How to solve ode of the form $ a_n(y')^n + a_n-1(y')^n-1 + cdots + a_1 y' + a_0 =0 $
Clash Royale CLAN TAG #URR8PPP 6 2 $begingroup$ $ a_n(y')^n + a_n-1(y')^n-1 + cdots + a_1 y'+a_0 =0 $ I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before. ordinary-differential-equations share | cite | improve this question edited Feb 14 at 16:02 user47475 asked Feb 14 at 4:05 user47475 user47475 65 6 $endgroup$ 1 $begingroup$ I'm not really sure, but it kind of looks like you should solve the polynomial for $y'$, then solve for $y$ $endgroup$ – Seth Feb 14 at 4:07 3 $begingroup$ Do you mean $cdots + a_1 y^prime + a_0$ at the end? $endgroup$ – parsiad Feb 14 at 4:19 $begingroup$ Would make sense @parsiad $endgroup$ – Neo Darwin Feb 14 at 4:22 2 $begingroup$ @NeoDarwin: encouraging OP to edit the post. $endgroup$ – parsiad Feb 14 at 4:23 add