2.22.1 Algorithm
ode internal name "first_order_ode_chini"
This ode is normally generated when we get an Abel ode of first kind \(f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\) and then remove the square
term \(f_{2}\) using the transformation \(y=u\left ( x\right ) -\frac {f_{2}}{3f_{3}}\). Again as mentioned above, this is done when the Abel invariant is
constant. See above section.
Now we check if the Chini invariant is also constant or not. The Chini invariant is given by
\[ \Delta =f^{-n-1}h^{-2n+1}\left ( fh^{\prime }-f^{\prime }h-ngfh\right ) ^{n}n^{-n}\]
And if
this comes out to be constant (i.e. do not depend on
\(x\)), then we can now solve the Chini ode using
method given in Kamke page 303.
Otherwise there is no general method to solve it. This below is my translation of Kamke 1.55,
page 303 on Chini ode. He says, given ode
\begin{equation} y^{\prime }=f\left ( x\right ) y^{n}+g\left ( x\right ) y+h\left ( x\right ) \tag {1}\end{equation}
If for a suitable constants
\(\alpha ,\beta \)\begin{equation} \left ( \frac {h}{f}\right ) ^{\frac {1}{n}}=e^{\int gdx}\left ( \beta +\alpha \int he^{-\int gdx}dx\right ) \tag {2}\end{equation}
Let
\begin{equation} z=\left ( \frac {h}{f}\right ) ^{\frac {1}{n}} \tag {3}\end{equation}
Then
\[ \alpha =\frac {z^{\prime }-gz}{h}\]
We get the
solution of the original ode as
\begin{align} y & =\left ( \frac {h}{f}\right ) ^{\frac {1}{n}}u\left ( x\right ) \nonumber \\ u & =y\left ( \frac {h}{f}\right ) ^{\frac {-1}{n}} \tag {5}\end{align}
The book here was not clear. But it seems we have two cases. If \(\Delta =0\) then solution is
\begin{equation} \int \frac {du}{u^{n}-\alpha u+1}+c_{1}-\int \left ( \frac {h}{f}\right ) ^{\frac {1}{n}}hdx=0 \tag {6}\end{equation}
And if
\(\Delta \neq 0\) then
solution is
\begin{equation} \int \frac {du}{\frac {u^{n}}{\Delta }-u+1}+c_{1}-\alpha \int \left ( \frac {h}{f}\right ) ^{\frac {1}{n}}hdx=0 \tag {7}\end{equation}
For
\(h=0\) the ode is Bernoulli. Lets try to figure how the above works on number of
examples. In the above
\(\Delta \) is Chini invariant. The book also was not clear what
\(\beta \) is. It seems it is
always zero. This is what is done in all examples below and solutions found were verified correct by
Maple.