Examples

y^{'} = - x e^{- x} - y + x e^{2 x} y^{3}

Comparing to

y^{'} = f_{0} + f_{1} y + f_{2} y^{2} + f_{3} y^{3}

Shows that

\begin{aligned} f_{0} & = - x e^{- x} \\ f_{1} & = - 1 \\ f_{2} & = 0 \\ f_{3} & = x e^{2 x} \end{aligned}

Since $f_{2} = 0$ then we check is if the invariant depends on $x$ or not.

\begin{aligned} Δ & = - \frac{{(- f_{0}^{'} f_{3} + f_{0} f_{3}^{'} + 3 f_{0} f_{3} f_{1})}^{3}}{27 f_{3}^{4} f_{0}^{5}} \\ = - \frac{{(- (- e^{- x} + x e^{- x}) (x e^{2 x}) + (- x e^{- x}) (e^{2 x} + 2 x e^{2 x}) + 3 (- x e^{- x}) (x e^{2 x}) (- 1))}^{3}}{27 {(x e^{2 x})}^{4} {(- x e^{- x})}^{5}} \\ = 0 \end{aligned}

Since $Δ$ does not depend on $x$ , then this is the easy case. We can convert the ode to separable using

\begin{aligned} y & = {(\frac{f_{0}}{f_{3}})}^{\frac{1}{3}} u \\ = {(\frac{- x e^{- x}}{x e^{2 x}})}^{\frac{1}{3}} u \\ = {(- e^{- 3 x})}^{\frac{1}{3}} u \\ = - e^{- x} u \end{aligned}

Applying this change of variable to the original ode results in

\begin{aligned} e^{- x} (u^{'} - u) & = - x e^{- x} + x u^{3} e^{- x} - e^{- x} u \\ u^{'} - u & = - x + x u^{3} - u \\ u^{'} & = - x + x u^{3} \\ = x (u^{3} - 1) \end{aligned}

Which is separable. Solving and simplifying gives

3 \sqrt{3} x^{2} - \sqrt{3} \ln (\frac{4}{3 (\frac{{(1 + 2 u)}^{2}}{3} + 1)}) - 2 \sqrt{3} \ln (u - 1) + 6 \sqrt{3} c_{1} + 6 \arctan (\frac{\sqrt{3} (2 u + 1)}{3}) = 0

But $u = - y e^{x}$ . Hence the solution to the original Abel ode is

3 \sqrt{3} x^{2} - \sqrt{3} \ln (\frac{4}{3 (\frac{{(1 - 2 y e^{x})}^{2}}{3} + 1)}) - 2 \sqrt{3} \ln (- y e^{x} - 1) + 6 \sqrt{3} c_{1} + 6 \arctan (\frac{\sqrt{3} (- 2 y e^{x} + 1)}{3}) = 0