Second-order linear differential equation of the form $x^2 y'' + (ax-b)y' - ay =0$ I need to solve the following differential equation
\begin{equation}
x^2 y'' + (ax-b)y' - ay =0
\end{equation}
with $a,b>0$, $x\geq 0$ and $y(0)=0$. The power series method will fail since there is a singularity at $x=0$, while the form of the equation does not conform with the Frobenius method. 
What other methods can I try in order to solve this?
 A: $$x^2 y'' +(ax-b)y'-ay=0
$$
Start by removing the leading behavior for small $x$.  Near $x=0$ the leading terms in the equation are
$$
-by'-ay \approx 0
$$
(this is valid as long as in the end, as $x\to 0$, $y''$ does not grow as fast as $y'/x^2$ or $y/x^2$, and $y'$ does not grow as fast as $\frac{y}x$).  
The solution to the approximate equation is $y=e^{-\frac{a}{b}x}$ which motivates the substitution
$$
y = u(x) e^{-\frac{a}{b}x} \\
y' = u' e^{-\frac{a}{b}x} - \frac{a}{b}ue^{-\frac{a}{b}x} \\
y'' = u''e^{-\frac{a}{b}x}-2\frac{a}{b}u'e^{-\frac{a}{b}x}+\frac{a^2}{b^2}ue^{-\frac{a}{b}x}
$$
and the differential equation becomes 
$$
x^2u'' -  \left( 2\frac{a}{b}x^2 -ax+b\right)u'
+\left( \frac{a^2}{b^2}x^2-\frac{a^2}{b}x+a-a \right)u=0
$$
with $u(0)=0$.
Now if we let $u(x) = \sum_0^\infty u_kx^k$
$$ \sum_0^\infty k(k-1)u_kx^k-2\frac{a}{b}\sum_0^\infty ku_kx^{k+1}
+ a\sum_0^\infty ku_kx^k-b\sum_0^\infty ku_kx^{k-1} \\
+\frac{a^2}{b^2}\sum_0^\infty u_kx^{k+2}-\frac{a^2}{b}\sum_0^\infty u_kx^{k+1}=0
$$
And in each term we can substitute an offset index to always get $x^k$; this gives 
$$
\left[ k(k-1) +ak\right]u_k  -\left[2\frac{a}{b}(k-1)+\frac{a^2}{b}\right]u_{k-1} -b(k+1)u_{k+1}+
\frac{a^2}{b^2}u_{k-2}=0
$$
and this is to be solved with $u_{-1}=u_0=0$, there is one free parameter $u_1$.
$$
u_2 = \frac{au_1}{b} \\
$$
and so forth.  So the solutions are this series, which is well behaved near zero, times $e^{-\frac{a}{b}x}$.
All of the coefficients are proportional to $u_1$.  The overall solution has one free parameter (a scale factor that we will continue to call $u_1$) and is then
$$
y =u_1 e^{-ax/b} \left[ 
x+\frac{a}{b}x^2+\frac23 \frac{2^2}{b^2} x^3+ \frac{5a^3+3a^2}{12b^2}x^4
+\frac{17a^4+39a^3+36}{60b^4}x^5 + \cdots\right]\\ =
u_1 \left[ x + \frac{a^2}{6b^2}x^3+\frac{a^3+3a^2}{12b^3}x^4 
+ \frac{3a^4+16a^3+24a^2}{60b^4}x^5+ \right]
$$
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 $b$.  
A: It has a trivial solution $y=ax-b$ .
Let $y=(ax-b)u$ ,
Then $y'=(ax-b)u'+au$
$y''=(ax-b)u''+au'+au'=(ax-b)u''+2au'$
$\therefore x^2((ax-b)u''+2au')+(ax-b)((ax-b)u'+au)-a(ax-b)u=0$
$x^2(ax-b)u''+2ax^2u'+(ax-b)^2u'+a(ax-b)u-a(ax-b)u=0$
$x^2(ax-b)u''+(2ax^2+(ax-b)^2)u'=0$
$x^2(ax-b)u''=-(2ax^2+(ax-b)^2)u'$
$\dfrac{u''}{u'}=-\dfrac{2a}{ax-b}-\dfrac{a}{x}+\dfrac{b}{x^2}$
$\ln u'=-2\ln(ax-b)-a\ln x-\dfrac{b}{x}+c$
$u'=\dfrac{Ce^{-\frac{b}{x}}}{x^a(ax-b)^2}$
$u=C_1+C_2\int^x\dfrac{e^{-\frac{b}{x}}}{x^a(ax-b)^2}~dx$
$\therefore y=C_1(ax-b)+C_2(ax-b)\int^x\dfrac{e^{-\frac{b}{x}}}{x^a(ax-b)^2}~dx$
A: With Laplace transforms this equation will be
$$s^2F''(s)+(4-a-b)sF'(s)+2(1-a)F(s)=0$$
which is Euler's equation.
