2.1.589 Problem 605

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9761]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 605
Date solved : Wednesday, March 05, 2025 at 07:58:36 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

2x2(2+3x)y+x(4+21x)y(19x)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.213 (sec)

Writing the ode as

(1)(6x3+4x2)y+(21x2+4x)y+(9x1)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=6x3+4x2(3)B=21x2+4xC=9x1

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=27x4816x(2+3x)2

Comparing the above to (5) shows that

s=27x48t=16x(2+3x)2

Therefore eq. (4) becomes

(7)z(x)=(27x4816x(2+3x)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 2.589: 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)=31=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=16x(2+3x)2. There is a pole at x=0 of order 1. There is a pole at x=23 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 1. For the pole at x=0 of order 1 then

[r]c=0αc+=1αc=1

Looking at poles of order 2. The partial fractions decomposition of r is

r=34x+516(x+23)2+34(x+23)

For the pole at x=23 let b be the coefficient of 1(x+23)2 in the partial fractions decomposition of r given above. Therefore b=516. Hence

[r]c=0αc+=12+1+4b=54αc=121+4b=14

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=27x4816x(2+3x)2

Since the gcd(s,t)=1. This gives b=316. Hence

[r]=0α+=12+1+4b=34α=121+4b=14

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

r=27x4816x(2+3x)2

pole c location pole order [r]c αc+ αc
0 1 0 0 1
23 2 0 54 14

Order of r at [r] α+ α
2 0 34 14

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+=34 then

d=α+(αc1+αc2)=34(34)=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]

Substituting the above values in the above results in

ω=(()[r]c1+αc1xc1)+(()[r]c2+αc2xc2)+(+)[r]=1x14(x+23)+(0)=1x14(x+23)=8+9x12x2+8x

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(1x14(x+23))(0)+((1x2+14(x+23)2)+(1x14(x+23))2(27x4816x(2+3x)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(1x14(x+23))dx=x(2+3x)1/4

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

y1=z1e12BAdx=z1e1221x2+4x6x3+4x2dx=z1eln(x)25ln(2+3x)4=z1(1x(2+3x)5/4)

Which simplifies to

y1=x(2+3x)3/2

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e21x2+4x6x3+4x2dx(y1)2dx=y1eln(x)5ln(2+3x)2(y1)2dx=y1(2+3xx32arctanh(2+3x22)2)

Therefore the solution is

y=c1y1+c2y2=c1(x(2+3x)3/2)+c2(x(2+3x)3/2(2+3xx32arctanh(2+3x22)2))

Will add steps showing solving for IC soon.

Maple. Time used: 0.043 (sec). Leaf size: 48
ode:=2*x^2*(2+3*x)*diff(diff(y(x),x),x)+x*(4+21*x)*diff(y(x),x)-(1-9*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=2+3x2c2+c1x+3arctanh(22+3x2)c2x(2+3x)3/2x

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 solve2x2(3x+2)(d2dx2y(x))+x(4+21x)(ddxy(x))(19x)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(1+9x)y(x)2(3x+2)x2(4+21x)(ddxy(x))2x(3x+2)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+(4+21x)(ddxy(x))2x(3x+2)+(1+9x)y(x)2(3x+2)x2=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=4+21x2x(3x+2),P3(x)=1+9x2(3x+2)x2]xP2(x)is analytic atx=0(xP2(x))|x=0=1x2P3(x)is analytic atx=0(x2P3(x))|x=0=14x=0is a regular singular pointCheck to see ifx0is a regular singular pointx0=0Multiply by denominators2x2(3x+2)(d2dx2y(x))+x(4+21x)(ddxy(x))+(1+9x)y(x)=0Assume series solution fory(x)y(x)=k=0akxk+rRewrite ODE with series expansionsConvertxmy(x)to series expansion form=0..1xmy(x)=k=0akxk+r+mShift index usingk>kmxmy(x)=k=makmxk+rConvertxm(ddxy(x))to series expansion form=1..2xm(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..3xm(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+2r)(1+2r)xr+(k=1(ak(2k+2r+1)(2k+2r1)+3ak1(2k+2r+1)(k+r))xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(1+2r)(1+2r)=0Values of r that satisfy the indicial equationr{12,12}Each term in the series must be 0, giving the recursion relation4((k+r12)ak+3ak1(k+r)2)(k+r+12)=0Shift index usingk>k+14((k+r+12)ak+1+3ak(k+r+1)2)(k+32+r)=0Recursion relation that defines series solution to ODEak+1=3ak(k+r+1)2k+2r+1Recursion relation forr=12ak+1=3ak(k+12)2kSolution forr=12[y(x)=k=0akxk12,ak+1=3ak(k+12)2k]Recursion relation forr=12ak+1=3ak(k+32)2k+2Solution forr=12[y(x)=k=0akxk+12,ak+1=3ak(k+32)2k+2]Combine solutions and rename parameters[y(x)=(k=0akxk12)+(k=0bkxk+12),ak+1=3ak(k+12)2k,bk+1=3bk(k+32)2k+2]
Mathematica. Time used: 0.297 (sec). Leaf size: 102
ode=2*x^2*(2+3*x)*D[y[x],{x,2}]+x*(4+21*x)*D[y[x],x]-(1-9*x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1x(1K[1]312K[1]+8)dK[1]121x(156K[2]+4+1K[2])dK[2])(c21xexp(21K[3]9K[1]+812K[1]2+8K[1]dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**2*(3*x + 2)*Derivative(y(x), (x, 2)) + x*(21*x + 4)*Derivative(y(x), x) - (1 - 9*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False