Constants at the end of the Frobenius method

[ABearon]

TL;DR

are c1 and c2 just random coefficients

I'm having a hard time grasping the concept of reducing the two recursive relations at the end of the frobenius method.

For example, 2xy''+y'+y=0
after going through all the math i get
y1(x) = C1[1-x+1/6*x^2-1/90*x^3+...]
y2(x) = C2x^1/2[1-1/3*x+1/30*x^2-1/630*x^3+...]

I know those are right, and I know we solve for what's inside the bracket by taking a C0 out. I'm just trying to clarify that, since c0 is different for each term and since it is arbitrary we can just write c1 for y1 and c2 for y2. I want to make sure this is where the c1 and c2 come from and not from trying to take out a c1 in the y1 brackets and c2 in the y2 brackets.

 

[fresh_42]

The constants only depend on the initial or boundary conditions. A certain solution is a certain flow through a vector field: different initial conditions make different flows although the vector field does not change.

Example: $\left(y'\right)^2 -4y = 0$ has the solution $y=x^2$. But this is only half the truth. All functions $y(x)=x^2+c$ are solutions, too. And we find the value of $c$ by knowing some actual value of $y(x_0)$. More or higher derivatives result in more initial conditions. The point $x_0$ is often chosen to be $x_0=0$ but could by any other. Say in my example we have $y(-3)=16$, then $y(-3)=(-3)^2+c=16$ which get's us $c=7$.