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## Note on solving Abel ﬁrst order ODE

August 1, 2022   Compiled on August 1, 2022 at 2:06am

### 1 Solving Abel ode

This is a note about solving Abel ode. The purpose of this note is show to solve this ode using few examples.

#### 1.1 Abel ode of ﬁrst kind

This ODE has the form \begin {equation} y^{\prime }(x)=f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}\tag {1} \end {equation}

Any one of the following forms is called an Abel ode of ﬁrst kind \begin {align*} y^{\prime } & =f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\\ y^{\prime } & =f_{1}y+f_{2}y^{2}+f_{3}y^{3}\\ y^{\prime } & =f_{2}y^{2}+f_{3}y^{3}\\ y^{\prime } & =f_{0}+f_{2}y^{2}+f_{3}y^{3}\\ y^{\prime } & =f_{0}+f_{3}y^{3}\\ y^{\prime } & =f_{0}+f_{1}y+f_{3}y^{3}\\ y^{\prime } & =f_{2}y^{2}+f_{3}y^{3} \end {align*}

The case for both $$f_{0}(x)=0,f_{2}(x)=0$$ is not allowed, else it becomes Bernoulli ode. Either $$f_{0}=0$$ or $$f_{2}=0$$ is allowed but not both at same time. The term $$f_{3}(x)$$ must be there in all cases. When $$f_{2}=0$$ then Abel invariant is given by

$\Delta =-\frac {\left ( -f_{0}^{\prime }f_{3}+f_{0}f_{3}^{\prime }+3f_{0}f_{3}f_{1}\right ) ^{3}}{27f_{3}^{4}f_{0}^{5}}$

In the case when $$f_{2}\neq 0$$, then $$f_{2}$$ is removed from the original ode using the change of dependent variable $$y=u\left ( x\right ) -\frac {f_{2}}{3f_{3}}$$. Now the new ode will not $$f_{2}$$ in it, and the above invariant applied to it.

There are two possibilities. $$\Delta$$ can be constant (does not depend on $$x$$) or not constant (function of $$x$$). The constant invariant is the easier case and can be solved. The non constant case is not fully solved and only few cases can be solved analytically.

##### 1.1.1 Solution method

Find what is called the abel invariant and check if constant.

$\Delta =-\frac {\left ( -f_{0}^{\prime }f_{3}+f_{0}f_{3}^{\prime }+3f_{0}f_{3}f_{1}\right ) ^{3}}{27f_{3}^{4}f_{0}^{5}}$

Then use the substitution $$y=\frac {1}{u}$$. Hence $$y^{\prime }=-\frac {1}{u^{2}}u^{\prime }$$. Substituting this in (1) gives

\begin {align} -\frac {1}{u^{2}}u^{\prime } & =f_{0}(x)+f_{1}(x)\frac {1}{u}+f_{2}(x)\frac {1}{u^{2}}+f_{3}(x)\frac {1}{u^{3}}\nonumber \\ -uu^{\prime } & =u^{3}f_{0}(x)+u^{2}f_{1}(x)+uf_{2}(x)+f_{3}(x)\nonumber \\ uu^{\prime } & =-u^{3}f_{0}(x)-u^{2}f_{1}(x)-uf_{2}(x)-f_{3}(x) \tag {2} \end {align}

Next, using substitution $$u=\frac {1}{E}\left ( y+\frac {f_{2}}{3f_{3}}\right )$$ where $$E=\exp \left ( \int f_{1}-\frac {f_{2}^{2}}{3f_{3}}dx\right )$$ in the above gives

$\frac {1}{E}\left ( y+\frac {f_{2}}{3f_{3}}\right ) u^{\prime }=-u^{3}f_{0}(x)-u^{2}f_{1}(x)-uf_{2}(x)-f_{3}(x)$

\begin {align*} u^{\prime } & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( y+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{E}\left ( y^{\prime }+\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \\ & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{E}\left ( -\frac {1}{u^{2}}u^{\prime }+\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \\ u^{\prime }+\frac {u^{\prime }}{Eu^{2}} & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3E}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\\ u^{\prime }\left ( 1+\frac {1}{Eu^{2}}\right ) & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3E}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\\ u^{\prime } & =\frac {Eu^{2}}{1+Eu^{2}}\left ( \frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3E}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \\ u^{\prime } & =\frac {u^{2}}{1+Eu^{2}}\left ( \frac {1}{E}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \end {align*}

Substituting the above into (2) gives

$u\frac {u^{2}}{1+Eu^{2}}\left ( \frac {1}{E}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) =-u^{3}f_{0}-u^{2}f_{1}-uf_{2}-f_{3}$

\begin {align*} E & =\exp \left ( \int f_{1}\left ( x\right ) -\frac {f_{2}^{2}\left ( x\right ) }{3f_{3}\left ( x\right ) }dx\right ) \\ \xi & =\int f_{3}\left ( x\right ) E^{2}dx\\ u & =\frac {1}{E}\left ( y+\frac {f_{2}\left ( x\right ) }{3f_{3}\left ( x\right ) }\right ) \end {align*}

The above are used to convert the ﬁrst kind Abel ode to canonical form. (To ﬁnish).