Internal
problem
ID
[9205]
Book
:
Collection
of
Kovacic
problems
Section
:
section
1
Problem
number
:
34
Date
solved
:
Wednesday, March 05, 2025 at 07:37:49 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
Time used: 0.273 (sec)
Writing the ode as
Comparing (1) and (2) shows that
Applying the Liouville transformation on the dependent variable gives
Then (2) becomes
Where
Substituting the values of
Comparing the above to (5) shows that
Therefore eq. (4) becomes
Equation (7) is now solved. After finding
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
Case |
Allowed pole order for |
Allowed value for |
1 |
|
|
2 |
Need to have at least one pole
that is either order |
no condition |
3 |
|
|
The order of
The poles of
Attempting to find a solution using case
Looking at poles of order 2. The partial fractions decomposition of
For the pole at
Since the order of
Let
Comparing Eq. (9) with Eq. (8) shows that
From Eq. (9) the sum up to
Now we need to find
This shows that the coefficient of
Where
Since the degree of
Hence
The following table summarizes the findings so far for poles and for the order of
pole |
pole order | |
|
|
| | | | |
Order of |
|
|
|
|
| | |
Now that the all
Where
Since
The above gives
Now that
Let
Substituting the above in eq. (1A) gives
The equation is satisfied since both sides are zero. Therefore the first solution to the ode
The first solution to the original ode in
Which simplifies to
The second solution
Substituting gives
Therefore the solution is
Will add steps showing solving for IC soon.
ode:=(2*x+1)*diff(diff(y(x),x),x)-2*diff(y(x),x)-(2*x+3)*y(x) = 0; dsolve(ode,y(x), singsol=all);
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
Maple step by step
ode=(2*x+1)*D[y[x],{x,2}]-2*D[y[x],x]-(2*x+3)*y[x]==0; ic={}; DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
from sympy import * x = symbols("x") y = Function("y") ode = Eq((2*x + 1)*Derivative(y(x), (x, 2)) - (2*x + 3)*y(x) - 2*Derivative(y(x), x),0) ics = {} dsolve(ode,func=y(x),ics=ics)
False