Find a differential equation that $y$ satisfies. The ODE should not contain any arbitrary constants.
$$y(x)=C_1(x^2-1)+C_2x$$
I proceeded with finding the two derivatives (since it is second-order):
$$y'(x)=2C_1x+C_2$$
$$y''(x)=2C_1$$
Then, using the general formula $y''+Ay'+By = 2C_1+A2C_1x+AC_2+BC_1x^2-BC_1+BC_2x$
Which results in $C_1(2+2Ax+Bx^2-B)+C_2(A+Bx)$
Solving the systems of equations, I found that $A=-2x$, $B=2$. Which means the ODE required is $$y''-2xy'+2y=0$$
Can anyone kindly advise if this approach is correct? Thanks.
Edit: Sorry, I found a computational error afterwards. Hence,
$$A=-\frac{2x}{x^2+1}$$
$$B=\frac{2}{x^2+1}$$
And it would then arrive at the recommended solution below.
Best Answer
No. I think what is required is to find the expressions for the integrals of motion $C_1$, $C_2$ in terms of $y$, $y'$ and then substitute them into the expression for $y''$. This gives the equation $$ y''=2\frac{xy'-y}{x^2+1}.\tag{1}$$ This is the ''minimal'' equation satisfied by $y(x)$. Of course there are more equations that will be consequences of this one. For example, differentiating it w.r.t. $x$ and substituting $y''(x)$ on the right by (1), you will get the equation $y'''=0$, which was proposed in the comment of @copper.hat.