2.1.346 Problem 353

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9518]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 353
Date solved : Wednesday, March 05, 2025 at 07:50:22 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

4(t23t+2)y2y+y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.306 (sec)

Writing the ode as

(1)(4t212t+8)y2y+y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=4t212t+8(3)B=2C=1

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=4t2+20t1916(t23t+2)2

Comparing the above to (5) shows that

s=4t2+20t19t=16(t23t+2)2

Therefore eq. (4) becomes

(7)z(t)=(4t2+20t1916(t23t+2)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 2.346: 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=16(t23t+2)2. There is a pole at t=2 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=38(t1)316(t1)238(t2)+516(t2)2

For the pole at t=2 let b be the coefficient of 1(t2)2 in the partial fractions decomposition of r given above. Therefore b=516. Hence

[r]c=0αc+=12+1+4b=54αc=121+4b=14

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=316. Hence

[r]c=0αc+=12+1+4b=34αc=121+4b=14

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=4t2+20t1916(t23t+2)2

Since the gcd(s,t)=1. This gives b=14. Hence

[r]=0α+=12+1+4b=12α=121+4b=12

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

r=4t2+20t1916(t23t+2)2

pole c location pole order [r]c αc+ αc
2 2 0 54 14
1 2 0 34 14

Order of r at [r] α+ α
2 0 12 12

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=12 then

d=α(αc1+αc2+)=12(12)=0

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]

The above gives

ω=(()[r]c1+αc1tc1)+((+)[r]c2+αc2+tc2)+()[r]=14(t2)+34(t1)+()(0)=14(t2)+34(t1)=2t54(t1)(t2)

Now that ω is determined, the next step is find a corresponding minimal polynomial p(t) of degree d=0 to solve the ode. The polynomial p(t) needs to satisfy the equation

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(t)=1

Substituting the above in eq. (1A) gives

(0)+2(14(t2)+34(t1))(0)+((14(t2)234(t1)2)+(14(t2)+34(t1))2(4t2+20t1916(t23t+2)2))=00=0

The equation is satisfied since both sides are zero. Therefore the first solution to the ode z=rz is

z1(t)=peωdt=e(14(t2)+34(t1))dt=(t1)3/4(t2)1/4

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

y1=z1e12BAdt=z1e1224t212t+8dt=z1eln(t1)4+ln(t2)4=z1((t2)1/4(t1)1/4)

Which simplifies to

y1=t1

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

y2=y1eBAdty12dt

Substituting gives

y2=y1e24t212t+8dt(y1)2dt=y1eln(t1)2+ln(t2)2(y1)2dt=y1(2t2t1+ln(32+t+t23t+2)(t1)(t2)t2t1)

Therefore the solution is

y=c1y1+c2y2=c1(t1)+c2(t1(2t2t1+ln(32+t+t23t+2)(t1)(t2)t2t1))

Will add steps showing solving for IC soon.

Maple. Time used: 0.019 (sec). Leaf size: 56
ode:=4*(t^2-3*t+2)*diff(diff(y(t),t),t)-2*diff(y(t),t)+y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 
y=c1t1+c2(t23t+2(ln(2)+ln(3+2t+2(t1)(t2)))2+t2)t2

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 solve4(t23t+2)(d2dt2y(t))2ddty(t)+y(t)=0Highest derivative means the order of the ODE is2d2dt2y(t)Isolate 2nd derivatived2dt2y(t)=y(t)4(t23t+2)+ddty(t)2(t23t+2)Group terms withy(t)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dt2y(t)ddty(t)2(t23t+2)+y(t)4(t23t+2)=0Check to see ift0is a regular singular pointDefine functions[P2(t)=12(t23t+2),P3(t)=14(t23t+2)](t1)P2(t)is analytic att=1((t1)P2(t))|t=1=12(t1)2P3(t)is analytic att=1((t1)2P3(t))|t=1=0t=1is a regular singular pointCheck to see ift0is a regular singular pointt0=1Multiply by denominators(4t212t+8)(d2dt2y(t))2ddty(t)+y(t)=0Change variables usingt=u+1so that the regular singular point is atu=0(4u24u)(d2du2y(u))2dduy(u)+y(u)=0Assume series solution fory(u)y(u)=k=0akuk+rRewrite ODE with series expansionsConvertdduy(u)to series expansiondduy(u)=k=0ak(k+r)uk+r1Shift index usingk>k+1dduy(u)=k=1ak+1(k+1+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 expansions2a0r(1+2r)u1+r+(k=0(2ak+1(k+1+r)(2k+1+2r)+ak(2k+2r1)2)uk+r)=0a0cannot be 0 by assumption, giving the indicial equation2r(1+2r)=0Values of r that satisfy the indicial equationr{0,12}Each term in the series must be 0, giving the recursion relationak(2k+2r1)24(k+1+r)ak+1(k+r+12)=0Recursion relation that defines series solution to ODEak+1=ak(2k+2r1)22(k+1+r)(2k+1+2r)Recursion relation forr=0ak+1=ak(2k1)22(k+1)(2k+1)Solution forr=0[y(u)=k=0akuk,ak+1=ak(2k1)22(k+1)(2k+1)]Revert the change of variablesu=t1[y(t)=k=0ak(t1)k,ak+1=ak(2k1)22(k+1)(2k+1)]Recursion relation forr=12ak+1=2akk2(k+32)(2k+2)Solution forr=12[y(u)=k=0akuk+12,ak+1=2akk2(k+32)(2k+2)]Revert the change of variablesu=t1[y(t)=k=0ak(t1)k+12,ak+1=2akk2(k+32)(2k+2)]Combine solutions and rename parameters[y(t)=(k=0ak(t1)k)+(k=0bk(t1)k+12),ak+1=ak(2k1)22(k+1)(2k+1),bk+1=2bkk2(k+32)(2k+2)]
Mathematica. Time used: 0.176 (sec). Leaf size: 112
ode=4*(t^2-3*t+2)*D[y[t],{t,2}]-2*D[y[t],t]+y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 
y(t)exp(1t2K[1]54(K[1]23K[1]+2)dK[1]121t12(K[2]23K[2]+2)dK[2])(c21texp(21K[3]2K[1]54(K[1]23K[1]+2)dK[1])dK[3]+c1)
Sympy. Time used: 0.471 (sec). Leaf size: 44
from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq((4*t**2 - 12*t + 8)*Derivative(y(t), (t, 2)) + y(t) - 2*Derivative(y(t), t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 
y(t)=t2(C1t1t21F0(0|t1t2)+C22F1(12,112|t1t2))