# Non-linear first order ODE $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{Axy \ + \ By^2 \ + \ Cy}{Dxy \ + Ey \ +\ Fx \ + G}$

I am trying to solve an ODE which has the following form: $$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \frac{Axy \ + \ By^2 \ + \ Cy}{Dxy \ + Ey \ +\ Fx \ + G}$$ with an initial condition $y(x_0) = y_0 \\$.

The approaches I could think of to solve this equation were to

1. Approximate it to a Darboux equation, but the approximations are not desirable.
2. Write $x = \dfrac{x}{z}$ and $y = \dfrac{y}{z}$ to get homogeneous degree on the right side, but I am unable to progress further.

Are there any methods to find explicit closed form solutions for such equations?

• With arbitrary $A,B,C,D,E,F,G$ it is hopeless. – Alexandre Eremenko Jan 9 '18 at 1:13
• Maple calls this an Abel equation of the second type, class B. In general closed-form solutions for these equations are not known. Even in simple particular cases, e.g. with all parameters $=1$, Maple does not find a closed-form solution. I suspect there is none. – Robert Israel Jan 9 '18 at 2:00
• They are not only "not known" but they don't exist with any reasonable definition of "closed form". – Alexandre Eremenko Jan 9 '18 at 2:56
• If $A=0$ and $C=-B y_0$, then $y=y_0=\mathrm{const}$ is a solution. – mo-user Aug 21 '18 at 18:20
• Also, if $B (B-D)y_0+A E-B F+F D+C (B-D)=0$, then $y=-Ax/(B-D)+y_0$ is a solution. – mo-user Aug 21 '18 at 18:25

The problem and its difficulty depend on the coefficients. For instance if all coefficients are zero except for $C=G=1$ then the original equation reduces to $$\frac{dy}{dx}=y\Rightarrow \frac{dy}{y}=\,dx$$ which is relatively easy to solve. In many cases when the coefficients are such that the numerator and denominator on the right side factor into products of simpler univariate polynomials then the situation becomes possible to get analytic solutions. That is if $A,B,C,D,E,F,G$ are such that $P(x,y):=Axy+By^2+Cy$ and $Q(x,y)=Dxy+Ey+Fx+G$ can be factored into $P(x,y)=P_1(x)P_2(y)$ and $Q(x,y)=Q_1(x)Q_2(y)$ then the original equation can be written as $$\frac{Q_2(y)}{P_2(y)}\,dy=\frac{P_1(x)}{Q_1(x)}\,dx$$ From the degree of $P,Q$ we get that $P_1,Q_1,Q_2$ are at most linear and $P_2$ at most quadratic. Therefore the question reduces to solving integrals of rational functions of one variable. For many cases there are analytic solutions. Again for general constants I don't think there is a unifying approach in getting solutions.