2.2.3 Problem 3

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9996]
Book : Collection of Kovacic problems
Section : section 2. Solution found using all possible Kovacic cases
Problem number : 3
Date solved : Wednesday, March 05, 2025 at 08:01:57 AM
CAS classification : [[_Emden, _Fowler]]

Solve

y=12yx2

Solved as second order ode using Kovacic algorithm

Time used: 0.092 (sec)

Writing the ode as

(1)y12yx2=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=1(3)B=0C=12x2

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=12x2

Comparing the above to (5) shows that

s=12t=x2

Therefore eq. (4) becomes

(7)z(x)=(12x2)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 2.824: 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=x2. 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=12x2

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=12. Hence

[r]c=0αc+=12+1+4b=4αc=121+4b=3

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=12x2

Since the gcd(s,t)=1. This gives b=12. Hence

[r]=0α+=12+1+4b=4α=121+4b=3

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

r=12x2

pole c location pole order [r]c αc+ αc
0 2 0 4 3

Order of r at [r] α+ α
2 0 4 3

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=3 then

d=α(αc1)=3(3)=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]=3x+()(0)=3x=3x

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(3x)(0)+((3x2)+(3x)2(12x2))=00=0

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

z1(x)=peωdx=e3xdx=1x3

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

y1=z1e12BAdx

Since B=0 then the above reduces to

y1=z1=1x3

Which simplifies to

y1=1x3

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

y2=y1eBAdxy12dx

Since B=0 then the above becomes

y2=y11y12dx=1x311x6dx=1x3(x77)

Therefore the solution is

y=c1y1+c2y2=c1(1x3)+c2(1x3(x77))

Will add steps showing solving for IC soon.

Maple. Time used: 0.003 (sec). Leaf size: 15
ode:=diff(diff(y(x),x),x) = 12*y(x)/x^2; 
dsolve(ode,y(x), singsol=all);
 
y=c2x7+c1x3

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 
<- LODE of Euler type successful`
 

Maple step by step

Let’s solved2dx2y(x)=12y(x)x2Highest derivative means the order of the ODE is2d2dx2y(x)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)12y(x)x2=0Multiply by denominators of the ODEx2(d2dx2y(x))12y(x)=0Make a change of variablest=ln(x)Substitute the change of variables back into the ODECalculate the1stderivative ofywith respect tox, using the chain ruleddxy(x)=(ddty(t))(ddxt(x))Compute derivativeddxy(x)=ddty(t)xCalculate the2ndderivative ofywith respect tox, using the chain ruled2dx2y(x)=(d2dt2y(t))(ddxt(x))2+(d2dx2t(x))(ddty(t))Compute derivatived2dx2y(x)=d2dt2y(t)x2ddty(t)x2Substitute the change of variables back into the ODEx2(d2dt2y(t)x2ddty(t)x2)12y(t)=0Simplifyd2dt2y(t)ddty(t)12y(t)=0Characteristic polynomial of ODEr2r12=0Factor the characteristic polynomial(r+3)(r4)=0Roots of the characteristic polynomialr=(3,4)1st solution of the ODEy1(t)=e3t2nd solution of the ODEy2(t)=e4tGeneral solution of the ODEy(t)=C1y1(t)+C2y2(t)Substitute in solutionsy(t)=C1e3t+C2e4tChange variables back usingt=ln(x)y(x)=C1x3+C2x4Simplifyy(x)=C1x3+C2x4
Mathematica. Time used: 0.011 (sec). Leaf size: 18
ode=D[y[x],{x,2}]==((4*(7/2)^2-1)/(4*x^2))*y[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)c2x7+c1x3
Sympy. Time used: 0.053 (sec). Leaf size: 12
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), (x, 2)) - 12*y(x)/x**2,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
y(x)=C1x3+C2x4