How do I solve this ODE with solution condition? I have:
$$(2x^2+3x)y''-6(x+1)y'+6y=6 \qquad \text{if}\ y_1\ \text{polynomial}$$
I was thinking of integral factor, but what does $y_1$ condition mean?
 A: hint
Look for the solution $y$ as
$$y(x)=\sum_{n=0}^\infty a_nx^n$$
plugg it in the ODE, and find recursive relation satisfied by $(a_n)$.
$$y'(x)=\sum_{n=0}^\infty (n+1)a_{n+1}x^n$$
$$y''(x)=\sum_{n=0}^\infty ( n+1)(n+2)a_{n+2}x^n$$
A: If $y$ is a polynomial, then the left hand side is a polynomial while the right hand side is a constant, i.e. a polynomial of degree $0$. This can only happen if each of the higher order terms are cancelled out on the left hand side. Take a term $y^k$ for $k > 0$. Substituting into the left hand side, and looking at the highest order terms, we get $2k(k-1)-6k+6$. Since we need that term to cancel, we need $k=1$ or $k=3$. That means that any solution has at most degree $3$.
So, choose $$y = ax^3+bx^2+cx+d \\ y' = 3ax^2+2bx+c \\ y'' = 6ax+2b$$
Substitute these into the equation to get $$ -2 b x^2 - 6 b x + 6(d-c)
 = 6$$
Since there's no $x^2$ or $x$ term on the right, we need $b = 0$. Then, that gives $d = c+1$, while $a$ and $c$ are free. Thus the solution is $$y = ax^3+cx+(c+1)$$
for any $a,c\in\Bbb{R}$
