Let [tex]y=C_1x+C_2x^3=C_1y_1+C_2y_2[/tex]. Then [tex]y_1[/tex] and [tex]y_2[/tex] are two fundamental, linearly independent solution that satisfy
[tex]f(x,y_1,{y_1}',{y_1}'')=0[/tex]
[tex]f(x,y_2,{y_2}',{y_2}'')=0[/tex]
Note that [tex]{y_1}'=1[/tex], so that [tex]x{y_1}'-y_1=0[/tex]. Adding [tex]y''[/tex] doesn't change this, since [tex]{y_1}''=0[/tex].
So if we suppose
[tex]f(x,y,y',y'')=y''+xy'-y=0[/tex]
then substituting [tex]y=y_2[/tex] would give
[tex]6x+x(3x^2)-x^3=6x+2x^3\neq0[/tex]
To make sure everything cancels out, multiply the second degree term by [tex]-\dfrac{x^2}3[/tex], so that
[tex]f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y[/tex]
Then if [tex]y=y_1+y_2[/tex], we get
[tex]-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0[/tex]
as desired. So one possible ODE would be
[tex]-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0[/tex]
(See "Euler-Cauchy equation" for more info)
