2.1.688 Problem 705

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9860]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 705
Date solved : Wednesday, March 05, 2025 at 08:00:05 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

x2y+(x2+x)y+(x9)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.313 (sec)

Writing the ode as

(1)x2y+(x2+x)y+(x9)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x2(3)B=x2+xC=x9

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=x22x+354x2

Comparing the above to (5) shows that

s=x22x+35t=4x2

Therefore eq. (4) becomes

(7)z(x)=(x22x+354x2)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 2.688: 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=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 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+354x212x

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=354. Hence

[r]c=0αc+=12+1+4b=72αc=121+4b=52

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)r1212x+172x2+172x32554x48334x5+32134x6+217094x7+

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=x22x+354x2=Q+R4x2=(14)+(2x+354x2)=14+2x+354x2

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

b=(12)(0)=12

Hence

[r]=12α+=12(bav)=12(12120)=12α=12(bav)=12(12120)=12

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

r=x22x+354x2

pole c location pole order [r]c αc+ αc
0 2 0 72 52

Order of r at [r] α+ α
0 12 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(52)=2

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]=52x+(12)=52x+12=5+x2x

Now that ω is determined, the next step is find a corresponding minimal polynomial p(x) of degree d=2 to solve the ode. The polynomial p(x) needs to satisfy the equation

(1A)p+2ωp+(ω+ω2r)p=0

Let

(2A)p(x)=x2+a1x+a0

Substituting the above in eq. (1A) gives

(2)+2(52x+12)(2x+a1)+((52x2)+(52x+12)2(x22x+354x2))=0(a18)x2a05a1x=0

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

{a0=20,a1=8}

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

p(x)=x28x+20

Therefore the first solution to the ode z=rz is

z1(x)=peωdx=(x28x+20)e(52x+12)dx=(x28x+20)ex25ln(x)2=(x28x+20)ex2x5/2

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

y1=z1e12BAdx=z1e12x2+xx2dx=z1ex2ln(x)2=z1(ex2x)

Which simplifies to

y1=x28x+20x3

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1ex2+xx2dx(y1)2dx=y1exln(x)(y1)2dx=y1((x3+9x2+36x+60)xexln(x)x28x+20)

Therefore the solution is

y=c1y1+c2y2=c1(x28x+20x3)+c2(x28x+20x3((x3+9x2+36x+60)xexln(x)x28x+20))

Will add steps showing solving for IC soon.

Maple. Time used: 0.006 (sec). Leaf size: 38
ode:=x^2*diff(diff(y(x),x),x)+(x^2+x)*diff(y(x),x)+(x-9)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c2(x3+9x2+36x+60)ex+c1(x28x+20)x3

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 solvex2(d2dx2y(x))+(x2+x)(ddxy(x))+(x9)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(x9)y(x)x2(x+1)(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)+(x+1)(ddxy(x))x+(x9)y(x)x2=0Check to see ifx0=0is a regular singular pointDefine functions[P2(x)=x+1x,P3(x)=x9x2]xP2(x)is analytic atx=0(xP2(x))|x=0=1x2P3(x)is analytic atx=0(x2P3(x))|x=0=9x=0is a regular singular pointCheck to see ifx0=0is a regular singular pointx0=0Multiply by denominatorsx2(d2dx2y(x))+x(x+1)(ddxy(x))+(x9)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=1..2xm(ddxy(x))=k=0ak(k+r)xk+r1+mShift index usingk>k+1mxm(ddxy(x))=k=1+mak+1m(k+1m+r)xk+rConvertx2(d2dx2y(x))to series expansionx2(d2dx2y(x))=k=0ak(k+r)(k+r1)xk+rRewrite ODE with series expansionsa0(3+r)(3+r)xr+(k=1(ak(k+r+3)(k+r3)+ak1(k+r))xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(3+r)(3+r)=0Values of r that satisfy the indicial equationr{3,3}Each term in the series must be 0, giving the recursion relationak(k+r+3)(k+r3)+ak1(k+r)=0Shift index usingk>k+1ak+1(k+4+r)(k2+r)+ak(k+r+1)=0Recursion relation that defines series solution to ODEak+1=ak(k+r+1)(k+4+r)(k2+r)Recursion relation forr=3; series terminates atk=2ak+1=ak(k2)(k+1)(k5)Apply recursion relation fork=0a1=2a05Apply recursion relation fork=1a2=a18Express in terms ofa0a2=a020Terminating series solution of the ODE forr=3. Use reduction of order to find the second linearly independent solutiony(x)=a0(125x+120x2)Recursion relation forr=3ak+1=ak(k+4)(k+7)(k+1)Solution forr=3[y(x)=k=0akxk+3,ak+1=ak(k+4)(k+7)(k+1)]Combine solutions and rename parameters[y(x)=a0(125x+120x2)+(k=0bkxk+3),bk+1=bk(4+k)(k+7)(k+1)]
Mathematica. Time used: 0.501 (sec). Leaf size: 96
ode=x^2*D[y[x],{x,2}]+(x+x^2)*D[y[x],x]+(x-9)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)(x28x+20)exp(1xK[1]52K[1]dK[1]x2)(c21xexp(21K[2]K[1]52K[1]dK[1])(K[2]28K[2]+20)2dK[2]+c1)x
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*Derivative(y(x), (x, 2)) + (x - 9)*y(x) + (x**2 + x)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False