show quadratic polynomial cannot solve differential equation I have a differential equation$ y\prime + 2xy = 1 $and I need to show that there is no quadratic polynomial that solves this equation. I set $y=Ax^2+Bx+C $ and solved for $y\prime$  plugged in $y$ and $y\prime$ into my differential equation. Is this the correct approach and if so, how can I show that my differential equation can never equal $1$? My final equation is: 
$$(Ax+B)+(2x(Ax^2+Bx+C))=1  $$ 
Thanks so much in advance.

You already received good answers.
I shall continue from your last equation; expanding and grouping terms, it leads to $$2 a x^3+2 b x^2+2  (a+c)x+b=0$$ which must be satisfied for any value of $x$. Starting from the highest power, this gives $a=0$, $b=0$,$c=0$. So, theonly $y$ which is a solution of the equation is $y=0$.
