Can, y=sin(x^2) be a solution to,
y''+py'+qy=0
With initial conditions, y(0)=A and y'(0)=B
Where p and q are continous in an open interval containing 0?
I solved it, thank you anyways.
If you are interested how it was solved:
The equation,
y''+py'+qy=0
Has basis of dimension 2 if p,q are continous in an neigborhood of x=0.
Thus, there exists two linearly independent solutions that span the set. One is given as sin(x^2). Let the other linearly independent solution be y.
Then we compute the Wronskian,
[sin(x^2) y]
[2xcos(x^2) y']
Thus,
y'sin(x^2)-2xycos(x^2)
But at x=0 the Wronskian is zero.
Thus, there does not exist a linearly independant solution with sin(x^2), a contradiction.