Solve differential equation: $y''-\dfrac{1}{x}y'+\dfrac{\alpha^2}{x^2}y=0$ The Bessel differential equation can be written like this $$y''+\dfrac{1}{x}y'+(1-\dfrac{\alpha^2}{x^2})y=0$$ and one of the linearly independent solutions is the bessel function of the first kind, $J_\alpha(x)$.
If we had the equation $$y''+\dfrac{1}{x}y'+(k^2-\dfrac{\alpha^2}{x^2})y=0$$ on of the independent solutions would be $J_\alpha(kx)$
Plugging $k = 0$ in the last equation would give me almost my differential equation:$$y''+\dfrac{1}{x}y'-\dfrac{\alpha^2}{x^2}y=0$$ But I am not sure how to handle the changes of sign. 
 A: Differential equations of this form are called Euler differential equations, and they can usually be transformed into equations with constant coefficients via exponential substitution (you can read more here)
If we use the substitution - $x=e^t\Rightarrow t=log(x)\Rightarrow y'=\frac{1}{x}y_t,y''=-\frac{1}{x^2}y_t+\frac{1}{x^2}y_{tt}$ we get an equation with constant coefficients:
$y_{tt}-2y_t+\alpha^2=0$
Can you take it from here?
A: This differential equation is of the Euler-Cauchy type. Use the ansatz $y = x^r$ to obtain
$$r(r-1)x^{r-2}+rx^{r-2}-\alpha^2x^{r-2}=0$$
If we assume $x\neq 0$ we obtain 
$$r(r-1)+r-\alpha^2=0.$$
This equation has two roots $r_{1,2}$ the general solution is given by
$$y = c_1x^{r_1} + c_2x^{r_2}.$$
A: Hint: Multiplying by $$x^2$$
$$x^2y''-xy'+\alpha^2y=0$$ and make the Ansatz:$$y=e^{\lambda x}$$
so you get $$\lambda_{1,2}=\pm\sqrt{1-\alpha^2}+1$$
Can you finish?
For your work: the solution is given by $$y \left( x \right) ={\it \_C1}\,{x}^{1+\sqrt {-{\alpha}^{2}+1}}+{\it 
\_C2}\,{x}^{1-\sqrt {-{\alpha}^{2}+1}}
$$
