2.1.378 Problem 385

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9550]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 385
Date solved : Wednesday, March 05, 2025 at 07:50:47 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

2x2(x2+x+1)y+x(11x2+11x+9)y+(7x2+10x+6)y=0

Solved as second order ode using Kovacic algorithm

Time used: 1.221 (sec)

Writing the ode as

(1)(2x4+2x3+2x2)y+(11x3+11x2+9x)y+(7x2+10x+6)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=2x4+2x3+2x2(3)B=11x3+11x2+9xC=7x2+10x+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=21x4+18x3+27x22x316(x3+x2+x)2

Comparing the above to (5) shows that

s=21x4+18x3+27x22x3t=16(x3+x2+x)2

Therefore eq. (4) becomes

(7)z(x)=(21x4+18x3+27x22x316(x3+x2+x)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 2.378: 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)=64=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=16(x3+x2+x)2. There is a pole at x=0 of order 2. There is a pole at x=12+i32 of order 2. There is a pole at x=12i32 of order 2. Since there is no odd order pole larger than 2 and the order at 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 is 2 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=316x2+14x+524+i324(x+12i32)2+524i324(x+12+i32)2+1843i372x+12i32+18+43i372x+12+i32

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=316. Hence

[r]c=0αc+=12+1+4b=34αc=121+4b=14

For the pole at x=12+i32 let b be the coefficient of 1(x+12i32)2 in the partial fractions decomposition of r given above. Therefore b=524+i324. Hence

[r]c=0αc+=12+1+4b=12+6+6i312αc=121+4b=126+6i312

For the pole at x=12i32 let b be the coefficient of 1(x+12+i32)2 in the partial fractions decomposition of r given above. Therefore b=524i324. Hence

[r]c=0αc+=12+1+4b=12+66i312αc=121+4b=1266i312

Since the order of r at is 2 then [r]=0. Let b be the coefficient of 1x2 in the Laurent series expansion of r at . which can be found by dividing the leading coefficient of s by the leading coefficient of t from

r=st=21x4+18x3+27x22x316(x3+x2+x)2

Since the gcd(s,t)=1. This gives b=2116. Hence

[r]=0α+=12+1+4b=74α=121+4b=34

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

r=21x4+18x3+27x22x316(x3+x2+x)2

pole c location pole order [r]c αc+ αc
0 2 0 34 14
12+i32 2 0 12+6+6i312 126+6i312
12i32 2 0 12+66i312 1266i312

Order of r at [r] α+ α
2 0 74 34

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+=74 then

d=α+(αc1+αc2++αc3+)=74(74)=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]

Substituting the above values in the above results in

ω=(()[r]c1+αc1xc1)+((+)[r]c2+αc2+xc2)+((+)[r]c3+αc3+xc3)+(+)[r]=14x+12+6+6i312x+12i32+12+66i312x+12+i32+(0)=14x+12+6+6i312x+12i32+12+66i312x+12+i32=7x2+3x+14x(x2+x+1)

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(14x+12+6+6i312x+12i32+12+66i312x+12+i32)(0)+((14x212+6+6i312(x+12i32)212+66i312(x+12+i32)2)+(14x+12+6+6i312x+12i32+12+66i312x+12+i32)2(21x4+18x3+27x22x316(x3+x2+x)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(14x+12+6+6i312x+12i32+12+66i312x+12+i32)dx=2(x2+x+1)3/42x1/4e3arctan((2x+1)33)6

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

y1=z1e12BAdx=z1e1211x3+11x2+9x2x4+2x3+2x2dx=z1e9ln(x)4ln(x2+x+1)43arctan((2x+1)33)6=z1(e3arctan((2x+1)33)6x9/4(x2+x+1)1/4)

Which simplifies to

y1=2x2+x+1e3arctan((2x+1)33)32x2

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e11x3+11x2+9x2x4+2x3+2x2dx(y1)2dx=y1e9ln(x)2ln(x2+x+1)23arctan((2x+1)33)3(y1)2dx=y1(e9ln(x)2ln(x2+x+1)23arctan((2x+1)33)3x4e23arctan((2x+1)33)38x2+8x+8dx)

Therefore the solution is

y=c1y1+c2y2=c1(2x2+x+1e3arctan((2x+1)33)32x2)+c2(2x2+x+1e3arctan((2x+1)33)32x2(e9ln(x)2ln(x2+x+1)23arctan((2x+1)33)3x4e23arctan((2x+1)33)38x2+8x+8dx))

Will add steps showing solving for IC soon.

