Question

Q:1 Verify that
$y = {x}^{2} sin \: x \: \: and \: y \: = 0 \: \\ \: both \: are \: solutions \: of \: the \: \\ initial \: value \: problem \: .$

(x^2)y''-4xy' + (x^2+6)y = 0
.
Does it Contradict the uniqueness.
and y(0) = y'(0) = 0.

Answer 1

Final Answer : Yes, it contradict the uniqueness.

Uniqueness is valid for (y"+ p(x)y'+q(x) = r (x))

where p(x) , q(x) , r(x) are continuous in an interval containing the point x(0) where the initial condition is satisfied.

Here, p(x) and q(x) are not defined at x = 0.
So, it contradicts the uniqueness.

Hope, you understand my answer and it may helps you.