2.1.152 Problem 154

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9324]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 154
Date solved : Wednesday, March 05, 2025 at 07:47:31 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

x2(x2+1)yx(9x2+1)y+(25x2+1)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.357 (sec)

Writing the ode as

(1)(x4+x2)y+(9x3x)y+(25x2+1)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x4+x2(3)B=9x3xC=25x2+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=x498x214(x3+x)2

Comparing the above to (5) shows that

s=x498x21t=4(x3+x)2

Therefore eq. (4) becomes

(7)z(x)=(x498x214(x3+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 2.152: 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+x)2. There is a pole at x=0 of order 2. There is a pole at x=i of order 2. There is a pole at x=i 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=14x2+6(xi)2+6(x+i)2+6ixi6ix+i

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=i let b be the coefficient of 1(xi)2 in the partial fractions decomposition of r given above. Therefore b=6. Hence

[r]c=0αc+=12+1+4b=3αc=121+4b=2

For the pole at x=i let b be the coefficient of 1(x+i)2 in the partial fractions decomposition of r given above. Therefore b=6. Hence

[r]c=0αc+=12+1+4b=3αc=121+4b=2

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=x498x214(x3+x)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=x498x214(x3+x)2

pole c location pole order [r]c αc+ αc
0 2 0 12 12
i 2 0 3 2
i 2 0 3 2

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(72)=4

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]=12x2xi2x+i+()(0)=12x2xi2x+i=12x4xx2+1

Now that ω is determined, the next step is find a corresponding minimal polynomial p(x) of degree d=4 to solve the ode. The polynomial p(x) needs to satisfy the equation

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(x)=x4+a3x3+a2x2+a1x+a0

Substituting the above in eq. (1A) gives

(12x2+6xa3+2a2)+2(12x2xi2x+i)(4x3+3x2a3+2a2x+a1)+((12x2+2(xi)2+2(x+i)2)+(12x2xi2x+i)2(x498x214(x3+x)2))=0(x4a3+4(4+a2)x3+9(a1+a3)x2+4(4a0+a2)x+a1)(x2+1)x(x+i)2(x+i)2=0

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

{a0=1,a1=0,a2=4,a3=0}

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

p(x)=x44x2+1

Therefore the first solution to the ode z=rz is

z1(x)=peωdx=(x44x2+1)e(12x2xi2x+i)dx=(x44x2+1)eln(x)22ln(x2+1)=(x44x2+1)x(x2+1)2

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

y1=z1e12BAdx=z1e129x3xx4+x2dx=z1eln(x)2+2ln(x2+1)=z1(x(x2+1)2)

Which simplifies to

y1=x54x3+x

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e9x3xx4+x2dx(y1)2dx=y1eln(x)+4ln(x2+1)(y1)2dx=y1(ln(x)+6x2+3x44x2+1)

Therefore the solution is

y=c1y1+c2y2=c1(x54x3+x)+c2(x54x3+x(ln(x)+6x2+3x44x2+1))

Will add steps showing solving for IC soon.

Maple. Time used: 0.007 (sec). Leaf size: 41
ode:=x^2*(x^2+1)*diff(diff(y(x),x),x)-x*(9*x^2+1)*diff(y(x),x)+(25*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=x(c2(x44x2+1)ln(x)+c1x4+(4c16c2)x2+c1+3c2)

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 
<- Kovacics algorithm successful`
 

Maple step by step

Let’s solvex2(x2+1)(d2dx2y(x))x(9x2+1)(ddxy(x))+(25x2+1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(25x2+1)y(x)x2(x2+1)+(9x2+1)(ddxy(x))x(x2+1)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)(9x2+1)(ddxy(x))x(x2+1)+(25x2+1)y(x)x2(x2+1)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=9x2+1x(x2+1),P3(x)=25x2+1x2(x2+1)]xP2(x)is analytic atx=0(xP2(x))|x=0=1x2P3(x)is analytic atx=0(x2P3(x))|x=0=1x=0is a regular singular pointCheck to see ifx0is a regular singular pointx0=0Multiply by denominatorsx2(x2+1)(d2dx2y(x))x(9x2+1)(ddxy(x))+(25x2+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=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+r)2xr+a1r2x1+r+(k=2(ak(k+r1)2+ak2(k7+r)2)xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(1+r)2=0Values of r that satisfy the indicial equationr=1Each term must be 0a1r2=0Solve for the dependent coefficient(s)a1=0Each term in the series must be 0, giving the recursion relationak(k+r1)2+ak2(k7+r)2=0Shift index usingk>k+2ak+2(k+1+r)2+ak(k+r5)2=0Recursion relation that defines series solution to ODEak+2=ak(k+r5)2(k+1+r)2Recursion relation forr=1; series terminates atk=4ak+2=ak(k4)2(k+2)2Solution forr=1[y(x)=k=0akxk+1,ak+2=ak(k4)2(k+2)2,a1=0]
Mathematica. Time used: 5.271 (sec). Leaf size: 138
ode=x^2*(1+x^2)*D[y[x],{x,2}]-x*(1+9*x^2)*D[y[x],x]+(1+25*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)(x44x2+1)exp(1x17K[1]22(K[1]3+K[1])dK[1]121x9K[2]2+1K[2]3+K[2]dK[2])(c21xexp(21K[3]17K[1]22(K[1]3+K[1])dK[1])(K[3]44K[3]2+1)2dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(x**2 + 1)*Derivative(y(x), (x, 2)) - x*(9*x**2 + 1)*Derivative(y(x), x) + (25*x**2 + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False