2.1.351 Problem 358

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9523]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 358
Date solved : Wednesday, March 05, 2025 at 07:50:26 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

xy+(12x)y+(x1)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.143 (sec)

Writing the ode as

(1)xy+(12x)y+(x1)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x(3)B=12xC=x1

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=14x2

Comparing the above to (5) shows that

s=1t=4x2

Therefore eq. (4) becomes

(7)z(x)=(14x2)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 2.351: 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)=20=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=4x2. 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 2 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. Since pole order is not larger than 2 and the order at is 2 then the necessary conditions for case three are met. Therefore

L=[1,2,4,6,12]

Attempting to find a solution using case n=1.

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

r=14x2

For the pole at x=0 let b be the coefficient of 1x2 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 2 then [r]=0. Let b be the coefficient of 1x2 in the Laurent series expansion of r at . which can be found by dividing the leading coefficient of s by the leading coefficient of t from

r=st=14x2

Since the gcd(s,t)=1. This gives b=14. Hence

[r]=0α+=12+1+4b=12α=121+4b=12

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

r=14x2

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

Order of r at [r] α+ α
2 0 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]=12x+()(0)=12x=12x

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(12x)(0)+((12x2)+(12x)2(14x2))=00=0

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

z1(x)=peωdx=e12xdx=x

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

y1=z1e12BAdx=z1e1212xxdx=z1exln(x)2=z1(exx)

Which simplifies to

y1=ex

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e12xxdx(y1)2dx=y1e2xln(x)(y1)2dx=y1(ln(x))

Therefore the solution is

y=c1y1+c2y2=c1(ex)+c2(ex(ln(x)))

Will add steps showing solving for IC soon.

Maple. Time used: 0.004 (sec). Leaf size: 13
ode:=diff(diff(y(x),x),x)*x+(1-2*x)*diff(y(x),x)+(-1+x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=ex(c2ln(x)+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 
<- Kovacics algorithm successful`
 

Maple step by step

Let’s solve(d2dx2y(x))x+(2x+1)(ddxy(x))+(x1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(x1)y(x)x+(2x1)(ddxy(x))xGroup terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)(2x1)(ddxy(x))x+(x1)y(x)x=0Check to see ifx0=0is a regular singular pointDefine functions[P2(x)=2x1x,P3(x)=x1x]xP2(x)is analytic atx=0(xP2(x))|x=0=1x2P3(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 denominators(d2dx2y(x))x+(2x+1)(ddxy(x))+(x1)y(x)=0Assume series solution fory(x)y(x)=k=0akxk+rRewrite ODE with series expansionsConvertxmy(x)to series expansion form=0..1xmy(x)=k=0akxk+r+mShift index usingk>kmxmy(x)=k=makmxk+rConvertxm(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 expansionsa0r2x1+r+(a1(1+r)2a0(1+2r))xr+(k=1(ak+1(k+1+r)2ak(2k+2r+1)+ak1)xk+r)=0a0cannot be 0 by assumption, giving the indicial equationr2=0Values of r that satisfy the indicial equationr=0Each term must be 0a1(1+r)2a0(1+2r)=0Each term in the series must be 0, giving the recursion relationak+1(k+1)2+(2k1)ak+ak1=0Shift index usingk>k+1ak+2(k+2)2+(2k3)ak+1+ak=0Recursion relation that defines series solution to ODEak+2=2kak+1ak+3ak+1(k+2)2Recursion relation forr=0ak+2=2kak+1ak+3ak+1(k+2)2Solution forr=0[y(x)=k=0akxk,ak+2=2kak+1ak+3ak+1(k+2)2,a1a0=0]
Mathematica. Time used: 0.025 (sec). Leaf size: 17
ode=x*D[y[x],{x,2}]+(1-2*x)*D[y[x],x]+(x-1)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)ex(c2log(x)+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*Derivative(y(x), (x, 2)) + (1 - 2*x)*Derivative(y(x), x) + (x - 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False