You solve an ODE to give you a function of the form you see with $C1, C2$ these are integration constants - you will then need the initial conditions and or boundary conditions to specify the constants that yield the specified IC. So ODE => Solve ODE => Fix parameters

the question of different coefficient values given the same IC comes down to uniqueness of your solution. Depends on what you mean, a limit cycle can have the same final form (shape of solution after some time period) regardless of initial conditions (subject to criterion for limit cycle behaviour)

yes fix parameters/coefficients based on the IC values - as the function can take any form. It is only once you fix the integration constant do you satisfy your IC (now not all IC can be satisfied by your function) but then you go back the drawing board then :)

The coefficients do not "vary from one case to the other". They must satisfy both equations simultaneously. The solution is $c_1=-2/9,c_2=2/9.$

$y(0)=0$ gives $C_1+C_2=\frac 12$ to me, not $C_1+C_2=0$.

@student91 it is a different ODE solution he is using as an example.

@Chinny84 oops, thanks!

@Luthier415Hz I saw that but I am talking about the second example.

I think you are confusing a few things - but the solution until you specify the IC can only be written in terms of coefficients. Once you pick ICs you then fix what the coefficients are. If you have too many ICs e.g. y'' = k then you may run into a problem as you have too many constraints for too few coefficients so you may not satisfy all the ICs.

Yes "the second example DOES satisfy the IC", and so will the first one if you correct your computation: $C_1=0,C_2=1/2.$

See also the picard lindelof theorem for why solutions exist. In general, if you can write your equation as $a(t)y''+b(t)y'+c(t)y=d(t)$, this problem can be written as a 2-dimensional single-order ODE and thus a unique solution exists onder certain conditions when y(0) and y'(0) are given.

You corrected your values of $C_1,C_2$ in the first example, but you forgot to erase "Then I see it does not satisfy the IC"

Still, why do you (in both examples) "assume that it does satisfy the IC"? Just check!

@Chinny84 one thing stroke me however, if a solution to an ODE is cos2x , which can never satisfy y(0)=0, then we have an example that does not satisfy an IC?

@Luthier415Hz True, but typically you specify IC because of the model (could be a Physical model). IC fixes the model, but the IC themselves have to mean something - an example, if we are are integrating a function with singularities then having a definite integral with that singularity between the bounds of integration is unlikely to be informative.

This is why I was not sure about the "simplicity" of this problem