2.1.749 Problem 771

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9921]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 771
Date solved : Wednesday, March 05, 2025 at 08:00:57 AM
CAS classification : [_Laguerre]

Solve

2xy(3+2x)y+y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.395 (sec)

Writing the ode as

(1)2xy+(32x)y+y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=2x(3)B=32xC=1

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+4x+2116x2

Comparing the above to (5) shows that

s=4x2+4x+21t=16x2

Therefore eq. (4) becomes

(7)z(x)=(4x2+4x+2116x2)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 2.749: 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=16x2. There is a pole at x=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=14+14x+2116x2

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=2116. Hence

[r]c=0αc+=12+1+4b=74αc=121+4b=34

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)r12+14x+54x258x354x4+3516x5+10564x61005128x7+

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

a=12

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

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

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=14

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+4x+2116x2=Q+R16x2=(14)+(4x+2116x2)=14+4x+2116x2

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 16 gives 14. Now b can be found.

b=(14)(0)=14

Hence

[r]=12α+=12(bav)=12(14120)=14α=12(bav)=12(14120)=14

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

r=4x2+4x+2116x2

pole c location pole order [r]c αc+ αc
0 2 0 74 34

Order of r at [r] α+ α
0 12 14 14

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+=14 then

d=α+(αc1)=14(34)=1

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]

Substituting the above values in the above results in

ω=(()[r]c1+αc1xc1)+(+)[r]=34x+(12)=34x+12=34x+12

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

(1A)p+2ωp+(ω+ω2r)p=0

Let

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

Substituting the above in eq. (1A) gives

(0)+2(34x+12)(1)+((34x2)+(34x+12)2(4x2+4x+2116x2))=032a02x=0

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

{a0=32}

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

p(x)=x32

Therefore the first solution to the ode z=rz is

z1(x)=peωdx=(x32)e(34x+12)dx=(x32)ex23ln(x)4=(3+2x)ex22x3/4

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

y1=z1e12BAdx=z1e1232x2xdx=z1ex2+3ln(x)4=z1(x3/4ex2)

Which simplifies to

y1=ex(3+2x)2

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e32x2xdx(y1)2dx=y1ex+3ln(x)2(y1)2dx=y1(4ex+3ln(x)2e2x(3+2x)2dx)

Therefore the solution is

y=c1y1+c2y2=c1(ex(3+2x)2)+c2(ex(3+2x)2(4ex+3ln(x)2e2x(3+2x)2dx))

Will add steps showing solving for IC soon.

Maple. Time used: 0.041 (sec). Leaf size: 24
ode:=2*diff(diff(y(x),x),x)*x-(2*x+3)*diff(y(x),x)+y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1hypergeom([2],[72],x)x5/22c2ex(x32)3

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 successful 
   <- special function solution successful 
      Solution using Kummer functions still has integrals. Trying a hypergeometric solution. 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
   <- Kovacics algorithm successful`
 

Maple step by step

Let’s solve2(d2dx2y(x))x(2x+3)(ddxy(x))+y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=y(x)2x+(2x+3)(ddxy(x))2xGroup terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)(2x+3)(ddxy(x))2x+y(x)2x=0Check to see ifx0=0is a regular singular pointDefine functions[P2(x)=2x+32x,P3(x)=12x]xP2(x)is analytic atx=0(xP2(x))|x=0=32x2P3(x)is analytic atx=0(x2P3(x))|x=0=0x=0is a regular singular pointCheck to see ifx0=0is a regular singular pointx0=0Multiply by denominators2(d2dx2y(x))x+(2x3)(ddxy(x))+y(x)=0Assume series solution fory(x)y(x)=k=0akxk+rRewrite ODE with series expansionsConvertxm(ddxy(x))to series expansion form=0..1xm(ddxy(x))=k=0ak(k+r)xk+r1+mShift index usingk>k+1mxm(ddxy(x))=k=1+mak+1m(k+1m+r)xk+rConvertx(d2dx2y(x))to series expansionx(d2dx2y(x))=k=0ak(k+r)(k+r1)xk+r1Shift index usingk>k+1x(d2dx2y(x))=k=1ak+1(k+1+r)(k+r)xk+rRewrite ODE with series expansionsa0r(5+2r)x1+r+(k=0(ak+1(k+1+r)(2k3+2r)ak(2k+2r1))xk+r)=0a0cannot be 0 by assumption, giving the indicial equationr(5+2r)=0Values of r that satisfy the indicial equationr{0,52}Each term in the series must be 0, giving the recursion relation2(k+1+r)(k+r32)ak+12ak(k+r12)=0Recursion relation that defines series solution to ODEak+1=ak(2k+2r1)(k+1+r)(2k3+2r)Recursion relation forr=0ak+1=ak(2k1)(k+1)(2k3)Solution forr=0[y(x)=k=0akxk,ak+1=ak(2k1)(k+1)(2k3)]Recursion relation forr=52ak+1=ak(2k+4)(k+72)(2k+2)Solution forr=52[y(x)=k=0akxk+52,ak+1=ak(2k+4)(k+72)(2k+2)]Combine solutions and rename parameters[y(x)=(k=0akxk)+(k=0bkxk+52),ak+1=ak(2k1)(k+1)(2k3),bk+1=bk(2k+4)(k+72)(2k+2)]
Mathematica. Time used: 0.225 (sec). Leaf size: 52
ode=2*x*D[y[x],{x,2}]-(3+2*x)*D[y[x],x]+y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)12ex(2x3)(c21x4eK[1]K[1]3/2(32K[1])2dK[1]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x*Derivative(y(x), (x, 2)) - (2*x + 3)*Derivative(y(x), x) + y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False