Maple. Time used: 0.692 (sec). Leaf size: 231
ode:=2*x^2*(x^2+x+1)*diff(diff(y(x),x),x)+x*(11*x^2+11*x+9)*diff(y(x),x)+(7*x^2+10*x+6)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=(2x+i3+1)53+3i63+6i(2x+i31)64i3+2368(3+i)3(i3)4(133+9i)e3arctan((2x+1)33)6(HeunG(3+ii3,0,0,52,12,53+3i33+3i,2x1+i3)c1x+HeunG(3+ii3,64(i31)3(i3)4,12,3,32,53+3i33+3i,2x1+i3)c2x)x5/2(x2+x+1)1/4

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) 
   Group is reducible, not completely reducible 
   Solution has integrals. Trying a special function solution free of integrals... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Kummer 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunG  ODE, case  a <> 0, e <> 0, g <> 0, c = 0 
   <- Kovacics algorithm successful`
 

Maple step by step

Let’s solve2x2(x2+x+1)(d2dx2y(x))+x(11x2+11x+9)(ddxy(x))+(7x2+10x+6)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(7x2+10x+6)y(x)2x2(x2+x+1)(11x2+11x+9)(ddxy(x))2x(x2+x+1)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+(11x2+11x+9)(ddxy(x))2x(x2+x+1)+(7x2+10x+6)y(x)2x2(x2+x+1)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=11x2+11x+92x(x2+x+1),P3(x)=7x2+10x+62x2(x2+x+1)]xP2(x)is analytic atx=0(xP2(x))|x=0=92x2P3(x)is analytic atx=0(x2P3(x))|x=0=3x=0is a regular singular pointCheck to see ifx0is a regular singular pointx0=0Multiply by denominators2x2(x2+x+1)(d2dx2y(x))+x(11x2+11x+9)(ddxy(x))+(7x2+10x+6)y(x)=0Assume series solution fory(x)y(x)=k=0akxk+rRewrite ODE with series expansionsConvertxmy(x)to series expansion form=0..2xmy(x)=k=0akxk+r+mShift index usingk>kmxmy(x)=k=makmxk+rConvertxm(ddxy(x))to series expansion form=1..3xm(ddxy(x))=k=0ak(k+r)xk+r1+mShift index usingk>k+1mxm(ddxy(x))=k=1+mak+1m(k+1m+r)xk+rConvertxm(d2dx2y(x))to series expansion form=2..4xm(d2dx2y(x))=k=0ak(k+r)(k+r1)xk+r2+mShift index usingk>k+2mxm(d2dx2y(x))=k=2+mak+2m(k+2m+r)(k+1m+r)xk+rRewrite ODE with series expansionsa0(2+r)(3+2r)xr+(a1(3+r)(5+2r)+a0(5+2r)(2+r))x1+r+(k=2(ak(k+r+2)(2k+2r+3)+ak1(2k+2r+3)(k+r+1)+ak2(2k+2r+3)(k+r1))xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(2+r)(3+2r)=0Values of r that satisfy the indicial equationr{2,32}Each term must be 0a1(3+r)(5+2r)+a0(5+2r)(2+r)=0Solve for the dependent coefficient(s)a1=(2+r)a03+rEach term in the series must be 0, giving the recursion relation2(k+r+32)((ak+ak2+ak1)k+(ak+ak2+ak1)r+2akak2+ak1)=0Shift index usingk>k+22(k+72+r)((ak+2+ak+ak+1)(k+2)+(ak+2+ak+ak+1)r+2ak+2ak+ak+1)=0Recursion relation that defines series solution to ODEak+2=kak+kak+1+rak+rak+1+ak+3ak+1k+4+rRecursion relation forr=2ak+2=kak+kak+1ak+ak+1k+2Solution forr=2[y(x)=k=0akxk2,ak+2=kak+kak+1ak+ak+1k+2,a1=0]Recursion relation forr=32ak+2=kak+kak+112ak+32ak+1k+52Solution forr=32[y(x)=k=0akxk32,ak+2=kak+kak+112ak+32ak+1k+52,a1=a03]Combine solutions and rename parameters[y(x)=(k=0akxk2)+(k=0bkxk32),ak+2=kak+kak+1ak+ak+1k+2,a1=0,bk+2=kbk+kbk+112bk+32bk+1k+52,b1=b03]
Mathematica. Time used: 0.394 (sec). Leaf size: 135
ode=2*x^2*(1+x+x^2)*D[y[x],{x,2}] + x*(9+11*x+11*x^2)*D[y[x],x] + (6+10*x+7*x^2)*y[x] == 0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1xK[1](7K[1]+3)+14K[1](K[1]2+K[1]+1)dK[1]121x(K[2]+1K[2]2+K[2]+1+92K[2])dK[2])(c21xexp(21K[3]7K[1]2+3K[1]+14K[1](K[1]2+K[1]+1)dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**2*(x**2 + x + 1)*Derivative(y(x), (x, 2)) + x*(11*x**2 + 11*x + 9)*Derivative(y(x), x) + (7*x**2 + 10*x + 6)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False