# 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.

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Does the $y(x^2)$ on the right side mean $y \cdot x^2$, or on the other hand that the function $y$ is evaluated at input $x^2$? – coffeemath Mar 7 '13 at 17:44
It is the function $y$ which is evaluated at input $x^2.$ – user52976 Mar 7 '13 at 17:46
The only polynomial that could fit such a description is quadratic. So try $y=ax^2+bx+c$ and see the resulting system of equations by setting the coefficients of both sides equal. – Maesumi Mar 7 '13 at 17:51

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).
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.