Solve Bessel equation with order $\frac{1}{2}$ using Frobenius method Consider the Bessel equation of index $n= \frac{1}{2}$
$x^2y''(x) +xy'(x)+(x^2-\frac{1}{4})y(x) = 0$ $, x>0$             
$(i)$ Show that $y(x) = u(x)x^{\frac{-1}{2}}$ solves the equation above if and only if $u$ satiesfies a familiar differential equation.
$(ii)$ Find the general solution for the Bessel equation valid when $x>0$
$(iii)$ What condition(s) must the arbitrary constants in $(ii)$ satisfy if the corresponding solution of the Bessel equation is to be bounded on $(0,\infty)$
I solved $(ii)$ and found the solutions to be $y_1(x) =x^{\frac{-1}{2}}sin(x)$ and $y_2(x) = a_0x^{\frac{-1}{2}}cos(x)+a_1x^{\frac{-1}{2}}sin(x)$, where $a_0$ and $a_1$ are arbitrary.
But I'm not sure how to approach $(i)$ and $(iii)$, can anyone help me?
Thank you very much!

As Robert Israel told, just substitute $y(x) = u(x) x^{-1/2}$ in the differential equation and simplify. You will arrive to a simple expression. This end point (i).  
Once point (i) has been solved, solve the new differential equation for $u(x)$. The general solution is what you name y2. This end point (ii).   
For (iii), look at each of the pieces which compose y2. You know that $cos(x)$ and $sin(x)$ are finite and bounded for all values of $x$ and that $x^{-1/2}$ is defined for ???. Is there any problem when $x$ approaches $0$ ?   
I am sure you can take from here.
