2.1.431 Problem 444

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9603]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 444
Date solved : Wednesday, March 05, 2025 at 07:51:31 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

(x22x)y+(x2+2)y+(2x2)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.353 (sec)

Writing the ode as

(1)(x22x)y+(x2+2)y+(2x2)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x22x(3)B=x2+2C=2x2

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=x48x3+24x224x+124(x22x)2

Comparing the above to (5) shows that

s=x48x3+24x224x+12t=4(x22x)2

Therefore eq. (4) becomes

(7)z(x)=(x48x3+24x224x+124(x22x)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 2.431: 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)=44=0

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=4(x22x)2. There is a pole at x=0 of order 2. There is a pole at x=2 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(x2)234x14(x2)+34x2

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=34. Hence

[r]c=0αc+=12+1+4b=32αc=121+4b=12

For the pole at x=2 let b be the coefficient of 1(x2)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)r121x+2x3+11x4+42x5+132x6+348x7+711x8+

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=x48x3+24x224x+124x416x3+16x2=Q+R4x416x3+16x2=(14)+(4x3+20x224x+124x416x3+16x2)=14+4x3+20x224x+124x416x3+16x2

Since the degree of t is 4, then we see that the coefficient of the term x3 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]=12α+=12(bav)=12(1120)=1α=12(bav)=12(1120)=1

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

r=x48x3+24x224x+124(x22x)2

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

Order of r at [r] α+ α
0 12 1 1

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+=1 then

d=α+(αc1+αc2)=1(1)=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]

Substituting the above values in the above results in

ω=(()[r]c1+αc1xc1)+(()[r]c2+αc2xc2)+(+)[r]=12x12(x2)+(12)=12x12(x2)+12=12x12x4+12

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(12x12(x2)+12)(0)+((12x2+12(x2)2)+(12x12(x2)+12)2(x48x3+24x224x+124(x22x)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(12x12(x2)+12)dx=ex2x2x

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

y1=z1e12BAdx=z1e12x2+2x22xdx=z1ex2+ln(x2)2+ln(x)2=z1(x2xex2)

Which simplifies to

y1=x2xexx(x2)

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1ex2+2x22xdx(y1)2dx=y1ex+ln(x2)+ln(x)(y1)2dx=y1(xex+ln(x2)+ln(x)e2xx2)

Therefore the solution is

y=c1y1+c2y2=c1(x2xexx(x2))+c2(x2xexx(x2)(xex+ln(x2)+ln(x)e2xx2))

Will add steps showing solving for IC soon.

Maple. Time used: 0.003 (sec). Leaf size: 14
ode:=(x^2-2*x)*diff(diff(y(x),x),x)+(-x^2+2)*diff(y(x),x)+(2*x-2)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1x2+exc2

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(x22x)(d2dx2y(x))+(x2+2)(ddxy(x))+(2x2)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=2(x1)y(x)x(x2)+(x22)(ddxy(x))x(x2)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)(x22)(ddxy(x))x(x2)+2(x1)y(x)x(x2)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=x22x(x2),P3(x)=2(x1)x(x2)]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 ifx0is a regular singular pointx0=0Multiply by denominatorsx(x2)(d2dx2y(x))+(x2+2)(ddxy(x))+(2x2)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..2xm(ddxy(x))=k=0ak(k+r)xk+r1+mShift index usingk>k+1mxm(ddxy(x))=k=1+mak+1m(k+1m+r)xk+rConvertxm(d2dx2y(x))to series expansion form=1..2xm(d2dx2y(x))=k=0ak(k+r)(k+r1)xk+r2+mShift index usingk>k+2mxm(d2dx2y(x))=k=2+mak+2m(k+2m+r)(k+1m+r)xk+rRewrite ODE with series expansions2a0r(2+r)x1+r+(2a1(1+r)(1+r)+a0(1+r)(2+r))xr+(k=1(2ak+1(k+r+1)(k+r1)+ak(k+r+1)(k+r2)ak1(k3+r))xk+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)+a0(1+r)(2+r)=0Each term in the series must be 0, giving the recursion relationak(k+r+1)(k+r2)2k2ak+1+(4rak+1ak1)k2r2ak+1ak1r+3ak1+2ak+1=0Shift index usingk>k+1ak+1(k+2+r)(k+r1)2(k+1)2ak+2+(4rak+2ak)(k+1)2r2ak+2rak+3ak+2ak+2=0Recursion relation that defines series solution to ODEak+2=k2ak+1+2krak+1+r2ak+1kak+kak+1rak+rak+1+2ak2ak+12(k2+2kr+r2+2k+2r)Recursion relation forr=0ak+2=k2ak+1kak+kak+1+2ak2ak+12(k2+2k)Series not valid forr=0, division by0in the recursion relation atk=0ak+2=k2ak+1kak+kak+1+2ak2ak+12(k2+2k)Recursion relation forr=2ak+2=k2ak+1kak+5kak+1+4ak+12(k2+6k+8)Solution forr=2[y(x)=k=0akxk+2,ak+2=k2ak+1kak+5kak+1+4ak+12(k2+6k+8),6a1=0]
Mathematica. Time used: 0.288 (sec). Leaf size: 115
ode=(x^2-2*x)*D[y[x],{x,2}]+(2-x^2)*D[y[x],x]+(2*x-2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1x(K[1]4)K[1]+22(K[1]2)K[1]dK[1]121x(1K[2]1+12K[2])dK[2])(c21xexp(21K[3]K[1]24K[1]+22(K[1]2)K[1]dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((2 - x**2)*Derivative(y(x), x) + (2*x - 2)*y(x) + (x**2 - 2*x)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False