Internal
problem
ID
[8527]
Book
:
Collection
of
Kovacic
problems
Section
:
section
1
Problem
number
:
400
Date
solved
:
Monday, October 21, 2024 at 05:09:14 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
Time used: 0.228 (sec)
Writing the ode as
Comparing (1) and (2) shows that
Applying the Liouville transformation on the dependent variable gives
Then (2) becomes
Where \(r\) is given by
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives
Comparing the above to (5) shows that
Therefore eq. (4) becomes
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation
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.
Case |
Allowed pole order for \(r\) |
Allowed value for \(\mathcal {O}(\infty )\) |
1 |
\(\left \{ 0,1,2,4,6,8,\cdots \right \} \) |
\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \) |
2 |
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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
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 \(t=4 x^{6}\). There is a pole at \(x=0\) of order \(6\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(4\) then the necessary conditions for case one are met. Therefore
Attempting to find a solution using case \(n=1\).
Looking at higher order poles of order \(2 v \)≥\( 4\) (must be even order for case one).Then for each pole \(c\), \([\sqrt r]_{c}\) is the sum of terms \(\frac {1}{(x-c)^i}\) for \(2 \leq i \leq v\) in the Laurent series expansion of \(\sqrt r\) expanded around each pole \(c\). Hence
Let \(a\) be the coefficient of the term \(\frac {1}{ (x-c)^v}\) in the above where \(v\) is the pole order divided by 2. Let \(b\) be the coefficient of \(\frac {1}{ (x-c)^{v+1}} \) in \(r\) minus the coefficient of \(\frac {1}{ (x-c)^{v+1}} \) in \([\sqrt r]_c\). Then
The partial fraction decomposition of \(r\) is
There is pole in \(r\) at \(x= 0\) of order \(6\), hence \(v=3\). Expanding \(\sqrt {r}\) as Laurent series about this pole \(c=0\) gives
Using eq. (1B), taking the sum up to \(v=3\) the above becomes
The above shows that the coefficient of \(\frac {1}{(x-0)^{3}}\) is
Now we need to find \(b\). let \(b\) be the coefficient of the term \(\frac {1}{(x-c)^{v+1}}\) in \(r\) minus the coefficient of the same term but in the sum \([\sqrt r]_c \) found in eq. (3B). Here \(c\) is current pole which is \(c=0\). This term becomes \(\frac {1}{x^{4}}\). The coefficient of this term in the sum \([\sqrt r]_c\) is seen to be \(0\) and the coefficient of this term \(r\) is found from the partial fraction decomposition from above to be \(-{\frac {1}{2}}\). Therefore
Hence
Since the order of \(r\) at \(\infty \) is \(4 > 2\) then
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is
pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
\(0\) | \(6\) | \(\frac {1}{2 x^{3}}\) | \(1\) | \(2\) |
Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
\(4\) | \(0\) | \(0\) | \(1\) |
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 \(\alpha _\infty ^{-} = 1\) then
Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using
The above gives
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
Let
Substituting the above in eq. (1A) gives
The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is
The first solution to the original ode in \(y\) is found from
Which simplifies to
The second solution \(y_2\) to the original ode is found using reduction of order
Substituting gives
Therefore the solution is
Will add steps showing solving for IC soon.
Methods for second order ODEs:
Solving time : 0.003
(sec)
Leaf size : 19
dsolve(x^3*diff(diff(y(x),x),x)+diff(y(x),x)-1/x*y(x) = 0, y(x),singsol=all)