I'm studying a course on ODEs and there is an example given where Legendre's equation is solved using reduction of order as follows:
$$(1-x^2)y''-2xy'+2y=0, \ \ \ P_1(x)=x \\ y(x)=xv, y' = v+xv',y''=2v'+xv'' \\\implies (1-x^2)(2v'+xv'')-2x(v+xv')+2xv=0 \\ \implies \frac{v''}{v'}=\frac{-2}{x}+\frac{1}{1-x}+\frac{1}{1+x}$$ after rearranging and solving the partial fraction decomposition. From here the integral is taken of both sides to give $$\log{v'} = -2\log{(x)}+\log{(1-x)}+\log{(1+x)} $$and it is stated that constants of integration do not matter at this point, which I do not understand. On exponentiation and integrating again the final answer is given as $$v=\frac{-1}{x}+\frac{1}{2}\log{\left(\frac{1+x}{1-x}\right)}\left[ + C \right] $$ Similarly here the constant is bracketed as if it is optional. If v was solved with constants of integration I believe the answer would be of the form $$ v=C_1\left[\frac{-1}{x}+\frac{1}{2}\log{\left(\frac{1+x}{1-x}\right)}\right]+C_2$$ Is this solution considered 'the same' because of being just a multiple of the previous solution (so not linearly independent)? Is it standard to ignore the constant of integration with this method or would it be safer to leave it in?
We get $v'=x^{-2}(1+x)(1-x)e^{C}$ if we add a constant $C$. When you integrate this you get another constant but the DE is not satisfied unless the second constant is $0$. So the solution we get is simply $C_1$ times the solution we get without adding a constant where $C_1=e^{C}$. It is obvious that if $y$ is a solution to the DE so is any constant multiple of it. So there is a no point in adding the constant during the process of finding $y$. We can always introduce the constant factor at the end.