2.1.63 Problem 65

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9235]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 65
Date solved : Wednesday, March 05, 2025 at 07:38:14 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

(x3+1)y+7x2y+9xy=0

Solved as second order ode using Kovacic algorithm

Time used: 0.438 (sec)

Writing the ode as

(1)(x3+1)y+7x2y+9xy=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x3+1(3)B=7x2C=9x

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=x(x3+8)4(x3+1)2

Comparing the above to (5) shows that

s=x(x3+8)t=4(x3+1)2

Therefore eq. (4) becomes

(7)z(x)=(x(x3+8)4(x3+1)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 2.63: 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=4(x3+1)2. There is a pole at x=1 of order 2. There is a pole at x=12i32 of order 2. There is a pole at x=12+i32 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=518(x+1)+736(x+1)2+736(x12i32)2+736(x12+i32)2+536+5i336x12i32+5365i336x12+i32

For the pole at x=1 let b be the coefficient of 1(x+1)2 in the partial fractions decomposition of r given above. Therefore b=736. Hence

[r]c=0αc+=12+1+4b=76αc=121+4b=16

For the pole at x=12i32 let b be the coefficient of 1(x12+i32)2 in the partial fractions decomposition of r given above. Therefore b=736. Hence

[r]c=0αc+=12+1+4b=76αc=121+4b=16

For the pole at x=12+i32 let b be the coefficient of 1(x12i32)2 in the partial fractions decomposition of r given above. Therefore b=736. Hence

[r]c=0αc+=12+1+4b=76αc=121+4b=16

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=x(x3+8)4(x3+1)2

Since the gcd(s,t)=1. This gives b=14. Hence

[r]=0α+=12+1+4b=12α=121+4b=12

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

r=x(x3+8)4(x3+1)2

pole c location pole order [r]c αc+ αc
1 2 0 76 16
12i32 2 0 76 16
12+i32 2 0 76 16

Order of r at [r] α+ α
2 0 12 12

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=12 then

d=α(αc1+αc2+αc3)=12(12)=1

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+αc2xc2)+(()[r]c3+αc3xc3)+()[r]=16(x+1)16(x12+i32)16(x12i32)+()(0)=16(x+1)16(x12+i32)16(x12i32)=x22x3+2

Now that ω is determined, the next step is find a corresponding minimal polynomial p(x) of degree d=1 to solve the ode. The polynomial p(x) needs to satisfy the equation

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(x)=x+a0

Substituting the above in eq. (1A) gives

(0)+2(16(x+1)16(x12+i32)16(x12i32))(1)+((16(x+1)2+16(x12+i32)2+16(x12i32)2)+(16(x+1)16(x12+i32)16(x12i32))2(x(x3+8)4(x3+1)2))=016a0x(x2x+1)(2x1+i3)2(i32x+1)2(x+1)=0

Solving for the coefficients ai in the above using method of undetermined coefficients gives

{a0=0}

Substituting these coefficients in p(x) in eq. (2A) results in

p(x)=x

Therefore the first solution to the ode z=rz is

z1(x)=peωdx=(x)e(16(x+1)16(x12+i32)16(x12i32))dx=(x)1((x+1)(2x1+i3)(i32x+1))1/6=x((x+1)(2x1+i3)(i32x+1))1/6

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

y1=z1e12BAdx=z1e127x2x3+1dx=z1e7ln(x3+1)6=z1(1(x3+1)7/6)

Which simplifies to

y1=x(x3+1)7/6(4x34)1/6

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e7x2x3+1dx(y1)2dx=y1e7ln(x3+1)3(y1)2dx=y1((4x34)1/3x2dx)

Therefore the solution is

y=c1y1+c2y2=c1(x(x3+1)7/6(4x34)1/6)+c2(x(x3+1)7/6(4x34)1/6((4x34)1/3x2dx))

Will add steps showing solving for IC soon.

