2.1.10 Problem 10

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9182]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 10
Date solved : Wednesday, March 05, 2025 at 07:37:30 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

(x26x+10)y4(x3)y+6y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.335 (sec)

Writing the ode as

(1)(x26x+10)y+(4x+12)y+6y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x26x+10(3)B=4x+12C=6

Applying the Liouville transformation on the dependent variable gives

z(x)=yeB2Adx

Then (2) becomes

(4)z(x)=rz(x)

Where r is given by

(5)r=st=2AB2BA+B24AC4A2

Substituting the values of A,B,C from (3) in the above and simplifying gives

(6)r=8(x26x+10)2

Comparing the above to (5) shows that

s=8t=(x26x+10)2

Therefore eq. (4) becomes

(7)z(x)=(8(x26x+10)2)z(x)

Equation (7) is now solved. After finding z(x) then y is found using the inverse transformation

y=z(x)eB2Adx

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 . The following table summarizes these cases.

Case

Allowed pole order for r

Allowed value for O()

1

{0,1,2,4,6,8,}

{,6,4,2,0,2,3,4,5,6,}

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

{1,2}

{2,3,4,5,6,7,}

Table 2.10: Necessary conditions for each Kovacic case

The order of r at is the degree of t minus the degree of s. Therefore

O()=deg(t)deg(s)=40=4

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=(x26x+10)2. There is a pole at x=3+i of order 2. There is a pole at x=3i of order 2. Since there is no odd order pole larger than 2 and the order at is 4 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 is 4 then the necessary conditions for case three are met. Therefore

L=[1,2,4,6,12]

Attempting to find a solution using case n=1.

Looking at poles of order 2. The partial fractions decomposition of r is

r=2(x3i)2+2(x3+i)2+2ix3i2ix3+i

For the pole at x=3+i let b be the coefficient of 1(x3i)2 in the partial fractions decomposition of r given above. Therefore b=2. Hence

[r]c=0αc+=12+1+4b=2αc=121+4b=1

For the pole at x=3i let b be the coefficient of 1(x3+i)2 in the partial fractions decomposition of r given above. Therefore b=2. Hence

[r]c=0αc+=12+1+4b=2αc=121+4b=1

Since the order of r at is 4>2 then

[r]=0α+=0α=1

The following table summarizes the findings so far for poles and for the order of r at where r is

r=8(x26x+10)2

pole c location pole order [r]c αc+ αc
3+i 2 0 2 1
3i 2 0 2 1

Order of r at [r] α+ α
4 0 0 1

Now that the all [r]c and its associated αc± have been determined for all the poles in the set Γ and [r] and its associated α± have also been found, the next step is to determine possible non negative integer d from these using

d=αs()cΓαcs(c)

Where s(c) is either + or and s() is the sign of α±. This is done by trial over all set of families s=(s(c))cΓ until such d is found to work in finding candidate ω. Trying α=1 then

d=α(αc1+αc2+)=1(1)=0

Since d an integer and d0 then it can be used to find ω using

ω=cΓ(s(c)[r]c+αcs(c)xc)+s()[r]

The above gives

ω=(()[r]c1+αc1xc1)+((+)[r]c2+αc2+xc2)+()[r]=1x3i+2x3+i+()(0)=1x3i+2x3+i=x33ix26x+10

Now that ω 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

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(x)=1

Substituting the above in eq. (1A) gives

(0)+2(1x3i+2x3+i)(0)+((1(x3i)22(x3+i)2)+(1x3i+2x3+i)2(8(x26x+10)2))=00=0

The equation is satisfied since both sides are zero. Therefore the first solution to the ode z=rz is

z1(x)=peωdx=e(1x3i+2x3+i)dx=(x26x+10)2(ix3i+1)3

The first solution to the original ode in y is found from

y1=z1e12BAdx=z1e124x+12x26x+10dx=z1eln(x26x+10)=z1(x26x+10)

Which simplifies to

y1=(x26x+10)3(ix3i+1)3

The second solution y2 to the original ode is found using reduction of order

y2=y1eBAdxy12dx

Substituting gives

y2=y1e4x+12x26x+10dx(y1)2dx=y1e2ln(x26x+10)(y1)2dx=y1(x26x+263(x3+i)3)

Therefore the solution is

y=c1y1+c2y2=c1((x26x+10)3(ix3i+1)3)+c2((x26x+10)3(ix3i+1)3(x26x+263(x3+i)3))

Will add steps showing solving for IC soon.

Maple. Time used: 0.003 (sec). Leaf size: 31
ode:=(x^2-6*x+10)*diff(diff(y(x),x),x)-4*(x-3)*diff(y(x),x)+6*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1x3+c2x2+6(5c1c2)x+60c1+26c23

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) 
<- Kovacics algorithm successful`
 

Mathematica. Time used: 0.314 (sec). Leaf size: 84
ode=(x^2-6*x+10)*D[y[x],{x,2}]-4*(x-3)*D[y[x],x]+6*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)(x26x+10)exp(1xK[1](33i)(K[1]6)K[1]+10dK[1])(c21xexp(21K[2]K[1](33i)(K[1]6)K[1]+10dK[1])dK[2]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((12 - 4*x)*Derivative(y(x), x) + (x**2 - 6*x + 10)*Derivative(y(x), (x, 2)) + 6*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False