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.
 A: You do not need to
equate coefficients,
and you can easily show that
no polynomial can be
a solution to
$y' + 2xy = 1
$.
Suppose $y$
is a polynomial
of degree $d$.
Then $2xy$
is a polynomial of
degree $d+1$
and
$y'$ is a polynomial
of degree $d-1$.
Therefore
$y' + 2xy
$
is a polynomial
of degree $d+1$,
since the term of degree
$d+1$
in $2xy$
can not be cancelled out
by $y'$
since its highest
degree term
has degree $d-1$.
Therefore
$y' + 2xy$
is a polynomial
of degree $d+1$
and therefore
can not be constant,
let alone $1$.
To actually solve the equation,
note that the integrating factor is
$e^{\int 2x}
=e^{x^2}
$.
Then
$e^{x^2}(y'+2xy)
=e^{x^2}y'+2xe^{x^2}y
=(e^{x^2}y)'
$
so
$1 = y'+2xy$
becomes
$e^{x^2}
= (e^{x^2}y)'
$.
Integrating,
$e^{x^2}y = \int e^{x^2}
$
(with the constant of integration implied)
or
$y = e^{-x^2}\int e^{x^2}
$.
To do the integration will involve the
error function or an equivalent.
A: Hint: two polynomials are equal iff their coefficients are equal. Equate coefficients.
A: 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 $y=0$.
