\(\displaystyle{x}^{{2}}{y}{''}+{\left({a}{x}-{b}\right)}{y}'-{a}{y}={0}\)

Start by removing the leading behavior for small x. Near x=0 the leading terms in the equation are

\(\displaystyle-{b}{y}'-{a}{y}\approx{0}\)

(this is valid as long as in the end, as \(\displaystyle{x}\to{0},{y}{''}\) does not grow as fast as \(\displaystyle{y}\frac{'}{{x}^{{2}}}\) or \(\displaystyle\frac{{y}}{{x}^{{2}}}\), does not grow as fast \(\displaystyle{\frac{{y}}{{x}}}\)).

The solution to the approximate equation is \(\displaystyle{y}={e}^{{-{\frac{{a}}{{b}}}{x}}}\) which motivates the substitution

\(\displaystyle{y}={u}{\left({x}\right)}{e}^{{-{\frac{{a}}{{b}}}{x}}}\)

\(\displaystyle{y}'={u}'{e}^{{-{\frac{{a}}{{b}}}{x}}}-{\frac{{a}}{{b}}}{u}{e}^{{-{\frac{{a}}{{b}}}{x}}}\)

\(\displaystyle{y}{''}={u}{''}{e}^{{-{\frac{{{a}}}{{{b}}}}{x}}}-{2}{\frac{{{a}{b}}}{{u}}}'{e}^{{-{\frac{{a}}{{b}}}{x}}}+{\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{u}{e}^{{-{\frac{{a}}{{b}}}{x}}}\)

and the differential equation becomes

\(\displaystyle{x}^{{2}}{u}{''}-{\left({2}{\frac{{{a}}}{{{b}}}}{x}^{{2}}-{a}{x}+{b}\right)}{u}'+{\left({\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{x}^{{2}}-{\frac{{{a}^{{2}}}}{{{b}}}}{x}+{a}-{a}\right)}{u}={0}\)

with u(0)=0

Now if we let \(\displaystyle{u}{\left({x}\right)}={\sum_{{0}}^{\infty}}{u}_{{k}}{x}^{{k}}\)

\(\displaystyle{\sum_{{0}}^{{\infty}}}{k}{\left({k}-{1}\right)}{u}_{{k}}{x}^{{k}}-{2}{\frac{{a}}{{b}}}{\sum_{{0}}^{\infty}}{k}{u}_{{k}}{x}^{{{k}+{1}}}+{a}{\sum_{{0}}^{{\infty}}}{k}{u}_{{k}}{x}^{{k}}\)-\(b\sum ku_kx^{k-1}+\frac{a^2}{b^2}\sum_0^\infty u_kx^{k+2}-\frac{a^2}{b}\sum_0^\infty u_kx^{k+1}=0\)

And in each term we can substitute an offset index to always get \(\displaystyle{x}^{{k}}\) this gives

\(\displaystyle{\left[{k}{\left({k}-{1}\right)}+{a}{k}\right)}{u}_{{k}}-{\left[{2}{\frac{{{a}}}{{{b}}}}{\left({k}-{1}\right)}+{\frac{{{a}^{{2}}}}{{{b}}}}\right]}{u}_{{{k}-{1}}}-{b}{\left({k}+{1}\right)}{u}_{{{k}+{1}}}+{\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{u}_{{{k}-{2}}}={0}\)

and this is to be solved with \(\displaystyle{u}_{{-{1}}}={u}_{{0}}={0}\)

there is on free parameter \(\displaystyle{u}_{{1}}\)

\(\displaystyle{u}_{{2}}={\frac{{{a}{u}_{{1}}}}{{{b}}}}\)

and so forth. So the solutions are this series, which is well behaved near zero, times \(\displaystyle{e}^{{-{\frac{{a}}{{b}}}{x}}}\)

All of the coefficients are proportional to \(\displaystyle{u}_{{1}}\). The overall solution has one free parameter and is then

\(\displaystyle{y}={u}_{{1}}{e}^{{-{a}\frac{{x}}{{b}}}}{\left[{x}+{\frac{{a}}{{b}}}{x}^{{2}}+{\frac{{23}}{{\frac{{{2}^{{2}}}}{{{b}^{{2}}}}}}}{x}^{{3}}+{\frac{{{5}{a}^{{3}}+{3}{a}^{{2}}}}{{{12}{b}^{{2}}}}}{x}^{{4}}+{\frac{{{17}{a}^{{4}}+{39}{a}^{{3}}+{36}}}{{{60}{b}^{{4}}}}}{x}^{{5}}+\ldots\right]}\)

\(\displaystyle={u}_{{1}}{\left[{x}+{\frac{{{a}^{{2}}}}{{{6}{b}^{{2}}}}}{x}^{{3}}+{\frac{{{a}^{{3}}+{3}{a}^{{2}}}}{{{12}{b}^{{3}}}}}{x}^{{4}}+{\frac{{{3}{a}^{{4}}+{16}{a}^{{3}}+{24}{a}^{{2}}}}{{{60}{b}^{{4}}}}}{x}^{{5}}+\ldots\right]}\)

As to whether this series is convergent for all positive x, or has a finite radius of convergence, that depends on the values of a and b.