Polynomial solution of an equation We consider the IVP $$y
'' −2xy' + 4y = 0, y(0) = −3, y'
(0) = 0$$
(the unknown is denoted by y(x)). Prove that the solution of this IVP is a
polynomial function. Can somebody give me some ideas, please?

Well, one can always try a polynomial of degree $2$, a quadratic:
$y = ax^2 + bx + c; \tag 1$
$c = y(0) = -3; \tag 2$
$b = y'(0) = 0; \tag 3$
then
$y = ax^2 - 3; \tag 4$
$y' = 2ax; \tag 5$
$y'' = 2a; \tag 6$
thus,
$y'' - 2xy' + 4y = 2a - 4ax^2 + 4ax^2 - 12 = 2a - 12 = 0, \tag 7$
and so
$a = 6, \tag 8$
whence
$y = 6x^2 - 3. \tag 9$
Check:
$y' = 12x, \tag{10}$
$y'' = 12, \tag{11}$
$12 - 2x(12x) + 4(6x^2 - 3) = 0✓. \tag{12}$
Note that
$y(0) = -3, \; y'(0) = 0. \tag{13}$
