For the equation $ xy'' + 2xy' + 6e^xy = 0 $, I need to find the first 3 nonzero terms in each of two linearly independent solutions about x=0. I changed this to the form of $$x^2y''+x[xp(x)]y'+[x^2q(x)]y=0$$ yielding $$x^2y''+2x^2y'+6xe^xy=0$$ I found the first answer to be $$ y = x-4x^2+17/3x^3-47/12x^4 $$ which matches the books answer. I found this using that that $ F(r) = r(r-1) + pr + q $, where $xp(x)=\sum_{n=0}^{inf}p_nx^n$ and $x^2q(x)=\sum_{n=0}^{inf}q_nx^n$ This gave me that $ r_1 =1, r_2=0 $. I plugged into the indicial equation $$ F(r+n)a_n + \sum_{k=0}^{n-1} a_k((r+k)p_{n-k} + q_{n-k}) = 0 $$(given in the book) and then applied this to $$y_{1}(x)=x^{r_1}(1+\sum_{n=1}^{inf} a_{n}(r_1)x^n)$$ using the $a_n$'s that I got from the indicial equation.
This is where I am stuck. My book (Boyce and DiPrima) says that $$y_2=ay_1(x)ln(x)+x^{r_2}[1+\sum_{n=1}^{inf}c_n(r_2)x^n]$$
It says that you can plug the form of the series solutions for y into the original equation but I'm not sure what they mean by that. I tried to take this form and differentiate twice then plug back in; it quickly got very complicated and I doubted that was the correct way to proceed. Any help would be great. Thanks.
Thanks for the guidance. Here is where I am now:
So I started with my eqn for $y_2$ and found y' and y'':$$y'=a(y_1'lnx+y_1/x)+\sum_{n=1}^{inf}nC_nx^{n-1}$$ $$y''=a(y_1''lnx+2y_1'/x-y_1/x^2)+\sum_{n=1}^{inf}(n-1)(n)x^{n-2}$$
I then set about plugging these into the original equation: $$xy''=a(y_1''xlnx+2y_1'-y_1/x)+\sum_{n=1}^{inf}(n-1)(n)x^{n-1}$$ $$2xy'=2a(y_1'xlnx+y_1)+2\sum_{n=1}^{inf}nC_nx^n$$ $$6e^xy=6e^x(ay_1lnx+1)+\sum_{n=1}^{inf}c_n/(n-1)! x^{2n-1}$$ How do I deal with the 2n in the last term? Am I correct in assuming that after I figure that out, I consolidate into one statement within the summation and one outside, set both equal to 0, and find a recurrence relation? Also, am I missing an easier way to be doing this?
References and personal experience suggest that the original author of this question, despite this temporary stall, has the extraordinary brainpower to overcome the difficulties of this problem!