2.1.125 Problem 127

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9297]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 127
Date solved : Wednesday, March 05, 2025 at 07:47:05 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

4x2(x2+x+1)y+12x2(1+x)y+(3x2+3x+1)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.842 (sec)

Writing the ode as

(1)(4x4+4x3+4x2)y+(12x3+12x2)y+(3x2+3x+1)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=4x4+4x3+4x2(3)B=12x3+12x2C=3x2+3x+1

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=2x24x14(x3+x2+x)2

Comparing the above to (5) shows that

s=2x24x1t=4(x3+x2+x)2

Therefore eq. (4) becomes

(7)z(x)=(2x24x14(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 2.125: 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)=62=4

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=4(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 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=12x14x2+38i38(x+12i32)2+38+i38(x+12+i32)2+145i312x+12i32+14+5i312x+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=14. Hence

[r]c=0αc+=12+1+4b=12αc=121+4b=12

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=38i38. Hence

[r]c=0αc+=12+1+4b=12+22i34αc=121+4b=1222i34

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=38+i38. Hence

[r]c=0αc+=12+1+4b=12+2+2i34αc=121+4b=122+2i34

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=2x24x14(x3+x2+x)2

pole c location pole order [r]c αc+ αc
0 2 0 12 12
12+i32 2 0 12+22i34 1222i34
12i32 2 0 12+2+2i34 122+2i34

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+αc3)=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+αc1+xc1)+(()[r]c2+αc2xc2)+(()[r]c3+αc3xc3)+()[r]=12x+1222i34x+12i32+122+2i34x+12+i32+()(0)=12x+1222i34x+12i32+122+2i34x+12+i32=2x2+12x(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(12x+1222i34x+12i32+122+2i34x+12+i32)(0)+((12x21222i34(x+12i32)2122+2i34(x+12+i32)2)+(12x+1222i34x+12i32+122+2i34x+12+i32)2(2x24x14(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(12x+1222i34x+12i32+122+2i34x+12+i32)dx=(x2+x+1)1/4x2e3arctan((2x+1)33)2

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

y1=z1e12BAdx=z1e1212x3+12x24x4+4x3+4x2dx=z1e3ln(x2+x+1)43arctan((2x+1)33)2=z1(e3arctan((2x+1)33)2(x2+x+1)3/4)

Which simplifies to

y1=e3arctan((2x+1)33)x2x2+x+1

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e12x3+12x24x4+4x3+4x2dx(y1)2dx=y1e3ln(x2+x+1)23arctan((2x+1)33)(y1)2dx=y1(e3ln(x2+x+1)23arctan((2x+1)33)(x2+x+1)e23arctan((2x+1)33)2xdx)

Therefore the solution is

y=c1y1+c2y2=c1(e3arctan((2x+1)33)x2x2+x+1)+c2(e3arctan((2x+1)33)x2x2+x+1(e3ln(x2+x+1)23arctan((2x+1)33)(x2+x+1)e23arctan((2x+1)33)2xdx))

Will add steps showing solving for IC soon.

Maple. Time used: 2.609 (sec). Leaf size: 143
ode:=4*x^2*(x^2+x+1)*diff(diff(y(x),x),x)+12*x^2*(x+1)*diff(y(x),x)+(3*x^2+3*x+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=e3arctan((2x+1)33)22x+i31x(c1(2ix+3i3+2ix+i)14i34+c2(2ix+3i3+2ix+i)34+i34hypergeom([1,12+i32],[i32+32],i3x+x+2i3x+x+2))(x2+x+1)3/4

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 
      -> Whittaker 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
         <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals 
   <- Kovacics algorithm successful`
 

Maple step by step

Let’s solve4x2(x2+x+1)(d2dx2y(x))+12x2(x+1)(ddxy(x))+(3x2+3x+1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(3x2+3x+1)y(x)4x2(x2+x+1)3(x+1)(ddxy(x))x2+x+1Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+3(x+1)(ddxy(x))x2+x+1+(3x2+3x+1)y(x)4x2(x2+x+1)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=3(x+1)x2+x+1,P3(x)=3x2+3x+14x2(x2+x+1)]xP2(x)is analytic atx=0(xP2(x))|x=0=0x2P3(x)is analytic atx=0(x2P3(x))|x=0=14x=0is a regular singular pointCheck to see ifx0is a regular singular pointx0=0Multiply by denominators4x2(x2+x+1)(d2dx2y(x))+12x2(x+1)(ddxy(x))+(3x2+3x+1)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=2..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(1+2r)2xr+(a1(1+2r)2+a0(3+2r)(1+2r))x1+r+(k=2(ak(2k+2r1)2+ak1(2k+2r+1)(2k+2r1)+ak2(2k+2r1)(2k3+2r))xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(1+2r)2=0Values of r that satisfy the indicial equationr=12Each term must be 0a1(1+2r)2+a0(3+2r)(1+2r)=0Solve for the dependent coefficient(s)a1=(3+2r)a01+2rEach term in the series must be 0, giving the recursion relation4((ak+ak2+ak1)k+(ak+ak2+ak1)rak23ak22+ak12)(k+r12)=0Shift index usingk>k+24((ak+2+ak+ak+1)(k+2)+(ak+2+ak+ak+1)rak+223ak2+ak+12)(k+32+r)=0Recursion relation that defines series solution to ODEak+2=2kak+2kak+1+2rak+2rak+1+ak+5ak+12k+2r+3Recursion relation forr=12ak+2=2kak+2kak+1+2ak+6ak+12k+4Solution forr=12[y(x)=k=0akxk+12,ak+2=2kak+2kak+1+2ak+6ak+12k+4,a1=2a0]
Mathematica. Time used: 0.512 (sec). Leaf size: 120
ode=4*x^2*(1+x+x^2)*D[y[x],{x,2}]+12*x^2*(1+x)*D[y[x],x]+(1+3*x+3*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1x2K[1]2+12K[1](K[1]2+K[1]+1)dK[1]121x3(K[2]+1)K[2]2+K[2]+1dK[2])(c21xexp(21K[3]2K[1]2+12K[1](K[1]2+K[1]+1)dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(12*x**2*(x + 1)*Derivative(y(x), x) + 4*x**2*(x**2 + x + 1)*Derivative(y(x), (x, 2)) + (3*x**2 + 3*x + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False