How to know if two power series solutions are linearly independent? I'm currently studying power series to solve differential equations. I would like to know if there's a way to tell whether two solutions are linearly independent or not. I think evaluating the Wronskian would be a tough task because the series may not be expressed in $\sum$ notation.
For instance if I'm working on a power series centered at a regular singular point, Frobenius theorem provides me a solution $y_1(x)$. If I were to find another one in the form 
$$y_2(x)=y_1(x)\int\frac{e^{-\int P(x)\,dx}}{y_1^2(x)}\,dx$$
I know that this solution is independent because I created it to be independent. However if I solve the system and find "two" solutions, how do I know if these solutions are  linearly independent?  I appreciate your thoughts.
 A: Edit: After clarifying in chat, it seems there are two questions here regarding finding power series solutions to second order, linear ODE.

  
*
  
*Will the two solutions I get always be linearly independent?
  
*After I've found two solutions, how can I check whether or not they are linearly independent?
  

Answer to Question 1:
When using the Method of Frobenius to obtain a power series solution of a second order linear ODE in the neighborhood of a regular singular point, in can happen that you end up finding only one linearly independent power series solution. (You do find two solutions, but the second is a constant multiple of the first.) 
This happens when solving Bessel's equation
$$
x^2y''(x)+xy'(x)+(x^2-\mu^2)y(x)=0
$$
about $x=0$. It is a long and tedious calculation, too long to type out, but you can study it here in Example 5.7.2 starting on page 345.
However, you can always fall back on the method you alluded to above, often called reduction of order/variation of constants, to find a second linearly independent solution $y_2$ once you have any one solution $y_1$. That applies the Bessel's equation as well.
Answer to Question 2:
Typically we would compute the Wronskian of $y_1$ and $y_2$, $W(y_1(x),y_2(x))(x)$ and appeal to the fact that $y_1$ and $y_2$ are linearly independent iff the Wronskian in nonzero. When $y_1$ and $y_2$ are power series this looks to be a cumbersome calculation. However, let's appeal to the full power of the Wronskian statement regarding linear independence:

If $W(y_1,y_2)(x_0)\not=0$ for some $x_0\in I$, then $y_1(x)$ and
  $y_2(x)$ are linearly independent on $I$.

Case 1: Ordinary points
Now just pick $x_0$ judiciously: say, $x_0$ is the ordinary point about which we found the two series solutions
$$
y_1(x)=\sum_{n=0}^\infty a_n(x-x_0)^n, \quad y_2(x)=\sum_{n=0}^\infty b_n(x-x_0)^n, 
$$
Then 
$$
W(y_1,y_2)(x_0)=y_1(x_0)y_2'(x_0)-y_2(x_0)y_1'(x_0)=a_0b_1-b_0a_1
$$
tells the tale (whether this is zero or not).
Case 2: (Regular) singular points
This takes quite a bit of typing. I will punt by pointing you here, page 14. If someone has a better "preserved" source than this and/or wants to edit my answer to give the full details in this case, please do.
