The number of non-trivial polynomial solutions of the differential equation $x^3y'(x)=y(x^2)$

Question

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.

Answer 1

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).

Answer 2

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.