2.1.764 Problem 786

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9936]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 786
Date solved : Wednesday, March 05, 2025 at 08:01:09 AM
CAS classification : [_Gegenbauer]

Solve

(x2+1)y2xy+2y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.298 (sec)

Writing the ode as

(1)(x2+1)y2xy+2y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x2+1(3)B=2xC=2

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=2x23(x21)2

Comparing the above to (5) shows that

s=2x23t=(x21)2

Therefore eq. (4) becomes

(7)z(x)=(2x23(x21)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 2.764: 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=(x21)2. There is a pole at x=1 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=14(x1)214(x+1)2+54(x1)54(x+1)

For the pole at x=1 let b be the coefficient of 1(x1)2 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=1 let b be the coefficient of 1(x+1)2 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=2x23(x21)2

Since the gcd(s,t)=1. This gives b=2. Hence

[r]=0α+=12+1+4b=2α=121+4b=1

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

r=2x23(x21)2

pole c location pole order [r]c αc+ αc
1 2 0 12 12
1 2 0 12 12

Order of r at [r] α+ α
2 0 2 1

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+=2 then

d=α+(αc1+αc2)=2(1)=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]

Substituting the above values in the above results in

ω=(()[r]c1+αc1xc1)+(()[r]c2+αc2xc2)+(+)[r]=12x2+12x+2+(0)=12x2+12x+2=xx21

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(12x2+12x+2)(1)+((12(x1)212(x+1)2)+(12x2+12x+2)2(2x23(x21)2))=02a0x21=0

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

{a0=0}

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

p(x)=x

Therefore the first solution to the ode z=rz is

z1(x)=peωdx=(x)e(12x2+12x+2)dx=(x)(x1)(x+1)=xx21

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

y1=z1e12BAdx=z1e122xx2+1dx=z1eln(x1)2ln(x+1)2=z1(1x1x+1)

Which simplifies to

y1=xx21x1x+1

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e2xx2+1dx(y1)2dx=y1eln(x1)ln(x+1)(y1)2dx=y1(1xln(x+1)2+ln(x1)2)

Therefore the solution is

y=c1y1+c2y2=c1(xx21x1x+1)+c2(xx21x1x+1(1xln(x+1)2+ln(x1)2))

Will add steps showing solving for IC soon.

Maple. Time used: 0.007 (sec). Leaf size: 25
ode:=(-x^2+1)*diff(diff(y(x),x),x)-2*x*diff(y(x),x)+2*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=ln(x+1)c2x2+c2ln(x1)x2+c1x+c2

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 solve(x2+1)(d2dx2y(x))2x(ddxy(x))+2y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=2y(x)x212(ddxy(x))xx21Group 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))xx212y(x)x21=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=2xx21,P3(x)=2x21](x+1)P2(x)is analytic atx=1((x+1)P2(x))|x=1=1(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 denominators(x21)(d2dx2y(x))+2x(ddxy(x))2y(x)=0Change variables usingx=u1so that the regular singular point is atu=0(u22u)(d2du2y(u))+(2u2)(dduy(u))2y(u)=0Assume series solution fory(u)y(u)=k=0akuk+rRewrite ODE with series expansionsConvertum(dduy(u))to series expansion form=0..1um(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..2um(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 expansions2a0r2u1+r+(k=0(2ak+1(k+1+r)2+ak(k+r+2)(k+r1))uk+r)=0a0cannot be 0 by assumption, giving the indicial equation2r2=0Values of r that satisfy the indicial equationr=0Each term in the series must be 0, giving the recursion relation2ak+1(k+1)2+ak(k+2)(k1)=0Recursion relation that defines series solution to ODEak+1=ak(k+2)(k1)2(k+1)2Recursion relation forr=0; series terminates atk=1ak+1=ak(k+2)(k1)2(k+1)2Apply recursion relation fork=0a1=a0Terminating series solution of the ODE forr=0. Use reduction of order to find the second linearly independent solutiony(u)=a0(u+1)Revert the change of variablesu=x+1[y(x)=a0x]
Mathematica. Time used: 0.02 (sec). Leaf size: 33
ode=(1-x^2)*D[y[x],{x,2}]-2*x*D[y[x],x]+2*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)c1x12c2(xlog(1x)xlog(x+1)+2)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-2*x*Derivative(y(x), x) + (1 - x**2)*Derivative(y(x), (x, 2)) + 2*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False