2.1.259 Problem 262

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9431]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 262
Date solved : Wednesday, March 05, 2025 at 07:49:09 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

(4x314x22x)y(6x27x+1)y+(6x1)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.416 (sec)

Writing the ode as

(1)(4x314x22x)y+(6x2+7x1)y+(6x1)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=4x314x22x(3)B=6x2+7x1C=6x1

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=12x4+156x3+297x278x316(2x37x2x)2

Comparing the above to (5) shows that

s=12x4+156x3+297x278x3t=16(2x37x2x)2

Therefore eq. (4) becomes

(7)z(x)=(12x4+156x3+297x278x316(2x37x2x)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 2.259: 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)=64=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=16(2x37x2x)2. There is a pole at x=0 of order 2. There is a pole at x=74+574 of order 2. There is a pole at x=74574 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=94x316x2+34(x74574)2+34(x74+574)2+982957152x74574+98+2957152x74+574

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=316. Hence

[r]c=0αc+=12+1+4b=34αc=121+4b=14

For the pole at x=74+574 let b be the coefficient of 1(x74574)2 in the partial fractions decomposition of r given above. Therefore b=34. Hence

[r]c=0αc+=12+1+4b=32αc=121+4b=12

For the pole at x=74574 let b be the coefficient of 1(x74+574)2 in the partial fractions decomposition of r given above. Therefore b=34. Hence

[r]c=0αc+=12+1+4b=32αc=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=12x4+156x3+297x278x316(2x37x2x)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=12x4+156x3+297x278x316(2x37x2x)2

pole c location pole order [r]c αc+ αc
0 2 0 34 14
74+574 2 0 32 12
74574 2 0 32 12

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=14 then

d=α(αc1+αc2+αc3)=14(34)=1

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+αc2xc2)+(()[r]c3+αc3xc3)+()[r]=14x12(x74574)12(x74+574)+()(0)=14x12(x74574)12(x74+574)=6x2+7x18x328x24x

Now that ω is determined, the next step is find a corresponding minimal polynomial p(x) of degree d=1 to solve the ode. The polynomial p(x) needs to satisfy the equation

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(x)=x+a0

Substituting the above in eq. (1A) gives

(0)+2(14x12(x74574)12(x74+574))(1)+((14x2+12(x74574)2+12(x74+574)2)+(14x12(x74574)12(x74+574))2(12x4+156x3+297x278x316(2x37x2x)2))=032(a0+1)(6x1)(2x27x1)(4x7+57)2(4x+7+57)2x=0

Solving for the coefficients ai in the above using method of undetermined coefficients gives

{a0=1}

Substituting these coefficients in p(x) in eq. (2A) results in

p(x)=x1

Therefore the first solution to the ode z=rz is

z1(x)=peωdx=(x1)e(14x12(x74574)12(x74+574))dx=(x1)e2ln(x)(7+57)(7+57)(57+757)57ln(4x757)2(399+5757)+(57+757)57ln(4x7+57)798+11457=(x1)x1/44x7574x7+57

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

y1=z1e12BAdx=z1e126x2+7x14x314x22xdx=z1eln(x)4+ln(2x27x1)2=z1(2x27x1x1/4)

Which simplifies to

y1=(x1)24

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e6x2+7x14x314x22xdx(y1)2dx=y1eln(x)2+ln(2x27x1)(y1)2dx=y1(16x(2x+1)eln(x)2+ln(2x27x1)(x1)(2x27x1))

Therefore the solution is

y=c1y1+c2y2=c1((x1)24)+c2((x1)24(16x(2x+1)eln(x)2+ln(2x27x1)(x1)(2x27x1)))

Will add steps showing solving for IC soon.