Maple. Time used: 0.168 (sec). Leaf size: 28
ode:=(x^3+1)*diff(diff(y(x),x),x)+7*x^2*diff(y(x),x)+9*x*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1hypergeom([1,1],[23],x3)+c2x(x3+1)4/3

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 
         <- 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 for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
   <- Kovacics algorithm successful`
 

Maple step by step

Let’s solve(x3+1)(d2dx2y(x))+7x2(ddxy(x))+9xy(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=9xy(x)x3+17x2(ddxy(x))x3+1Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+7x2(ddxy(x))x3+1+9xy(x)x3+1=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=7x2x3+1,P3(x)=9xx3+1](x+1)P2(x)is analytic atx=1((x+1)P2(x))|x=1=73(x+1)2P3(x)is analytic atx=1((x+1)2P3(x))|x=1=0x=1is a regular singular pointCheck to see ifx0is a regular singular pointx0=1Multiply by denominators(x3+1)(d2dx2y(x))+7x2(ddxy(x))+9xy(x)=0Change variables usingx=u1so that the regular singular point is atu=0(u33u2+3u)(d2du2y(u))+(7u214u+7)(dduy(u))+(9u9)y(u)=0Assume series solution fory(u)y(u)=k=0akuk+rRewrite ODE with series expansionsConvertumy(u)to series expansion form=0..1umy(u)=k=0akuk+r+mShift index usingk>kmumy(u)=k=makmuk+rConvertum(dduy(u))to series expansion form=0..2um(dduy(u))=k=0ak(k+r)uk+r1+mShift index usingk>k+1mum(dduy(u))=k=1+mak+1m(k+1m+r)uk+rConvertum(d2du2y(u))to series expansion form=1..3um(d2du2y(u))=k=0ak(k+r)(k+r1)uk+r2+mShift index usingk>k+2mum(d2du2y(u))=k=2+mak+2m(k+2m+r)(k+1m+r)uk+rRewrite ODE with series expansionsa0r(4+3r)u1+r+(a1(1+r)(7+3r)a0(3r2+11r+9))ur+(k=1(ak+1(k+1+r)(3k+7+3r)ak(3k2+6kr+3r2+11k+11r+9)+ak1(k+2+r)2)uk+r)=0a0cannot be 0 by assumption, giving the indicial equationr(4+3r)=0Values of r that satisfy the indicial equationr{0,43}Each term must be 0a1(1+r)(7+3r)a0(3r2+11r+9)=0Each term in the series must be 0, giving the recursion relationak+1(k+1+r)(3k+7+3r)ak(3k2+6kr+3r2+11k+11r+9)+ak1(k+2+r)2=0Shift index usingk>k+1ak+2(k+2+r)(3k+10+3r)ak+1(3(k+1)2+6(k+1)r+3r2+11k+20+11r)+ak(k+r+3)2=0Recursion relation that defines series solution to ODEak+2=k2ak3k2ak+1+2krak6krak+1+r2ak3r2ak+1+6kak17kak+1+6rak17rak+1+9ak23ak+1(k+2+r)(3k+10+3r)Recursion relation forr=0ak+2=k2ak3k2ak+1+6kak17kak+1+9ak23ak+1(k+2)(3k+10)Solution forr=0[y(u)=k=0akuk,ak+2=k2ak3k2ak+1+6kak17kak+1+9ak23ak+1(k+2)(3k+10),7a19a0=0]Revert the change of variablesu=x+1[y(x)=k=0ak(x+1)k,ak+2=k2ak3k2ak+1+6kak17kak+1+9ak23ak+1(k+2)(3k+10),7a19a0=0]Recursion relation forr=43ak+2=k2ak3k2ak+1+103kak9kak+1+259ak173ak+1(k+23)(3k+6)Solution forr=43[y(u)=k=0akuk43,ak+2=k2ak3k2ak+1+103kak9kak+1+259ak173ak+1(k+23)(3k+6),a1+a03=0]Revert the change of variablesu=x+1[y(x)=k=0ak(x+1)k43,ak+2=k2ak3k2ak+1+103kak9kak+1+259ak173ak+1(k+23)(3k+6),a1+a03=0]Combine solutions and rename parameters[y(x)=(k=0ak(x+1)k)+(k=0bk(x+1)k43),ak+2=k2ak3k2ak+1+6kak17kak+1+9ak23ak+1(k+2)(3k+10),7a19a0=0,bk+2=k2bk3k2bk+1+103kbk9kbk+1+259bk173bk+1(k+23)(3k+6),b1+b03=0]
Mathematica. Time used: 0.638 (sec). Leaf size: 39
ode=(1+x^3)*D[y[x],{x,2}]+7*x^2*D[y[x],x]+9*x*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)c1xc2Hypergeometric2F1(13,13,23,x3)(x3+1)4/3
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(7*x**2*Derivative(y(x), x) + 9*x*y(x) + (x**3 + 1)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False