I am having trouble solving this problem and could use some hints, ideas or a solid walk through that I could use to clear up the foggy areas.
$$x^2y''+3xy'+(1+x)y=0$$
I have proven that $X_0=0$ is a regular singular point by showing that $~P(0)=0~$ and that the limit as $~x\to 0~$ for both $~p(o),~ q(o)~$ are finite.
I know that the indicial equation is then:
$r(r-1)+3r+1=0$
which gives $r_1=r_2=-1$
This is where I get a bit unsure
Because we have repeated roots, I believe we look for a solution $Y_1$ of the form:
$$X^{r_1} {\sum^{\infty}_1} A_n(r_1)X^n$$
but I have two different ideas about the method of finding $A_n$. I think we find $A_n$ as:
$$\frac{1}{F(r+n)} {\sum^{n-1}_{k=o}A_k[(r+k)P_{n-k}}+q_{n-k}]$$
and here:
$P_0=3$ $P_n$=0 all else $q_0=q_1=1$ $q_n$=o all else
which comes directly from the expansion of $xp(x)$ and $X^2q(x)$ matched to p and q.
My solution: $$X^{-1}[1+x-\frac{1}{4}x^2+\frac{1}{36}x^3]$$
book solution: $$\frac{1}{x}{\sum^{\infty}_0}\frac{(-1)^nx^n}{(n!)^2}$$
that is for $Y_1(x)$. I obviously made a book keeping error with negative signs while calculating my values of $A_n$ so that isn't a big deal.
my problem is with the solutions of $y_2(x)$ which the book gives as: $$y_1(x)ln(x)-\frac{2}{x}[1-{\sum^{\infty}_{n=1}} \frac{(-1)^nH_n}{(n!)^2}x^n]$$
I just can't get my answer to look anything remotely like the provided solution. Can anyone help me out?
I want to add that I added the Bessel function tag because this came from the Bessel function section of our HW. Perhaps I am supposed to apply those methods? It just doesn't seem of the correct form.
may i add that if you you realise that $$ x^2y'' + 3xy' + (1+x)y = x^2y'' + 2xy' + xy' + y + xy =0 $$ or $$ \dfrac{d}{dx}\left(x^2y'\right) + \dfrac{d}{dx}(xy) + xy = 0 $$ let $v = yx$ we find
$$ \begin{align} \dfrac{d}{dx}\left(x^2\left(\frac{v}{x}\right)'\right) + v' + v &=& -v' + (v'x)' + v' + v\\ &=& xv'' + v' + v = 0 \end{align} $$ thus try solving $$ x\dfrac{d^2v}{dx^2} + \dfrac{dv}{dx} + v = 0 $$ can you take it from here?