1.4.1 Example 1

\[ y^{\prime }=-xe^{-x}-y+xe^{2x}y^{3}\]

Comparing to

\[ y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\]

Shows that

\begin{align*} f_{0} & =-xe^{-x}\\ f_{1} & =-1\\ f_{2} & =0\\ f_{3} & =xe^{2x}\end{align*}

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

\begin{align*} \Delta & =-\frac {\left ( -f_{0}^{\prime }f_{3}+f_{0}f_{3}^{\prime }+3f_{0}f_{3}f_{1}\right ) ^{3}}{27f_{3}^{4}f_{0}^{5}}\\ & =-\frac {\left ( -\left ( -e^{-x}+xe^{-x}\right ) \left ( xe^{2x}\right ) +\left ( -xe^{-x}\right ) \left ( e^{2x}+2xe^{2x}\right ) +3\left ( -xe^{-x}\right ) \left ( xe^{2x}\right ) \left ( -1\right ) \right ) ^{3}}{27\left ( xe^{2x}\right ) ^{4}\left ( -xe^{-x}\right ) ^{5}}\\ & =0 \end{align*}

Since \(\Delta \) does not depend on \(x\), then this is the easy case. We can convert the ode to separable using

\begin{align*} y & =\left ( \frac {f_{0}}{f_{3}}\right ) ^{\frac {1}{3}}u\\ & =\left ( \frac {-xe^{-x}}{xe^{2x}}\right ) ^{\frac {1}{3}}u\\ & =\left ( -e^{-3x}\right ) ^{\frac {1}{3}}u\\ & =-e^{-x}u \end{align*}

In the above surd was used to obtain the cubic real root assuming everything is real.

Applying this change of variable to the original ode results in

\begin{align*} e^{-x}\left ( u^{\prime }-u\right ) & =xe^{-x}+xu^{3}e^{-x}-e^{-x}u\\ u^{\prime }-u & =x+xu^{3}-u\\ u^{\prime } & =x+xu^{3}\\ & =x\left ( u^{3}+1\right ) \end{align*}

Which is separable. Solving and simplifying gives

\[ \ln \left ( \frac {\left ( u+1\right ) ^{\frac {1}{3}}}{\left ( u^{2}-u+1\right ) ^{\frac {1}{6}}}\right ) +\frac {\sqrt {3}}{3}\arctan \left ( \frac {\sqrt {3}}{3}\left ( 2u-1\right ) \right ) =\frac {x^{2}}{2}+c_{1}\]

But \(u=-ye^{x}\). Hence the solution to the original Abel ode is

\[ \ln \left ( \frac {\left ( -ye^{x}+1\right ) ^{\frac {1}{3}}}{\left ( y^{2}e^{2x}+ye^{x}+1\right ) ^{\frac {1}{6}}}\right ) +\frac {\sqrt {3}}{3}\arctan \left ( \frac {\sqrt {3}}{3}\left ( -2ye^{x}-1\right ) \right ) =\frac {x^{2}}{2}+c_{1}\]