2nd order odes with non constant coeff. I am trying to solve these two DE: $ y''+(2x)/(1+x^2)y'+1/(1+x^2)^2y=0 $ and $ xy''-y'-4x^3y=0 $ and I am looking for methods on how to find the solutions. Should I go with the series method or is there a simpler way?  

Hint: 
(2)  You have that $y_1 = \text {sinh}\  x^2 $ is a solution of your DE. Write $$\begin{align}y_2 &= v\ \text{sinh}\ x^2 \\ y'_2 &= v'\ \text{sinh}\ x^2 + \ v\ (2x\ \text{cosh}\ x^2) \\ y''_2 & = v''\text{sinh}\ x^2 + 2v' (2x\ \text{cosh}\ x^2) + v(2\ \text{cosh}\ x^2+ 4x^2\ \text{sinh}\ x^2)  \  \end{align}$$
Plugging in the differential equation we have 
$$xv''\text{sinh}\ x^2 + [4x^2\ (2x\ \text{cosh}\ x^2)-\ \text{sinh}\ x^2]v' = 0$$
Then set $w=v'$ and make separation of variables we have 
$$\frac{dw}{w} = (\frac{1}{x} - 4x\ \frac{coshx^2}{sinhx^2})\ dx$$
You take it from here. 
You will find $w = v' = \frac{x}{sinh^2(x^2)}$ , because $ln$ is one-to-one. Then $v = -\frac{1}{2}coth(x^2)$.
The general solution is 
$$y(x) = c_1\ \text{sinh}\ x^2 + c_2\ \text{cosh}\ x^2$$