Maple. Time used: 0.007 (sec). Leaf size: 21
ode:=(4*x^3-14*x^2-2*x)*diff(diff(y(x),x),x)-(6*x^2-7*x+1)*diff(y(x),x)+(6*x-1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c2x+c1(x1)+2c2x3/2

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) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful`
 

Maple step by step

Let’s solve(4x314x22x)(d2dx2y(x))(6x27x+1)(ddxy(x))+(6x1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(6x1)y(x)2x(2x27x1)+(6x27x+1)(ddxy(x))2x(2x27x1)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)(6x27x+1)(ddxy(x))2x(2x27x1)+(6x1)y(x)2x(2x27x1)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=6x27x+12x(2x27x1),P3(x)=6x12x(2x27x1)]xP2(x)is analytic atx=0(xP2(x))|x=0=12x2P3(x)is analytic atx=0(x2P3(x))|x=0=0x=0is a regular singular pointCheck to see ifx0is a regular singular pointx0=0Multiply by denominators2x(2x27x1)(d2dx2y(x))+(6x2+7x1)(ddxy(x))+(6x1)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=0..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=1..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 expansionsa0r(1+2r)x1+r+(a1(1+r)(1+2r)a0(14r221r+1))xr+(k=1(ak+1(k+1+r)(2k+1+2r)ak(14k2+28kr+14r221k21r+1)+2ak1(k2+r)(2k5+2r))xk+r)=0a0cannot be 0 by assumption, giving the indicial equationr(1+2r)=0Values of r that satisfy the indicial equationr{0,12}Each term must be 0a1(1+r)(1+2r)a0(14r221r+1)=0Each term in the series must be 0, giving the recursion relation(14ak+4ak12ak+1)k2+((28ak+8ak14ak+1)r+21ak18ak13ak+1)k+(14ak+4ak12ak+1)r2+(21ak18ak13ak+1)rak+20ak1ak+1=0Shift index usingk>k+1(14ak+1+4ak2ak+2)(k+1)2+((28ak+1+8ak4ak+2)r+21ak+118ak3ak+2)(k+1)+(14ak+1+4ak2ak+2)r2+(21ak+118ak3ak+2)rak+1+20akak+2=0Recursion relation that defines series solution to ODEak+2=4k2ak14k2ak+1+8krak28krak+1+4r2ak14r2ak+110kak7kak+110rak7rak+1+6ak+6ak+12k2+4kr+2r2+7k+7r+6Recursion relation forr=0ak+2=4k2ak14k2ak+110kak7kak+1+6ak+6ak+12k2+7k+6Solution forr=0[y(x)=k=0akxk,ak+2=4k2ak14k2ak+110kak7kak+1+6ak+6ak+12k2+7k+6,a1a0=0]Recursion relation forr=12ak+2=4k2ak14k2ak+16kak21kak+1+2akak+12k2+9k+10Solution forr=12[y(x)=k=0akxk+12,ak+2=4k2ak14k2ak+16kak21kak+1+2akak+12k2+9k+10,3a1+6a0=0]Combine solutions and rename parameters[y(x)=(k=0akxk)+(k=0bkxk+12),ak+2=4k2ak14k2ak+110kak7kak+1+6ak+6ak+12k2+7k+6,a1a0=0,bk+2=4k2bk14k2bk+16kbk21kbk+1+2bkbk+12k2+9k+10,3b1+6b0=0]
Mathematica. Time used: 1.932 (sec). Leaf size: 155
ode=(4*x^3-14*x^2-2*x)*D[y[x],{x,2}]-(6*x^2-7*x+1)*D[y[x],x]+(6*x-1)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)(x1)exp(1x6K[1]27K[1]+18K[1]3+28K[1]2+4K[1]dK[1]121x(74K[2]K[2](2K[2]7)1+12K[2])dK[2])(c21xexp(21K[3]6K[1]27K[1]+18K[1]3+28K[1]2+4K[1]dK[1])(K[3]1)2dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((6*x - 1)*y(x) - (6*x**2 - 7*x + 1)*Derivative(y(x), x) + (4*x**3 - 14*x**2 - 2*x)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False