Internal
problem
ID
[9817] Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948 Section
:
Chapter
1,
Additional
non-linear
first
order Problem
number
:
838 Date
solved
:
Thursday, October 17, 2024 at 10:12:30 PM CAS
classification
:
[_rational, _Riccati]
Equation (7) is now solved. After finding \(z(x)\) then \(u\) is found using the inverse transformation
\begin{align*} u &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end{align*}
The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3
cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table
summarizes these cases.
Need to have at least one pole
that is either order \(2\) or odd order
greater than \(2\). Any other pole order
is allowed as long as the above
condition is satisfied. Hence the
following set of pole orders are all
allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).
no condition
3
\(\left \{ 1,2\right \} \)
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)
Table 13: Necessary conditions for each Kovacic case
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 x^{2}\).
There is a pole at \(x=0\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is
\(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then
necessary conditions for case two are met. Since pole order is not larger than \(2\) and
the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore
\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is
\[
r = -\frac {1}{4 x^{2}}
\]
For the pole at \(x=0\) let \(b\)
be the coefficient of \(\frac {1}{ x^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {1}{4}}\). Hence
Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\)
at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\)
from
\begin{alignat*}{2} r &= \frac {s}{t} &&= -\frac {1}{4 x^{2}} \end{alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {1}{4}}\). Hence
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\)
is
\[ r=-\frac {1}{4 x^{2}} \]
pole \(c\) location
pole order
\([\sqrt r]_c\)
\(\alpha _c^{+}\)
\(\alpha _c^{-}\)
\(0\)
\(2\)
\(0\)
\(\frac {1}{2}\)
\(\frac {1}{2}\)
Order of \(r\) at \(\infty \)
\([\sqrt r]_\infty \)
\(\alpha _\infty ^{+}\)
\(\alpha _\infty ^{-}\)
\(2\)
\(0\)
\(\frac {1}{2}\)
\(\frac {1}{2}\)
Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \)
and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative
integer \(d\) from these using
Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until
such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = {\frac {1}{2}}\) then
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree
\(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {30 x^{3}+25 \sqrt {x}+25 y \left (x \right )^{2}-20 x^{3} y \left (x \right )-100 y \left (x \right ) \sqrt {x}+4 x^{6}+40 x^{{7}/{2}}+100 x}{25 x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {30 x^{3}+25 \sqrt {x}+25 y \left (x \right )^{2}-20 x^{3} y \left (x \right )-100 y \left (x \right ) \sqrt {x}+4 x^{6}+40 x^{{7}/{2}}+100 x}{25 x} \end {array} \]
2.262.3 Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryingChinidifferentialorder: 1; looking for linear symmetriestryingexactLookingfor potential symmetriestryingRiccatitryingRiccati sub-methods:<- Riccati particular case Kamke (d) successful`