=\frac{a_0}{4!\left(4-\frac{7}{3}\right)\left(4-\frac{4}{3}\right)\left(4-\frac{1}{3}\right)\left(4+\frac{2}{3}\right)}\\
&=\frac{a_0}{4!\left(4+\frac{2}{3}\right)_4}
\end{align*}

We obtain the following representation for $y_1(x)$:
  \begin{align*}
y_1(x)&=x^m\sum_{n=0}^{\infty}a_nx^n\\
&=x^0\sum_{n=0}^{\infty}\frac{(-1)^na_0}{n!{\left(n+\frac{2}{3}\right)}_n}x^n\\
&=a_0\sum_{n=0}^{\infty}\frac{(-1)^n}{{\left(n+\frac{2}{3}\right)}_n}\frac{x^n}{n!}
\end{align*}


[2015-06-04]: Additional note according to OPs comments

Since
\begin{align*}
\prod_{j=1}^{n}(3j+2)&=5\cdot8\cdot11\cdot\ldots\cdot(3n+2)\\
&=(3\cdot1+2)(3\cdot2+2)(3\cdot3+2)\cdot\ldots\cdot(3n+2)\\
&=3^n\left(1+\frac{2}{3}\right)\left(2+\frac{2}{3}\right)\left(3+\frac{2}{3}\right)\cdot\ldots\cdot\left(n+\frac{2}{3}\right)\\
&=3^n\left(n+\frac{2}{3}\right)_n
\end{align*}
  it's perfectly valid, to write
  \begin{align*}
a_n=\frac{(-1)^n3^na_0}{n!\prod_{j=1}^{n}(3j+2)}=\frac{(-1)^na_0}{n!\left(n+\frac{2}{3}\right)_n}
\end{align*}

Note: Usually,  it's more convenient to use the Pochhammer symbol, since consecutive factors have the nice difference $1$ and we do not need to write the product symbol together with factor $3^n$. It's similar to the situation that we prefer to write $n!$ instead of $\prod_{j=1}^nj$.

When looking at the coefficient
  \begin{align*}
a_n=\frac{-3a_{n-1}}{(n+m)(3n+3m+2)}\qquad n\geq 1,m\in \mathbb{R}
\end{align*}
  calculated by OP in one of his comments, we observe that $m=-\frac{2}{3}$ is a proper choice besides $m=0$ in order to get a second linearly independent solution, since in this case we have the simple recurrence relation
  \begin{align*}
a_n&=\frac{-3a_{n-1}}{(n-\frac{2}{3})3n}=\frac{-a_{n-1}}{(n-\frac{2}{3})n}\\
&=\frac{a_{n-2}}{\left(n-\frac{2}{3}\right)\left(n-\frac{5}{3}\right)n(n-1)}\\
&=\ldots\\
&=\frac{(-1)^na_0}{\left(n-\frac{2}{3}\right)\left((n-1)-\frac{2}{3}\right)\cdot\ldots\cdot\left((n-(n-1))-\frac{2}{3}\right)n!}\\
&=\frac{(-1)^na_0}{\left(n-\frac{2}{3}\right)\left(n-\frac{5}{3}\right)\cdot\ldots\cdot\left(\frac{1}{3}\right)n!}\\
&=\frac{(-1)^na_0}{n!\left(n-\frac{2}{3}\right)_n}\tag{3}
\end{align*}
  Of course, we can also use the product symbol in (3) instead of the Pochhammer symbol
  \begin{align*}
\left(n-\frac{2}{3}\right)_n&=\prod_{j=0}^{n-1}\left(n-j-\frac{2}{3}\right)=\prod_{j=0}^{n-1}\left(n-(n-1-j)-\frac{2}{3}\right)\\
&=\prod_{j=0}^{n-1}\left(j+\frac{1}{3}\right)=\frac{1}{3^n}\prod_{j=0}^{n-1}\left(3j+1\right)
=\frac{1}{3^n}\prod_{j=1}^{n}\left(3j-2\right)\\
\end{align*}

So, we get $y_2(x)$
\begin{align*}
y_2(x)&=a_0x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n}{\left(n-\frac{2}{3}\right)_n}\frac{x^n}{n!}\\
&=a_0x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n3^n}{\prod_{j=1}^{n}\left(3j-2\right)}\frac{x^n}{n!}
\end{align*}

Conclusion: Alternately to the representation of $y(x)$ via Bessel functions of the first kind we obtain as general solution for $y(x)$:
\begin{align*}
y(x)&=
c_0\sum_{n=0}^{\infty}\frac{(-1)^n}{{\left(n+\frac{2}{3}\right)}_n}\frac{x^n}{n!}+c_1x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n}{\left(n-\frac{2}{3}\right)_n}\frac{x^n}{n!}\\
&=c_0\sum_{n=0}^{\infty}\frac{(-1)^n3^n}{\prod_{j=1}^{n}(3j+2)}\frac{x^n}{n!}+c_1x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n3^n}{\prod_{j=1}^{n}\left(3j-2\right)}\frac{x^n}{n!}\qquad c_0,c_1\in\mathbb{R}\\
\end{align*}