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?
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$$a_n=(-1)^n \frac{a_0}{\prod_{i=0}^{n-1} (3i+5)}$$
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?
Best Answer
Hint: if you replace $y(x)$ with $x^{-1/3}g(2\sqrt{x})$, then $g$ satisfies a Bessel differential equation.