I know this DE is solvable... I need help with a seemingly simple looking diff equ
$$
x\frac {d^{2}y} {dx^{2}}+2y=0
$$

$$
\rightarrow \frac {d^{2}y} {dx^{2}}+2\frac {y} {x}=0
$$
$v= (\frac {y} {x})$ substitution isn't working as it eventually shows:
$$
xv''+v'+2v=0
$$
which isn't any easier.  That variable is making my blood pressure go up.
I know I'm looking past something stupid. 
 A: Don't feel bad about not being able to solve this.  Many (maybe, most) easily expressed differential equations cannot be solved in terms of elementary functions.  In spite of this, standard existence and uniqueness theorems often guarantee the existence of solutions.  When the solution of an important class of differential equations cannot be expressed in terms of elementary functions, we give the solution a name, in this case, the Bessel functions.  Functions generated in this fashion are generally called special functions.
Many mathematical software packages know about special functions.  For example, if you type this differential equation into WolframAlpha, you'll get all kinds of information about the solution.
Text entered: x*y'' + 2y = 0
Output:

Note that the solution is expressed in terms of functions denoted $J_1$ and $Y_1$, called Bessel functions.  Also, in spite of the fact that there's no simple formula, several plots are generated.
A: I thought the wolfram alpha page on Bessel function lacked definition, therefore I will provide the full derivation of it, so lurkers like me can get review from time to time, and since some applications require the full notation.

Starting with a function similar to:
$$
y''+f(x)y'+g(x)y=0
$$
You're looking for two functions$f(x)$ and $g(x)$ that satisfies the differential equation and $y$ is the expression.  
Starting with $y$:
$$
y=x^{a}J_{\pm}(bx^{c}) = x^{a}J(z);  z=bx^{c}
$$
Taking derivatives of $x$, in which $J$ is a function of $z$, therefore implementing the chain rule:
$$
\frac {dy} {dx}=ax^{a-1}J+x^{a}(\frac {dJ} {dz})(\frac {dz} {dx})
$$
$$
\Rightarrow \frac {dy} {dx}=ax^{a-1}J+x^{a}(\frac {dJ} {dz})bcx^{c-1}
$$
$$
\Rightarrow \frac {dy} {dx}=ax^{a-1}J+bcx^{a+c-1}(\frac {dJ} {dz})
$$
Now taking the second derivative
$$
\Rightarrow \frac {d^{2}y} {dx^{2}}=a(a-1)x^{a-2}J+bcx^{a+c-2}(2a+c-1)\frac {dJ} {dz} + b^{2}c^{2}x^{a+2c-2}\frac {d^{2}J} {dz^{2}}
$$
Now replace $\frac {d^{2}} {dz^{2}}$ in the last equation into the differential equation
$$
J''=-(\frac {1} {z}J'+(1-\frac {m^{2}} {z^{2}})J)
$$
Now determine$f(x)$ and $g(x)$ such that
$$
y''+f(x)y'+g(x)y=0
$$
$$
\Rightarrow [a(a-1)x^{a-2}J+bcx^{a+c-2}(2a+c-1)\frac {dJ} {dz} + b^{2}c^{2}x^{a+2c-2}\frac {d^{2}J} {dz^{2}}]+f(x)[ax^{a-1}J+bcx^{a+c-1}(\frac {dJ} {dz})]+g(x)[x^{a}J(z)]=0
$$
$$
\Rightarrow [a(a-1)x^{a-2}J+bcx^{a+c-2}(2a+c-1)\frac {dJ} {dz} + b^{2}c^{2}x^{a+2c-2}[-\frac{1}{bx^{c}}(\frac {dJ} {dz})-(1-\frac {m^{2}}{bx^{c}})J]+f(x)[ax^{a-1}J+bcx^{a+c-1}(\frac {dJ} {dz})]+g(x)[x^{a}J(z)]=0
$$
And group all of the terms into two term, ie:
$$
()\frac {dJ} {dz}+()J=0
$$
With each set $()=0$
This gives:
$$
f(x)=\frac {(1-2a)} {x}
$$
and
$$
g(x)=-(\frac {c^{2}m^{2}-a^{2}} {x^{2}})+b^{2}c^{2}x^{2c-2}
$$

And thus, once simplified, substitution reveals the famous Bessel Function:
$$
\therefore \frac {d^{2}y} {dx^{2}}+\frac {1-2a} {x}\frac {dy} {dx}-[(\frac {c^{2}m^{2}-a^{2}} {x^{2}})+b^{2}c^{2}x^{2c-2}]y=0
$$
with solutions:
$$
y=k_1x^{a}J_m(bx^{c}+k_2x^{a}(bx^{c})
$$

Therefore, given the DE:
$$
xy''-2y=0
$$
comparing the Bessel Function, we can see that
$$
1-2a=0; 
$$
$$
b^{2}c^{2}=2;
$$
$$
c^{2}m^{2}-a^{2}=0;
$$ 
$$
2c-2=-1
$$

Set up a system of equations using these, solve for the constants, and plug into the Bessel solution: $y=k_1x^{a}J_m(bx^{c}+k_2x^{a}(bx^{c})$, for the answer
