The number of non-trivial polynomial solutions of the differential equation $x^3y'(x)=y(x^2)$ I came across the following problem that says:   

The number of non-trivial polynomial solutions of the differential equation  $x^3y'(x)=y(x^2)$ is which of the following?
  $(1)0\space (2)1 \space (3)3 (4)\infty.$   

Can someone point me in the right direction? Thanks in advance for your time.
 A: For a polynomial of order $n$, this reads:
$$3+(n-1)=2n \to n=2$$
Write this out:
$$2ax^4+bx^3=ax^4+bx^2+c \to ax^4+bx^2(x-1)-c=0$$
So, the answer is zero, since no such non-trivial polynomials exist (non-trivial forth degree polynomials have at most 4 real roots).
A: Well, if you proceed using the ansatz (thanks @user33640 for the "correction")
$$
y(x) = \sum_{n=0}^N a_n x^n
$$
and substitute it directly in the equation, you end up with
$$
x^3 \sum_{n=1}^N n a_n x^{n-1} = \sum_{n=0}^N a_n x^{2n}.
$$
This means that
$$
\sum_{n=1}^N n a_n x^{n+2} - \sum_{n=0}^N a_n x^{2n} = 0
$$
which tells you that all $a_n = 0$ for $n=2m+1$. This in turn help us to propose an improved ansatz
$$
y(x) = \sum_{n=0}^N b_{2n} x^{2n}.
$$
Again, substituting in the ode, we have
$$
\sum_{n = 1}^N 2n b_{2n} x^{2n+2} = \sum_{n=0}^N b_{2n} x^{4n}
$$
Now, the only way to this equation to be satisfied, is if 
$$
2N + 2 = 4N \quad \Longrightarrow \quad N=1
$$
Then $y(x) = b_0 + b_2 x^2$ implies that
$$
2 b_2 x^4 = b_0 + b_2 x^4
$$
which means $b_0 = b_2 = 0$ hence no non-trivial solution exist.
