2.1.638 Problem 655

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9810]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 655
Date solved : Wednesday, March 05, 2025 at 07:59:19 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

zy+(2z3)y+4yz=0

Solved as second order ode using Kovacic algorithm

Time used: 0.322 (sec)

Writing the ode as

(1)zy+(2z3)y+4yz=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=z(3)B=2z3C=4z

Applying the Liouville transformation on the dependent variable gives

z(z)=yeB2Adz

Then (2) becomes

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

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=4z212z14z2

Comparing the above to (5) shows that

s=4z212z1t=4z2

Therefore eq. (4) becomes

(7)z(z)=(4z212z14z2)z(z)

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

y=z(z)eB2Adz

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 2.638: 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=4z2. There is a pole at z=0 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=114z23z

For the pole at z=0 let b be the coefficient of 1z2 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 Or()=0 then

v=Or()2=02=0

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

[r]=i=0vaizi(8)=i=00aizi

Let a be the coefficient of zv=z0 in the above sum. The Laurent series of r at is

(9)r132z54z2158z311532z449564z52285128z611055256z7+

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

a=1

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

[r]=i=00aizi(10)=1

Now we need to find b, where b be the coefficient of zv1=z1=1z in r minus the coefficient of same term but in ([r])2 where [r] was found above in Eq (10). Hence

([r])2=1

This shows that the coefficient of 1z in the above is 0. Now we need to find the coefficient of 1z 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 1z in r will be the coefficient in R of the term in z of degree of t minus one, divided by the leading coefficient in t. Doing long division gives

r=st=4z212z14z2=Q+R4z2=(1)+(12z14z2)=1+12z14z2

Since the degree of t is 2, then we see that the coefficient of the term z in the remainder R is 12. Dividing this by leading coefficient in t which is 4 gives 3. Now b can be found.

b=(3)(0)=3

Hence

[r]=1α+=12(bav)=12(310)=32α=12(bav)=12(310)=32

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

r=4z212z14z2

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

Order of r at [r] α+ α
0 1 32 32

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=32 then

d=α(αc1+)=32(12)=1

Since d an integer and d0 then it can be used to find ω using

ω=cΓ(s(c)[r]c+αcs(c)zc)+s()[r]

The above gives

ω=((+)[r]c1+αc1+zc1)+()[r]=12z+()(1)=12z1=12z1

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

(1A)p+2ωp+(ω+ω2r)p=0

Let

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

Substituting the above in eq. (1A) gives

(0)+2(12z1)(1)+((12z2)+(12z1)2(4z212z14z2))=01+2a0z=0

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

{a0=12}

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

p(z)=z12

Therefore the first solution to the ode z=rz is

z1(z)=peωdz=(z12)e(12z1)dz=(z12)ez+ln(z)2=(1+2z)zez2

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

y1=z1e12BAdz=z1e122z3zdz=z1ez+3ln(z)2=z1(z3/2ez)

Which simplifies to

y1=z2e2z(1+2z)2

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

y2=y1eBAdzy12dz

Substituting gives

y2=y1e2z3zdz(y1)2dz=y1e2z+3ln(z)(y1)2dz=y1(4Ei1(2z)4e2z1+2z)

Therefore the solution is

y=c1y1+c2y2=c1(z2e2z(1+2z)2)+c2(z2e2z(1+2z)2(4Ei1(2z)4e2z1+2z))

Will add steps showing solving for IC soon.

Maple. Time used: 0.008 (sec). Leaf size: 36
ode:=z*diff(diff(y(z),z),z)+(2*z-3)*diff(y(z),z)+4/z*y(z) = 0; 
dsolve(ode,y(z), singsol=all);
 
y(z)=2z2(e2zc2(12+z)Ei1(2z)+c1(12+z)e2z+c22)

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 solvez(d2dz2y(z))+(3+2z)(ddzy(z))+4y(z)z=0Highest derivative means the order of the ODE is2d2dz2y(z)Isolate 2nd derivatived2dz2y(z)=4y(z)z2(3+2z)(ddzy(z))zGroup terms withy(z)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dz2y(z)+(3+2z)(ddzy(z))z+4y(z)z2=0Check to see ifz0=0is a regular singular pointDefine functions[P2(z)=3+2zz,P3(z)=4z2]zP2(z)is analytic atz=0(zP2(z))|z=0=3z2P3(z)is analytic atz=0(z2P3(z))|z=0=4z=0is a regular singular pointCheck to see ifz0=0is a regular singular pointz0=0Multiply by denominators(d2dz2y(z))z2+z(3+2z)(ddzy(z))+4y(z)=0Assume series solution fory(z)y(z)=k=0akzk+rRewrite ODE with series expansionsConvertzm(ddzy(z))to series expansion form=1..2zm(ddzy(z))=k=0ak(k+r)zk+r1+mShift index usingk>k+1mzm(ddzy(z))=k=1+mak+1m(k+1m+r)zk+rConvertz2(d2dz2y(z))to series expansionz2(d2dz2y(z))=k=0ak(k+r)(k+r1)zk+rRewrite ODE with series expansionsa0(2+r)2zr+(k=1(ak(k+r2)2+2ak1(k+r1))zk+r)=0a0cannot be 0 by assumption, giving the indicial equation(2+r)2=0Values of r that satisfy the indicial equationr=2Each term in the series must be 0, giving the recursion relationak(k+r2)2+2ak1(k+r1)=0Shift index usingk>k+1ak+1(k+r1)2+2ak(k+r)=0Recursion relation that defines series solution to ODEak+1=2ak(k+r)(k+r1)2Recursion relation forr=2ak+1=2ak(k+2)(k+1)2Solution forr=2[y(z)=k=0akzk+2,ak+1=2ak(k+2)(k+1)2]
Mathematica. Time used: 0.225 (sec). Leaf size: 55
ode=z*D[y[z],{z,2}]+(2*z-3)*D[y[z],z]+4/z*y[z]==0; 
ic={}; 
DSolve[{ode,ic},y[z],z,IncludeSingularSolutions->True]
 
y(z)12e2zz2(2z1)(c21z4e2K[1](12K[1])2K[1]dK[1]+c1)
Sympy. Time used: 1.741 (sec). Leaf size: 416
from sympy import * 
z = symbols("z") 
y = Function("y") 
ode = Eq(z*Derivative(y(z), (z, 2)) + (2*z - 3)*Derivative(y(z), z) + 4*y(z)/z,0) 
ics = {} 
dsolve(ode,func=y(z),ics=ics)
 
Solution too large to show