2.1.15 Problem 15

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9187]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 15
Date solved : Wednesday, March 05, 2025 at 07:37:34 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

(1t)y+tyy=0

Solved as second order ode using Kovacic algorithm

Time used: 0.224 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

A=1t(3)B=tC=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=t24t+64(1+t)2

Comparing the above to (5) shows that

s=t24t+6t=4(1+t)2

Therefore eq. (4) becomes

(7)z(t)=(t24t+64(1+t)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 2.15: 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)=22=0

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=4(1+t)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 0 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. Therefore

L=[1,2]

Attempting to find a solution using case n=1.

Looking at poles of order 2. The partial fractions decomposition of r is

r=14+34(1+t)212(1+t)

For the pole at t=1 let b be the coefficient of 1(1+t)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 Or()=0 then

v=Or()2=02=0

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

[r]=i=0vaiti(8)=i=00aiti

Let a be the coefficient of tv=t0 in the above sum. The Laurent series of r at is

(9)r1212t+1t3+114t4+214t5+152t6+6t711716t8+

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

a=12

From Eq. (9) the sum up to v=0 gives

[r]=i=00aiti(10)=12

Now we need to find b, where b be the coefficient of tv1=t1=1t in r minus the coefficient of same term but in ([r])2 where [r] was found above in Eq (10). Hence

([r])2=14

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

r=Q+Rt

Where Q is the quotient and R is the remainder. Then the coefficient of 1t in r will be the coefficient in R of the term in t of degree of t minus one, divided by the leading coefficient in t. Doing long division gives

r=st=t24t+64t28t+4=Q+R4t28t+4=(14)+(2t+54t28t+4)=14+2t+54t28t+4

Since the degree of t is 2, then we see that the coefficient of the term t in the remainder R is 2. Dividing this by leading coefficient in t which is 4 gives 12. Now b can be found.

b=(12)(0)=12

Hence

[r]=12α+=12(bav)=12(12120)=12α=12(bav)=12(12120)=12

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

r=t24t+64(1+t)2

pole c location pole order [r]c αc+ αc
1 2 0 32 12

Order of r at [r] α+ α
0 12 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)=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]

Substituting the above values in the above results in

ω=(()[r]c1+αc1tc1)+(+)[r]=12(1+t)+(12)=12(1+t)+12=t22t2

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(12(1+t)+12)(0)+((12(1+t)2)+(12(1+t)+12)2(t24t+64(1+t)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(12(1+t)+12)dt=et21+t

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

y1=z1e12BAdt=z1e12t1tdt=z1et2+ln(1+t)2=z1(1+tet2)

Which simplifies to

y1=et

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

y2=y1eBAdty12dt

Substituting gives

y2=y1et1tdt(y1)2dt=y1et+ln(1+t)(y1)2dt=y1(tet+ln(1+t)e2t1+t)

Therefore the solution is

y=c1y1+c2y2=c1(et)+c2(et(tet+ln(1+t)e2t1+t))

Will add steps showing solving for IC soon.

Maple. Time used: 0.005 (sec). Leaf size: 12
ode:=(-t+1)*diff(diff(y(t),t),t)+t*diff(y(t),t)-y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 
y=c1t+c2et

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(1t)(d2dt2y(t))+t(ddty(t))y(t)=0Highest derivative means the order of the ODE is2d2dt2y(t)Isolate 2nd derivatived2dt2y(t)=y(t)t1+t(ddty(t))t1Group terms withy(t)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dt2y(t)t(ddty(t))t1+y(t)t1=0Check to see ift0=1is a regular singular pointDefine functions[P2(t)=tt1,P3(t)=1t1](t1)P2(t)is analytic att=1((t1)P2(t))|t=1=1(t1)2P3(t)is analytic att=1((t1)2P3(t))|t=1=0t=1is a regular singular pointCheck to see ift0=1is a regular singular pointt0=1Multiply by denominators(t1)(d2dt2y(t))t(ddty(t))+y(t)=0Change variables usingt=u+1so that the regular singular point is atu=0u(d2du2y(u))+(u1)(dduy(u))+y(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+rConvertu(d2du2y(u))to series expansionu(d2du2y(u))=k=0ak(k+r)(k+r1)uk+r1Shift index usingk>k+1u(d2du2y(u))=k=1ak+1(k+1+r)(k+r)uk+rRewrite ODE with series expansionsa0r(2+r)u1+r+(k=0(ak+1(k+1+r)(k+r1)ak(k+r1))uk+r)=0a0cannot be 0 by assumption, giving the indicial equationr(2+r)=0Values of r that satisfy the indicial equationr{0,2}Each term in the series must be 0, giving the recursion relation(k+r1)(ak+1(k+1+r)ak)=0Recursion relation that defines series solution to ODEak+1=akk+1+rRecursion relation forr=0ak+1=akk+1Solution forr=0[y(u)=k=0akuk,ak+1=akk+1]Revert the change of variablesu=t1[y(t)=k=0ak(t1)k,ak+1=akk+1]Recursion relation forr=2ak+1=akk+3Solution forr=2[y(u)=k=0akuk+2,ak+1=akk+3]Revert the change of variablesu=t1[y(t)=k=0ak(t1)k+2,ak+1=akk+3]Combine solutions and rename parameters[y(t)=(k=0ak(t1)k)+(k=0bk(t1)k+2),ak+1=akk+1,bk+1=bkk+3]
Mathematica. Time used: 0.17 (sec). Leaf size: 90
ode=(1-t)*D[y[t],{t,2}]+t*D[y[t],t]-y[t] == 0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 
y(t)exp(1tK[1]22(K[1]1)dK[1]121tK[2]K[2]1dK[2])(c21texp(21K[3]K[1]22(K[1]1)dK[1])dK[3]+c1)
Sympy. Time used: 0.820 (sec). Leaf size: 27
from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t*Derivative(y(t), t) + (1 - t)*Derivative(y(t), (t, 2)) - y(t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 
y(t)=C2(t424+t36+t22+1)+C1t+O(t6)