2.1.33 Problem 34

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9205]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 34
Date solved : Wednesday, March 05, 2025 at 07:37:49 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

(2x+1)y2y(2x+3)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.273 (sec)

Writing the ode as

(1)(2x+1)y2y+(2x3)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=2x+1(3)B=2C=2x3

Applying the Liouville transformation on the dependent variable gives

z(x)=yeB2Adx

Then (2) becomes

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

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=4x2+8x+6(2x+1)2

Comparing the above to (5) shows that

s=4x2+8x+6t=(2x+1)2

Therefore eq. (4) becomes

(7)z(x)=(4x2+8x+6(2x+1)2)z(x)

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

y=z(x)eB2Adx

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 2.33: 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=(2x+1)2. There is a pole at x=12 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=1+34(x+12)2+1x+12

For the pole at x=12 let b be the coefficient of 1(x+12)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 xi for 0iv in the Laurent series for r at . Therefore

[r]=i=0vaixi(8)=i=00aixi

Let a be the coefficient of xv=x0 in the above sum. The Laurent series of r at is

(9)r1+12x14x3+1132x42164x5+1564x6332x71172048x8+

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

a=1

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

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

Now we need to find b, where b be the coefficient of xv1=x1=1x 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 1x in the above is 0. Now we need to find the coefficient of 1x 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 1x in r will be the coefficient in R of the term in x of degree of t minus one, divided by the leading coefficient in t. Doing long division gives

r=st=4x2+8x+64x2+4x+1=Q+R4x2+4x+1=(1)+(4x+54x2+4x+1)=1+4x+54x2+4x+1

Since the degree of t is 2, then we see that the coefficient of the term x in the remainder R is 4. Dividing this by leading coefficient in t which is 4 gives 1. Now b can be found.

b=(1)(0)=1

Hence

[r]=1α+=12(bav)=12(110)=12α=12(bav)=12(110)=12

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

r=4x2+8x+6(2x+1)2

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

Order of r at [r] α+ α
0 1 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)xc)+s()[r]

The above gives

ω=(()[r]c1+αc1xc1)+()[r]=12(x+12)+()(1)=12(x+12)1=2(x+1)2x+1

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

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(x)=1

Substituting the above in eq. (1A) gives

(0)+2(12(x+12)1)(0)+((12(x+12)2)+(12(x+12)1)2(4x2+8x+6(2x+1)2))=00=0

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

z1(x)=peωdx=e(12(x+12)1)dx=ex2x+1

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

y1=z1e12BAdx=z1e1222x+1dx=z1eln(2x+1)2=z1(2x+1)

Which simplifies to

y1=ex

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e22x+1dx(y1)2dx=y1eln(2x+1)(y1)2dx=y1(xe2x)

Therefore the solution is

y=c1y1+c2y2=c1(ex)+c2(ex(xe2x))

Will add steps showing solving for IC soon.

Maple. Time used: 0.006 (sec). Leaf size: 16
ode:=(2*x+1)*diff(diff(y(x),x),x)-2*diff(y(x),x)-(2*x+3)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1ex+c2exx

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(2x+1)(d2dx2y(x))2ddxy(x)(2x+3)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(2x+3)y(x)2x+1+2(ddxy(x))2x+1Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)2(ddxy(x))2x+1(2x+3)y(x)2x+1=0Check to see ifx0=12is a regular singular pointDefine functions[P2(x)=22x+1,P3(x)=2x+32x+1](x+12)P2(x)is analytic atx=12((x+12)P2(x))|x=12=1(x+12)2P3(x)is analytic atx=12((x+12)2P3(x))|x=12=0x=12is a regular singular pointCheck to see ifx0=12is a regular singular pointx0=12Multiply by denominators(2x+1)(d2dx2y(x))2ddxy(x)+(2x3)y(x)=0Change variables usingx=u12so that the regular singular point is atu=02u(d2du2y(u))2dduy(u)+(2u2)y(u)=0Assume series solution fory(u)y(u)=k=0akuk+rRewrite ODE with series expansionsConvertumy(u)to series expansion form=0..1umy(u)=k=0akuk+r+mShift index usingk>kmumy(u)=k=makmuk+rConvertdduy(u)to series expansiondduy(u)=k=0ak(k+r)uk+r1Shift index usingk>k+1dduy(u)=k=1ak+1(k+1+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 expansions2a0r(2+r)u1+r+(2a1(1+r)(1+r)2a0)ur+(k=1(2ak+1(k+1+r)(k+r1)2ak2ak1)uk+r)=0a0cannot be 0 by assumption, giving the indicial equation2r(2+r)=0Values of r that satisfy the indicial equationr{0,2}Each term must be 02a1(1+r)(1+r)2a0=0Each term in the series must be 0, giving the recursion relation2ak+1(k+1+r)(k+r1)2ak2ak1=0Shift index usingk>k+12ak+2(k+2+r)(k+r)2ak+12ak=0Recursion relation that defines series solution to ODEak+2=ak+1+ak(k+2+r)(k+r)Recursion relation forr=0ak+2=ak+1+ak(k+2)kSeries not valid forr=0, division by0in the recursion relation atk=0ak+2=ak+1+ak(k+2)kRecursion relation forr=2ak+2=ak+1+ak(k+4)(k+2)Solution forr=2[y(u)=k=0akuk+2,ak+2=ak+1+ak(k+4)(k+2),6a12a0=0]Revert the change of variablesu=x+12[y(x)=k=0ak(x+12)k+2,ak+2=ak+1+ak(k+4)(k+2),6a12a0=0]
Mathematica. Time used: 0.358 (sec). Leaf size: 69
ode=(2*x+1)*D[y[x],{x,2}]-2*D[y[x],x]-(2*x+3)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)2x+1exp(1x(12K[1]11)dK[1])(c21xexp(21K[2](12K[1]11)dK[1])dK[2]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((2*x + 1)*Derivative(y(x), (x, 2)) - (2*x + 3)*y(x) - 2*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False