2.1.577 Problem 593

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9749]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 593
Date solved : Wednesday, March 05, 2025 at 07:58:26 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

9x2(3+x)y+3x(3+7x)y+(3+4x)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.201 (sec)

Writing the ode as

(1)(9x3+27x2)y+(21x2+9x)y+(3+4x)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=9x3+27x2(3)B=21x2+9xC=3+4x

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=14x2

Comparing the above to (5) shows that

s=1t=4x2

Therefore eq. (4) becomes

(7)z(x)=(14x2)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 2.577: 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)=20=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=4x2. There is a pole at x=0 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

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

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=14x2

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=14x2

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

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)=12(12)=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+αc1xc1)+()[r]=12x+()(0)=12x=12x

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)(0)+((12x2)+(12x)2(14x2))=00=0

The equation is satisfied since both sides are zero. Therefore the first solution to the ode z=rz is

z1(x)=peωdx=e12xdx=x

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

y1=z1e12BAdx=z1e1221x2+9x9x3+27x2dx=z1eln(x)6ln(3+x)=z1(1x1/6(3+x))

Which simplifies to

y1=x1/33+x

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e21x2+9x9x3+27x2dx(y1)2dx=y1eln(x)32ln(3+x)(y1)2dx=y1(x29+3x+2x3+x+33+x2ln(3+x)x32ln(3+x)+ln(x)+2ln(3+x)(3+x)3x32)

Therefore the solution is

y=c1y1+c2y2=c1(x1/33+x)+c2(x1/33+x(x29+3x+2x3+x+33+x2ln(3+x)x32ln(3+x)+ln(x)+2ln(3+x)(3+x)3x32))

Will add steps showing solving for IC soon.

Maple. Time used: 0.031 (sec). Leaf size: 19
ode:=9*x^2*(x+3)*diff(diff(y(x),x),x)+3*x*(3+7*x)*diff(y(x),x)+(3+4*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=x1/3(c2ln(x)+c1)x+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 
<- Kovacics algorithm successful`
 

Maple step by step

Let’s solve9x2(x+3)(d2dx2y(x))+3x(7x+3)(ddxy(x))+(3+4x)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(3+4x)y(x)9x2(x+3)(7x+3)(ddxy(x))3x(x+3)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+(7x+3)(ddxy(x))3x(x+3)+(3+4x)y(x)9x2(x+3)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=7x+33x(x+3),P3(x)=3+4x9x2(x+3)](x+3)P2(x)is analytic atx=3((x+3)P2(x))|x=3=2(x+3)2P3(x)is analytic atx=3((x+3)2P3(x))|x=3=0x=3is a regular singular pointCheck to see ifx0is a regular singular pointx0=3Multiply by denominators9x2(x+3)(d2dx2y(x))+3x(7x+3)(ddxy(x))+(3+4x)y(x)=0Change variables usingx=u3so that the regular singular point is atu=0(9u354u2+81u)(d2du2y(u))+(21u2117u+162)(dduy(u))+(9+4u)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 expansions81a0r(1+r)u1+r+(81a1(1+r)(2+r)9a0(1+r)(1+6r))ur+(k=1(81ak+1(k+r+1)(k+2+r)9ak(k+r+1)(6k+6r+1)+ak1(3k1+3r)2)uk+r)=0a0cannot be 0 by assumption, giving the indicial equation81r(1+r)=0Values of r that satisfy the indicial equationr{1,0}Each term must be 081a1(1+r)(2+r)9a0(1+r)(1+6r)=0Each term in the series must be 0, giving the recursion relation81ak+1(k+r+1)(k+2+r)54(k+r+16)(k+r+1)ak+ak1(3k1+3r)2=0Shift index usingk>k+181ak+2(k+2+r)(k+3+r)54(k+76+r)(k+2+r)ak+1+ak(3k+3r+2)2=0Recursion relation that defines series solution to ODEak+2=9k2ak54k2ak+1+18krak108krak+1+9r2ak54r2ak+1+12kak171kak+1+12rak171rak+1+4ak126ak+181(k+2+r)(k+3+r)Recursion relation forr=1ak+2=9k2ak54k2ak+16kak63kak+1+ak9ak+181(k+1)(k+2)Solution forr=1[y(u)=k=0akuk1,ak+2=9k2ak54k2ak+16kak63kak+1+ak9ak+181(k+1)(k+2),0=0]Revert the change of variablesu=x+3[y(x)=k=0ak(x+3)k1,ak+2=9k2ak54k2ak+16kak63kak+1+ak9ak+181(k+1)(k+2),0=0]Recursion relation forr=0ak+2=9k2ak54k2ak+1+12kak171kak+1+4ak126ak+181(k+2)(k+3)Solution forr=0[y(u)=k=0akuk,ak+2=9k2ak54k2ak+1+12kak171kak+1+4ak126ak+181(k+2)(k+3),162a19a0=0]Revert the change of variablesu=x+3[y(x)=k=0ak(x+3)k,ak+2=9k2ak54k2ak+1+12kak171kak+1+4ak126ak+181(k+2)(k+3),162a19a0=0]Combine solutions and rename parameters[y(x)=(k=0ak(x+3)k1)+(k=0bk(x+3)k),ak+2=9k2ak54k2ak+16kak63kak+1+ak9ak+181(k+1)(k+2),0=0,bk+2=9k2bk54k2bk+1+12kbk171kbk+1+4bk126bk+181(k+2)(k+3),162b19b0=0]
Mathematica. Time used: 0.286 (sec). Leaf size: 49
ode=9*x^2*(3+x)*D[y[x],{x,2}]+3*x*(3+7*x)*D[y[x],x]+(3+4*x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)x(c2log(x)+c1)exp(121x(2K[1]+3+13K[1])dK[1])
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(9*x**2*(x + 3)*Derivative(y(x), (x, 2)) + 3*x*(7*x + 3)*Derivative(y(x), x) + (4*x + 3)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False