While studying power series and special functions in my textbook I encountered a problem which asked me to find the second independent solution besides the one from the frobenius power series (which in this case transformed to a regular power series soln as the order is zero).
Using a theorem given in my textbook that states the second solution is of the type $y_1 logx + x^p \sum_{n=0}^\infty c_n x^n$ I substituted this into the zero order bessel ODE and after multiple tries I am missing the $c_n$ by factor of 2. Note here while substituting I took
$p=0$(obviously) $c_0=0$ for simplicity.
Note:- I do realise that in further discussions on special functions there might be many other ways to evaluate the other solution near zero but currently I would appreciate solution the above type. Edit1:-Solved
Edit 2:- the answer involved substituting above form of solution into the zero order bessel ode finding a recursion equation for coefficients that pointed to odd coefficients being zero since $a1=0$ and the first solution is a frobenius series has only even powers of x. Then general Induction leads to the correct solution of second type.
You are looking for solutions to $$ xy''+y'+xy=0 $$ as the $n=0$ case of the general Bessel equation $x^2y''+xy'+(x^2-n^2)y=0$.
The coefficients for the power series satisfy ($a_n=0$ for $n<0$) $$ [n(n-1)+n]a_n+a_{n-2}=0 $$ so that with $a_0=1$ one gets the solution $$ y_1(x)=1-\frac14x^2+\frac1{64}x^4\mp\dots $$ Order reduction tells us that the second solution is of the form $y_2=uy_1$ where $$ x(y_1u''+2y_1'u')+y_1u'=0\implies \frac{u''}{u'}=-\frac{2xy_1'+y_1}{xy_1} \\ \implies \ln|u'|=-\ln|xy_1^2|+C\qquad\text{(set $C=0$)} \\ u'=\frac{1}{xy_1^2}=\frac1x+\frac12x+\dots $$ so that $$ y_2=\ln(x)y_1+\frac12x+\sum_{k=1}^\infty c_{2k+1}x^{2k+1} $$ Does that conform with your solution?