# Can anyone transform this ODE into a different form?

Does the first-order ODE $$\bigl(g_2(x)y^2+g_1(x)y+g_0(x)\bigr)y'=f_3(x)y^3+f_2(x)y^2+f_1(x)y+f_0(x),$$ where $g_2(x)\neq0$ and $f_0(x)$ , $f_1(x)$ , $f_2(x)$ , $f_3(x)$ are not both equal to $0$, be possible to transform to a first-order ODE which is of the form $y_1'=\sum_{k=0}^nh_k(x_1)y_1^k$ , where $n$ is as small as possible?

I start my motivation by considering the first-order ODE $$\bigl(g_1(x)y+g_0(x)\bigr)y'=f_3(x)y^3+f_2(x)y^2+f_1(x)y+f_0(x),$$ where $g_1(x)\neq0$ and $f_0(x)$ , $f_1(x)$ , $f_2(x)$ , $f_3(x)$ are not both equal to $0$ first, the effect is quite directly perceived through the senses since just let $u=y+\dfrac{g_0(x)}{g_1(x)}$ and then let $u=\dfrac{1}{y_1}$ , $x=x_1$ is OK.

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Write the original DE as $Q(x,y) y' = P(x,y)$ where $P$ and $Q$ are polynomials in $y$; we may assume they are relatively prime. Consider a new independent variable $v$ defined by $v = G(x,y)$, where we may try for $G$ that is a nonconstant polynomial in $y$. We want this to satisfy a differential equation $v' = R(x,v)$ where $R$ is a polynomial in $v$.

Let's work backwards: start with the DE $v' = R(x,v)$ and define $y$ implicitly by $v = G(x,y)$ where $G$ is a nonconstant polynomial in $y$.
Then $v' = G_x + G_y y' = R(x,G(x,y))$ so $y$ satisfies the differential equation $y = \dfrac{(R(x,G(x,y)) - G_x(x,y))}{G_y(x,y)} = \dfrac{P(x,y)}{Q(x,y)}$ where for some $H(x,y)$, $H(x,y) P(x,y) = R(x,G(x,y)) - G_x(x,y)$ and $H(x,y) Q(x,y) = G_y(x,y)$.

Can we take $H(x,y) = 1$? If $G_y = Q$ is a quadratic in $y$, $G$ is a cubic in $y$, and $G_x$ is either cubic or quadratic. Thus for $P$ to be cubic in $y$, $R(x,G(x,y))$ is at most cubic in $y$. But since $G$ is cubic in $y$, $R(x,v)$ must be at most linear in $v$, say $R(x,v) = \alpha(x) + \beta(x) v$. Then $P(x,y) = \alpha(x) + \beta(x) G(x,y) - G_x(x,y)$ and $Q(x,y) = G_y(x,y)$. This defines one family of solutions. For example, with $\alpha(x) = \beta(x) = 1$ and $G(x,y) = x y + y^3$, we have $P(x,y) = 1 + (x-1) y + y^3$, $Q(x,y) = x + 3 y^2$; the substitution $v = x y + y^3$ transforms $y' = \dfrac{1+(x-1) y + y^3}{x+3y^2}$ into $v' = v + 1$. So at least in some cases this transformation is possible.

More generally, if $H$ is of degree $d$ in $y$, then $G_y$ is of degree $2+d$ and $G$ is of degree $3+d$, and $R(x,G(x,y))$ is of degree at most $3+d$, so again $R$ must be at most linear. If $R(x,v) = \alpha(x) + \beta(x) v$, $H(x,y)$ must divide both $G_y(x,y)$ and $\alpha(x) + \beta(x) G(x,y) - G_x(x,y)$. For example, we might try $H(x,y) = y$ with $G(x,y) = g_0(x) + g_2(x) y^2 + g_3(x) y^3 + g_4(x) y^4$ where $\alpha(x) + \beta(x) g_0(x) - g_0'(x) = 0$, obtaining $P(x,y) = (\beta(x) g_4(x) - g_4'(x)) y^3 + (\beta(x) g_3(x) - g_3'(x)) y^2 + (\beta(x) g_2(x) - g_2'(x)) y$ and $Q(x,y) = 4 g_4(x) y^2 + 3 g_3(x) y + 2 g_2(x)$.

EDIT: If there is a transformation to $v' = R(x,v)$ where $v$ is linear, a further transformation $u = a(x) + b(x) v$ will make the differential equation $u' = 0$, so we may as well assume $R(x,v) = 0$. But that is equivalent to saying $H(x,y)$ is an integrating factor for the differential equation $P(x,y) - Q(x,y) y' = 0$. The necessary and sufficient condition for $H(x,y)$ to be an integrating factor is that $(P H)_y = -(Q H)_x$. There is an integrating factor of the form $H(x)$ iff $(P_y + Q_x)/Q$ is a function of $x$ only. I don't know of a nice necessary and sufficient condition for having an integrating factor of the form $\sum_{j=0}^d h_j(x) y^j$.

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Your idea is really great, but can you reverse your process so that to determine the relationships between $f_0(x)$ , $f_1(x)$ , $f_2(x)$ , $f_3(x)$ , $g_0(x)$ , $g_1(x)$ , $g_2(x)$ in the orginal ODE which is at most possible to transform to linear ODE, Riccati Equation, Abel equation of the first kind, ...... respectively? – doraemonpaul Jun 9 '12 at 1:09