2.1.576 Problem 592

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9748]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 592
Date solved : Wednesday, March 05, 2025 at 07:58:25 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

4x2(1+x)y+8x2y+(1+x)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.135 (sec)

Writing the ode as

(1)(4x3+4x2)y+8x2y+(1+x)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=4x3+4x2(3)B=8x2C=1+x

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 2.576: 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=z1e128x24x3+4x2dx=z1eln(1+x)=z1(11+x)

Which simplifies to

y1=x1+x

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e8x24x3+4x2dx(y1)2dx=y1e2ln(1+x)(y1)2dx=y1(ln(x))

Therefore the solution is

y=c1y1+c2y2=c1(x1+x)+c2(x1+x(ln(x)))

Will add steps showing solving for IC soon.

Maple. Time used: 0.042 (sec). Leaf size: 19
ode:=4*x^2*(x+1)*diff(diff(y(x),x),x)+8*x^2*diff(y(x),x)+y(x)*(x+1) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=x(c2ln(x)+c1)x+1

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 solve4x2(x+1)(d2dx2y(x))+8x2(ddxy(x))+(x+1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=y(x)4x22(ddxy(x))x+1Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+2(ddxy(x))x+1+y(x)4x2=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=2x+1,P3(x)=14x2](x+1)P2(x)is analytic atx=1((x+1)P2(x))|x=1=2(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 denominators4x2(x+1)(d2dx2y(x))+8x2(ddxy(x))+(x+1)y(x)=0Change variables usingx=u1so that the regular singular point is atu=0(4u38u2+4u)(d2du2y(u))+(8u216u+8)(dduy(u))+uy(u)=0Assume series solution fory(u)y(u)=k=0akuk+rRewrite ODE with series expansionsConvertuy(u)to series expansionuy(u)=k=0akuk+r+1Shift index usingk>k1uy(u)=k=1ak1uk+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 expansions4a0r(1+r)u1+r+(4a1(1+r)(2+r)8a0r(1+r))ur+(k=1(4ak+1(k+r+1)(k+2+r)8ak(k+r)(k+r+1)+ak1(2k1+2r)2)uk+r)=0a0cannot be 0 by assumption, giving the indicial equation4r(1+r)=0Values of r that satisfy the indicial equationr{1,0}Each term must be 04a1(1+r)(2+r)8a0r(1+r)=0Each term in the series must be 0, giving the recursion relationak1(2k1+2r)28(k+r+1)((k2r21)ak+1+ak(k+r))=0Shift index usingk>k+1ak(2k+2r+1)28(k+2+r)((k232r2)ak+2+ak+1(k+r+1))=0Recursion relation that defines series solution to ODEak+2=4k2ak8k2ak+1+8krak16krak+1+4r2ak8r2ak+1+4kak24kak+1+4rak24rak+1+ak16ak+14(k+2+r)(k+3+r)Recursion relation forr=1ak+2=4k2ak8k2ak+14kak8kak+1+ak4(k+1)(k+2)Solution forr=1[y(u)=k=0akuk1,ak+2=4k2ak8k2ak+14kak8kak+1+ak4(k+1)(k+2),0=0]Revert the change of variablesu=x+1[y(x)=k=0ak(x+1)k1,ak+2=4k2ak8k2ak+14kak8kak+1+ak4(k+1)(k+2),0=0]Recursion relation forr=0ak+2=4k2ak8k2ak+1+4kak24kak+1+ak16ak+14(k+2)(k+3)Solution forr=0[y(u)=k=0akuk,ak+2=4k2ak8k2ak+1+4kak24kak+1+ak16ak+14(k+2)(k+3),8a1=0]Revert the change of variablesu=x+1[y(x)=k=0ak(x+1)k,ak+2=4k2ak8k2ak+1+4kak24kak+1+ak16ak+14(k+2)(k+3),8a1=0]Combine solutions and rename parameters[y(x)=(k=0ak(x+1)k1)+(k=0bk(x+1)k),ak+2=4k2ak8k2ak+14kak8kak+1+ak4(k+1)(k+2),0=0,bk+2=4k2bk8k2bk+1+4kbk24kbk+1+bk16bk+14(k+2)(k+3),8b1=0]
Mathematica. Time used: 0.031 (sec). Leaf size: 24
ode=4*x^2*(1+x)*D[y[x],{x,2}]+8*x^2*D[y[x],x]+(1+x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)x(c2log(x)+c1)x+1
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*x**2*(x + 1)*Derivative(y(x), (x, 2)) + 8*x**2*Derivative(y(x), x) + (x + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False