I get the following DE:
Plugging back I get:
Though I might have mistaken somewhere.
It seemed obvious to me that there should be solutions of the form , for some values of . The reason is that when you differentiate a power of y, it reduces the power by 1 (and when you differentiate twice, the power goes down by 2). So when you put in the expression , each term will be a multiple of either or .
In fact, if then and . Thus
If that is 0 for all y then (putting the coefficients of and equal to 0), you get , from which or ; and , from which or .