2.1.591 Problem 607

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9763]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 607
Date solved : Wednesday, March 05, 2025 at 07:58:37 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

4x2(1+x)y+4x(3+8x)y(549x)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.276 (sec)

Writing the ode as

(1)(4x3+4x2)y+(32x2+12x)y+(49x5)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=4x3+4x2(3)B=32x2+12xC=49x5

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=x28x+84(x2+x)2

Comparing the above to (5) shows that

s=x28x+8t=4(x2+x)2

Therefore eq. (4) becomes

(7)z(x)=(x28x+84(x2+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 2.591: 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)=42=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=4(x2+x)2. There is a pole at x=0 of order 2. There is a pole at x=1 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=6x+61+x+2x2+154(1+x)2

For the pole at x=1 let b be the coefficient of 1(1+x)2 in the partial fractions decomposition of r given above. Therefore b=154. Hence

[r]c=0αc+=12+1+4b=52αc=121+4b=32

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=2. Hence

[r]c=0αc+=12+1+4b=2αc=121+4b=1

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=x28x+84(x2+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=x28x+84(x2+x)2

pole c location pole order [r]c αc+ αc
1 2 0 52 32
0 2 0 2 1

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+)=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]c2+αc2+xc2)+()[r]=32(1+x)+2x+()(0)=32(1+x)+2x=x+42x(1+x)

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(32(1+x)+2x)(0)+((32(1+x)22x2)+(32(1+x)+2x)2(x28x+84(x2+x)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(32(1+x)+2x)dx=x2(1+x)3/2

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

y1=z1e12BAdx=z1e1232x2+12x4x3+4x2dx=z1e3ln(x)25ln(1+x)2=z1(1x3/2(1+x)5/2)

Which simplifies to

y1=x(1+x)4

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e32x2+12x4x3+4x2dx(y1)2dx=y1e3ln(x)5ln(1+x)(y1)2dx=y1(13x3+ln(x)3x32x2)

Therefore the solution is

y=c1y1+c2y2=c1(x(1+x)4)+c2(x(1+x)4(13x3+ln(x)3x32x2))

Will add steps showing solving for IC soon.

Maple. Time used: 0.042 (sec). Leaf size: 40
ode:=4*x^2*(x+1)*diff(diff(y(x),x),x)+4*x*(3+8*x)*diff(y(x),x)-(5-49*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1x3+6c2ln(x)x318c2x29c2x2c2(x+1)4x5/2

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))+4x(3+8x)(ddxy(x))(549x)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(5+49x)y(x)4(x+1)x2(3+8x)(ddxy(x))x(x+1)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+(3+8x)(ddxy(x))x(x+1)+(5+49x)y(x)4(x+1)x2=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=3+8xx(x+1),P3(x)=5+49x4(x+1)x2](x+1)P2(x)is analytic atx=1((x+1)P2(x))|x=1=5(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))+4x(3+8x)(ddxy(x))+(5+49x)y(x)=0Change variables usingx=u1so that the regular singular point is atu=0(4u38u2+4u)(d2du2y(u))+(32u252u+20)(dduy(u))+(54+49u)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 expansions4a0r(4+r)u1+r+(4a1(1+r)(5+r)2a0(4r2+22r+27))ur+(k=1(4ak+1(k+1+r)(k+5+r)2ak(4k2+8kr+4r2+22k+22r+27)+ak1(2k+5+2r)2)uk+r)=0a0cannot be 0 by assumption, giving the indicial equation4r(4+r)=0Values of r that satisfy the indicial equationr{4,0}Each term must be 04a1(1+r)(5+r)2a0(4r2+22r+27)=0Each term in the series must be 0, giving the recursion relation4ak+1(k+1+r)(k+5+r)2ak(4k2+8kr+4r2+22k+22r+27)+ak1(2k+5+2r)2=0Shift index usingk>k+14ak+2(k+2+r)(k+6+r)2ak+1(4(k+1)2+8(k+1)r+4r2+22k+49+22r)+ak(2k+2r+7)2=0Recursion relation that defines series solution to ODEak+2=4k2ak8k2ak+1+8krak16krak+1+4r2ak8r2ak+1+28kak60kak+1+28rak60rak+1+49ak106ak+14(k+2+r)(k+6+r)Recursion relation forr=4ak+2=4k2ak8k2ak+14kak+4kak+1+ak+6ak+14(k2)(k+2)Series not valid forr=4, division by0in the recursion relation atk=2ak+2=4k2ak8k2ak+14kak+4kak+1+ak+6ak+14(k2)(k+2)Recursion relation forr=0ak+2=4k2ak8k2ak+1+28kak60kak+1+49ak106ak+14(k+2)(k+6)Solution forr=0[y(u)=k=0akuk,ak+2=4k2ak8k2ak+1+28kak60kak+1+49ak106ak+14(k+2)(k+6),20a154a0=0]Revert the change of variablesu=x+1[y(x)=k=0ak(x+1)k,ak+2=4k2ak8k2ak+1+28kak60kak+1+49ak106ak+14(k+2)(k+6),20a154a0=0]
Mathematica. Time used: 0.194 (sec). Leaf size: 104
ode=4*x^2*(1+x)*D[y[x],{x,2}]+4*x*(3+8*x)*D[y[x],x]-(5-49*x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1xK[1]+42K[1]2+2K[1]dK[1]121x8K[2]+3K[2]2+K[2]dK[2])(c21xexp(21K[3]K[1]+42K[1]2+2K[1]dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*x**2*(x + 1)*Derivative(y(x), (x, 2)) + 4*x*(8*x + 3)*Derivative(y(x), x) - (5 - 49*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False