2.1.131 Problem 133

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9303]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 133
Date solved : Wednesday, March 05, 2025 at 07:47:12 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

36x2(12x)y+24x(19x)y+(170x)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.360 (sec)

Writing the ode as

(1)(72x3+36x2)y+(216x2+24x)y+(170x)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=72x3+36x2(3)B=216x2+24xC=170x

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=32x2+48x936(2x2x)2

Comparing the above to (5) shows that

s=32x2+48x9t=36(2x2x)2

Therefore eq. (4) becomes

(7)z(x)=(32x2+48x936(2x2x)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 2.131: 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=36(2x2x)2. There is a pole at x=0 of order 2. There is a pole at x=12 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=736(x12)213(x12)14x2+13x

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=12 let b be the coefficient of 1(x12)2 in the partial fractions decomposition of r given above. Therefore b=736. Hence

[r]c=0αc+=12+1+4b=76αc=121+4b=16

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=32x2+48x936(2x2x)2

Since the gcd(s,t)=1. This gives b=29. Hence

[r]=0α+=12+1+4b=23α=121+4b=13

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

r=32x2+48x936(2x2x)2

pole c location pole order [r]c αc+ αc
0 2 0 12 12
12 2 0 76 16

Order of r at [r] α+ α
2 0 23 13

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=13 then

d=α(αc1++αc2)=13(13)=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+αc1+xc1)+(()[r]c2+αc2xc2)+()[r]=12x16(x12)+()(0)=12x16(x12)=3+4x12x26x

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(12x16(x12))(0)+((12x2+16(x12)2)+(12x16(x12))2(32x2+48x936(2x2x)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(12x16(x12))dx=x(1+2x)1/6

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

y1=z1e12BAdx=z1e12216x2+24x72x3+36x2dx=z1e7ln(1+2x)6ln(x)3=z1(1(1+2x)7/6x1/3)

Which simplifies to

y1=x1/6(1+2x)4/3

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e216x2+24x72x3+36x2dx(y1)2dx=y1e7ln(1+2x)32ln(x)3(y1)2dx=y1(3(1+2x)1/3ln((1+2x)1/3+1)+ln((1+2x)2/3(1+2x)1/3+1)23arctan((1+2(1+2x)1/3)33))

Therefore the solution is

y=c1y1+c2y2=c1(x1/6(1+2x)4/3)+c2(x1/6(1+2x)4/3(3(1+2x)1/3ln((1+2x)1/3+1)+ln((1+2x)2/3(1+2x)1/3+1)23arctan((1+2(1+2x)1/3)33)))

Will add steps showing solving for IC soon.

Maple. Time used: 0.150 (sec). Leaf size: 93
ode:=36*x^2*(1-2*x)*diff(diff(y(x),x),x)+24*x*(1-9*x)*diff(y(x),x)+(1-70*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=x1/6(23arctan(3(1+2x)1/32+(1+2x)1/3)c22ln(1+(1+2x)1/3)c2+ln(1(1+2x)1/3+(1+2x)2/3)c2+6c2(1+2x)1/3+3c1)3(1+2x)4/3

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 
   Solution has integrals. Trying a special function solution free of integrals... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Kummer 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         <- heuristic approach successful 
         -> solution has integrals; searching for one without integrals... 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
            <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
         <- hypergeometric solution without integrals succesful 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
   <- Kovacics algorithm successful`
 

Maple step by step

Let’s solve36x2(2x+1)(d2dx2y(x))+24x(19x)(ddxy(x))+(170x)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(1+70x)y(x)36x2(2x1)2(1+9x)(ddxy(x))3x(2x1)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+2(1+9x)(ddxy(x))3x(2x1)+(1+70x)y(x)36x2(2x1)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=2(1+9x)3x(2x1),P3(x)=1+70x36x2(2x1)]xP2(x)is analytic atx=0(xP2(x))|x=0=23x2P3(x)is analytic atx=0(x2P3(x))|x=0=136x=0is a regular singular pointCheck to see ifx0is a regular singular pointx0=0Multiply by denominators36x2(2x1)(d2dx2y(x))+24x(1+9x)(ddxy(x))+(1+70x)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+6r)2xr+(k=1(ak(6k+6r1)2+2ak1(6k+1+6r)(6k+6r1))xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(1+6r)2=0Values of r that satisfy the indicial equationr=16Each term in the series must be 0, giving the recursion relation36((2k2r13)ak1+ak(k+r16))(k+r16)=0Shift index usingk>k+136((2k732r)ak+ak+1(k+56+r))(k+56+r)=0Recursion relation that defines series solution to ODEak+1=2(6k+6r+7)ak6k+6r+5Recursion relation forr=16ak+1=2(6k+8)ak6k+6Solution forr=16[y(x)=k=0akxk+16,ak+1=2(6k+8)ak6k+6]
Mathematica. Time used: 0.296 (sec). Leaf size: 112
ode=36*x^2*(1-2*x)*D[y[x],{x,2}]+24*x*(1-9*x)*D[y[x],x]+(1-70*x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1x34K[1]6K[1]12K[1]2dK[1]121x218K[2]3K[2]6K[2]2dK[2])(c21xexp(21K[3]34K[1]6K[1]12K[1]2dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(36*x**2*(1 - 2*x)*Derivative(y(x), (x, 2)) + 24*x*(1 - 9*x)*Derivative(y(x), x) + (1 - 70*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False