2.1.200 Problem 202

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9372]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 202
Date solved : Wednesday, March 05, 2025 at 07:48:14 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

y+(t2+2t+1)y(4+4t)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.322 (sec)

Writing the ode as

(1)y+(1+t)2y+(44t)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=1(3)B=(1+t)2C=44t

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=t4+4t3+6t2+24t+214

Comparing the above to (5) shows that

s=t4+4t3+6t2+24t+21t=4

Therefore eq. (4) becomes

(7)z(t)=(214+6t+14t4+t3+32t2)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 2.200: 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)=04=4

There are no poles in r. Therefore the set of poles Γ is empty. Since there is no odd order pole larger than 2 and the order at is 4 then the necessary conditions for case one are met. Therefore

L=[1]

Attempting to find a solution using case n=1.

Since the order of r at is Or()=4 then

v=Or()2=42=2

[r] is the sum of terms involving ti for 0iv in the Laurent series for r at . Therefore

[r]=i=0vaiti(8)=i=02aiti

Let a be the coefficient of tv=t2 in the above sum. The Laurent series of r at is

(9)rt22+t+12+5t5t2+5t330t4+105t5+

Comparing Eq. (9) with Eq. (8) shows that

a=12

From Eq. (9) the sum up to v=2 gives

[r]=i=02aiti(10)=12t2+t+12

Now we need to find b, where b be the coefficient of tv1=t1=t in r minus the coefficient of same term but in ([r])2 where [r] was found above in Eq (10). Hence

([r])2=14t4+t3+32t2+t+14

This shows that the coefficient of t in the above is 1. Now we need to find the coefficient of t in r. How this is done depends on if v=0 or not. Since v=2 which is not zero, then starting r=st, we do long division and write this in the form

r=Q+Rt

Where Q is the quotient and R is the remainder. Then the coefficient of t in r will be the coefficient this term in the quotient. Doing long division gives

r=st=t4+4t3+6t2+24t+214=Q+R4=(214+6t+14t4+t3+32t2)+(0)=214+6t+14t4+t3+32t2

We see that the coefficient of the term 1t in the quotient is 6. Now b can be found.

b=(6)(1)=5

Hence

[r]=12t2+t+12α+=12(bav)=12(5122)=4α=12(bav)=12(5122)=6

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

r=214+6t+14t4+t3+32t2

Order of r at [r] α+ α
4 12t2+t+12 4 6

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+=4, and since there are no poles, then

d=α+=4

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]=0+(12t2+t+12)=12t2+t+12=(1+t)22

Now that ω is determined, the next step is find a corresponding minimal polynomial p(t) of degree d=4 to solve the ode. The polynomial p(t) needs to satisfy the equation

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(t)=t4+a3t3+a2t2+a1t+a0

Substituting the above in eq. (1A) gives

(12t2+6ta3+2a2)+2(12t2+t+12)(4t3+3t2a3+2ta2+a1)+((1+t)+(12t2+t+12)2(214+6t+14t4+t3+32t2))=0(a3+4)t4+2(2a2+a3)t3+3(4a1+a3)t2+2(2a0a1+a2+3a3)t4a0+a1+2a2=0

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

{a0=5,a1=8,a2=6,a3=4}

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

p(t)=t4+4t3+6t2+8t+5

Therefore the first solution to the ode z=rz is

z1(t)=peωdt=(t4+4t3+6t2+8t+5)e(12t2+t+12)dt=(t4+4t3+6t2+8t+5)e(1+t)36=(1+t)(t3+3t2+3t+5)e(1+t)36

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

y1=z1e12BAdt=z1e12(1+t)21dt=z1e(1+t)36=z1(e(1+t)36)

Which simplifies to

y1=(1+t)(t3+3t2+3t+5)

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

y2=y1eBAdty12dt

Substituting gives

y2=y1e(1+t)21dt(y1)2dt=y1e(1+t)33(y1)2dt=y1(e(1+t)33(1+t)2(t3+3t2+3t+5)2dt)

Therefore the solution is

y=c1y1+c2y2=c1((1+t)(t3+3t2+3t+5))+c2((1+t)(t3+3t2+3t+5)(e(1+t)33(1+t)2(t3+3t2+3t+5)2dt))

Will add steps showing solving for IC soon.

Maple. Time used: 1.993 (sec). Leaf size: 60
ode:=diff(diff(y(t),t),t)+(t^2+2*t+1)*diff(y(t),t)-(4+4*t)*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 
y=(t+1)(t3+3t2+3t+5)(c2(et(t2+3t+3)3(t+1)2(t3+3t2+3t+5)2dt)+c1)

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 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunT  ODE, case  c = 0 
   Special function solution also has integrals. Returning default Liouvillian solution. 
<- Kovacics algorithm successful`
 

Maple step by step

Let’s solved2dt2y(t)+(t2+2t+1)(ddty(t))(4t+4)y(t)=0Highest derivative means the order of the ODE is2d2dt2y(t)Isolate 2nd derivatived2dt2y(t)=(t2+2t+1)(ddty(t))+(4t+4)y(t)Group terms withy(t)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dt2y(t)+(t2+2t+1)(ddty(t))+(4t4)y(t)=0Assume series solution fory(t)y(t)=k=0aktkRewrite ODE with series expansionsConverttmy(t)to series expansion form=0..1tmy(t)=k=max(0,m)aktk+mShift index usingk>kmtmy(t)=k=max(0,m)+makmtkConverttm(ddty(t))to series expansion form=0..2tm(ddty(t))=k=max(0,1m)akktk1+mShift index usingk>k+1mtm(ddty(t))=k=max(0,1m)+m1ak+1m(k+1m)tkConvertd2dt2y(t)to series expansiond2dt2y(t)=k=2akk(k1)tk2Shift index usingk>k+2d2dt2y(t)=k=0ak+2(k+2)(k+1)tkRewrite ODE with series expansions2a2+a14a0+(k=1(ak+2(k+2)(k+1)+ak+1(k+1)+2ak(k2)+ak1(k5))tk)=0Each term must be 02a2+a14a0=0Each term in the series must be 0, giving the recursion relationk2ak+2+(2ak+ak1+ak+1+3ak+2)k4ak5ak1+ak+1+2ak+2=0Shift index usingk>k+1(k+1)2ak+3+(2ak+1+ak+ak+2+3ak+3)(k+1)4ak+15ak+ak+2+2ak+3=0Recursion relation that defines the series solution to the ODE[y(t)=k=0aktk,ak+3=akk+2ak+1k+kak+24ak2ak+1+2ak+2k2+5k+6,2a2+a14a0=0]
Mathematica. Time used: 4.118 (sec). Leaf size: 78
ode=D[y[t],{t,2}]+(t^2+2*t+1)*D[y[t],t]-(4+4*t)*y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 
y(t)(t+1)(t3+3t2+3t+5)(c21te13K[1](K[1]2+3K[1]+3)(K[1]+1)2(K[1]3+3K[1]2+3K[1]+5)2dK[1]+c1)
Sympy. Time used: 0.890 (sec). Leaf size: 32
from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq((-4*t - 4)*y(t) + (t**2 + 2*t + 1)*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 
y(t)=C2(2t2+1)+C1t(t38+t22t2+1)+O(t6)