Internal
problem
ID
[8365]
Book
:
Collection
of
Kovacic
problems
Section
:
section
1
Problem
number
:
230
Date
solved
:
Monday, October 21, 2024 at 05:06:40 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
Time used: 0.259 (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
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. 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. Therefore
Attempting to find a solution using case \(n=1\).
Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = -2\) then
\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore
Let \(a\) be the coefficient of \(x^v=x^1\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is
Comparing Eq. (9) with Eq. (8) shows that
From Eq. (9) the sum up to \(v=1\) gives
Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{0}=1\) in \(r\) minus the coefficient of same term but in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence
This shows that the coefficient of \(1\) in the above is \(0\). Now we need to find the coefficient of \(1\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=1\) which is not zero, then starting \(r=\frac {s}{t}\), we do long division and write this in the form
Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient of \(1\) in \(r\) will be the coefficient this term in the quotient. Doing long division gives
We see that the coefficient of the term \(\frac {1}{x}\) in the quotient is \(\frac {9}{2}\). Now \(b\) can be found.
Hence
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is
Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
\(-2\) | \(\frac {x}{2}\) | \(4\) | \(-5\) |
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 \(\alpha _\infty ^{+} = 4\), and since there are no poles, then
Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using
Substituting the above values in the above results in
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=4\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation
Let
Substituting the above in eq. (1A) gives
Solving for the coefficients \(a_i\) in the above using method of undetermined coefficients gives
Substituting these coefficients in \(p(x)\) in eq. (2A) results in
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 : 0.022
(sec)
Leaf size : 47
dsolve(diff(diff(y(x),x),x)+x*diff(y(x),x)-4*y(x) = 0, y(x),singsol=all)
Solving time : 0.028
(sec)
Leaf size : 43
DSolve[{D[y[x],{x,2}]+x*D[y[x],x]-4*y[x]==0,{}}, y[x],x,IncludeSingularSolutions->True]