2.1.614 Problem 630

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9786]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 630
Date solved : Wednesday, March 05, 2025 at 07:58:58 AM
CAS classification : [_Gegenbauer]

Solve

(t2+1)y2ty+2y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.297 (sec)

Writing the ode as

(1)(t2+1)y2ty+2y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=t2+1(3)B=2tC=2

Applying the Liouville transformation on the dependent variable gives

z(t)=yeB2Adt

Then (2) becomes

(4)z(t)=rz(t)

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=2t23(t21)2

Comparing the above to (5) shows that

s=2t23t=(t21)2

Therefore eq. (4) becomes

(7)z(t)=(2t23(t21)2)z(t)

Equation (7) is now solved. After finding z(t) then y is found using the inverse transformation

y=z(t)eB2Adt

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 2.614: 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=(t21)2. There is a pole at t=1 of order 2. There is a pole at t=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=54(t1)14(t1)214(t+1)254(t+1)

For the pole at t=1 let b be the coefficient of 1(t1)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 t=1 let b be the coefficient of 1(t+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 1t2 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=2t23(t21)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=2t23(t21)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)tc)+s()[r]

Substituting the above values in the above results in

ω=(()[r]c1+αc1tc1)+(()[r]c2+αc2tc2)+(+)[r]=12t2+12t+2+(0)=12t2+12t+2=tt21

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

(1A)p+2ωp+(ω+ω2r)p=0

Let

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

Substituting the above in eq. (1A) gives

(0)+2(12t2+12t+2)(1)+((12(t1)212(t+1)2)+(12t2+12t+2)2(2t23(t21)2))=02a0t21=0

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

{a0=0}

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

p(t)=t

Therefore the first solution to the ode z=rz is

z1(t)=peωdt=(t)e(12t2+12t+2)dt=(t)(t1)(t+1)=tt21

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

y1=z1e12BAdt=z1e122tt2+1dt=z1eln(t1)2ln(t+1)2=z1(1t1t+1)

Which simplifies to

y1=tt21t1t+1

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

y2=y1eBAdty12dt

Substituting gives

y2=y1e2tt2+1dt(y1)2dt=y1eln(t1)ln(t+1)(y1)2dt=y1(ln(t1)2ln(t+1)2+1t)

Therefore the solution is

y=c1y1+c2y2=c1(tt21t1t+1)+c2(tt21t1t+1(ln(t1)2ln(t+1)2+1t))

Will add steps showing solving for IC soon.

Maple. Time used: 0.006 (sec). Leaf size: 25
ode:=(-t^2+1)*diff(diff(y(t),t),t)-2*t*diff(y(t),t)+2*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 
y=c2ln(t+1)t2+c2ln(t1)t2+c1t+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(t2+1)(d2dt2y(t))2t(ddty(t))+2y(t)=0Highest derivative means the order of the ODE is2d2dt2y(t)Isolate 2nd derivatived2dt2y(t)=2y(t)t212(ddty(t))tt21Group terms withy(t)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dt2y(t)+2(ddty(t))tt212y(t)t21=0Check to see ift0is a regular singular pointDefine functions[P2(t)=2tt21,P3(t)=2t21](t+1)P2(t)is analytic att=1((t+1)P2(t))|t=1=1(t+1)2P3(t)is analytic att=1((t+1)2P3(t))|t=1=0t=1is a regular singular pointCheck to see ift0is a regular singular pointt0=1Multiply by denominators(t21)(d2dt2y(t))+2t(ddty(t))2y(t)=0Change variables usingt=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=t+1[y(t)=a0t]
Mathematica. Time used: 0.02 (sec). Leaf size: 33
ode=(1-t^2)*D[y[t],{t,2}]-2*t*D[y[t],t]+2*y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 
y(t)c1t12c2(tlog(1t)tlog(t+1)+2)
Sympy. Time used: 0.869 (sec). Leaf size: 20
from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-2*t*Derivative(y(t), t) + (1 - t**2)*Derivative(y(t), (t, 2)) + 2*y(t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 
y(t)=C2(t43t2+1)+C1t+O(t6)