2.1.103 Problem 105

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9275]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 105
Date solved : Wednesday, March 05, 2025 at 07:46:46 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

3x2(x2+2)y+x(11x2+1)y+(5x2+1)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.421 (sec)

Writing the ode as

(1)(3x4+6x2)y+(11x3+x)y+(5x2+1)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=3x4+6x2(3)B=11x3+xC=5x2+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=5x44x23536(x32x)2

Comparing the above to (5) shows that

s=5x44x235t=36(x32x)2

Therefore eq. (4) becomes

(7)z(x)=(5x44x23536(x32x)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 2.103: 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=36(x32x)2. There is a pole at x=0 of order 2. There is a pole at x=2 of order 2. There is a pole at x=2 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=35144x2764(x2)2764(x+2)2+312384(x2)312384(x+2)

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=35144. Hence

[r]c=0αc+=12+1+4b=712αc=121+4b=512

For the pole at x=2 let b be the coefficient of 1(x2)2 in the partial fractions decomposition of r given above. Therefore b=764. Hence

[r]c=0αc+=12+1+4b=78αc=121+4b=18

For the pole at x=2 let b be the coefficient of 1(x+2)2 in the partial fractions decomposition of r given above. Therefore b=764. Hence

[r]c=0αc+=12+1+4b=78αc=121+4b=18

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=5x44x23536(x32x)2

Since the gcd(s,t)=1. This gives b=536. Hence

[r]=0α+=12+1+4b=56α=121+4b=16

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

r=5x44x23536(x32x)2

pole c location pole order [r]c αc+ αc
0 2 0 712 512
2 2 0 78 18
2 2 0 78 18

Order of r at [r] α+ α
2 0 56 16

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+=56 then

d=α+(αc1++αc2+αc3)=56(56)=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+αc1+xc1)+(()[r]c2+αc2xc2)+(()[r]c3+αc3xc3)+(+)[r]=712x+18x82+18x+82+(0)=712x+18x82+18x+82=5x276x312x

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(712x+18x82+18x+82)(0)+((712x218(x2)218(x+2)2)+(712x+18x82+18x+82)2(5x44x23536(x32x)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(712x+18x82+18x+82)dx=x7/12(x+2)1/8(x2)1/8

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

y1=z1e12BAdx=z1e1211x3+x3x4+6x2dx=z1eln(x)127ln(x22)8=z1(1x1/12(x22)7/8)

Which simplifies to

y1=x(x22)3/4

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e11x3+x3x4+6x2dx(y1)2dx=y1eln(x)67ln(x22)4(y1)2dx=y1(eln(x)67ln(x22)4(x22)3/2xdx)

Therefore the solution is

y=c1y1+c2y2=c1(x(x22)3/4)+c2(x(x22)3/4(eln(x)67ln(x22)4(x22)3/2xdx))

Will add steps showing solving for IC soon.

Maple. Time used: 0.082 (sec). Leaf size: 35
ode:=3*x^2*(-x^2+2)*diff(diff(y(x),x),x)+x*(-11*x^2+1)*diff(y(x),x)+(-5*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1x(2x2+4)3/4+c2x1/3hypergeom([23,1],[1112],x22)

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 
         <- heuristic approach successful 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
   <- Kovacics algorithm successful`
 

Maple step by step

Let’s solve3x2(x2+2)(d2dx2y(x))+x(11x2+1)(ddxy(x))+(5x2+1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(5x21)y(x)3x2(x22)(11x21)(ddxy(x))3x(x22)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+(11x21)(ddxy(x))3x(x22)+(5x21)y(x)3x2(x22)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=11x213x(x22),P3(x)=5x213x2(x22)]xP2(x)is analytic atx=0(xP2(x))|x=0=16x2P3(x)is analytic atx=0(x2P3(x))|x=0=16x=0is a regular singular pointCheck to see ifx0is a regular singular pointx0=0Multiply by denominators3x2(x22)(d2dx2y(x))+x(11x21)(ddxy(x))+(5x21)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(1+3r)(1+2r)xra1(2+3r)(1+2r)x1+r+(k=2(ak(3k+3r1)(2k+2r1)+ak2(3k+3r1)(k+r1))xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(1+3r)(1+2r)=0Values of r that satisfy the indicial equationr{12,13}Each term must be 0a1(2+3r)(1+2r)=0Solve for the dependent coefficient(s)a1=0Each term in the series must be 0, giving the recursion relation6((kr+1)ak22+(k+r12)ak)(k13+r)=0Shift index usingk>k+26((k1r)ak2+(k+32+r)ak+2)(k+53+r)=0Recursion relation that defines series solution to ODEak+2=(k+r+1)ak2k+3+2rRecursion relation forr=12ak+2=(k+32)ak2k+4Solution forr=12[y(x)=k=0akxk+12,ak+2=(k+32)ak2k+4,a1=0]Recursion relation forr=13ak+2=(k+43)ak2k+113Solution forr=13[y(x)=k=0akxk+13,ak+2=(k+43)ak2k+113,a1=0]Combine solutions and rename parameters[y(x)=(k=0akxk+12)+(k=0bkxk+13),ak+2=(k+32)ak2k+4,a1=0,bk+2=(k+43)bk2k+113,b1=0]
Mathematica. Time used: 0.298 (sec). Leaf size: 118
ode=3*x^2*(2-x^2)*D[y[x],{x,2}]+x*(1-11*x^2)*D[y[x],x]+(1-5*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1x75K[1]212K[1]6K[1]3dK[1]121x111K[2]26K[2]3K[2]3dK[2])(c21xexp(21K[3]75K[1]212K[1]6K[1]3dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(3*x**2*(2 - x**2)*Derivative(y(x), (x, 2)) + x*(1 - 11*x**2)*Derivative(y(x), x) + (1 - 5*x**2)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False