2.1.107 Problem 109

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9279]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 109
Date solved : Wednesday, March 05, 2025 at 07:46:50 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solve

x(x2+1)y+(7x2+4)y+8xy=0

Solved as second order ode using Kovacic algorithm

Time used: 0.331 (sec)

Writing the ode as

(1)(x3+x)y+(7x2+4)y+8xy=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x3+x(3)B=7x2+4C=8x

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=3x4+14x2+84(x3+x)2

Comparing the above to (5) shows that

s=3x4+14x2+8t=4(x3+x)2

Therefore eq. (4) becomes

(7)z(x)=(3x4+14x2+84(x3+x)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 2.107: 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)=64=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=4(x3+x)2. There is a pole at x=0 of order 2. There is a pole at x=i of order 2. There is a pole at x=i 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=2x2316(xi)2316(x+i)2+7i16(xi)7i16(x+i)

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=2. Hence

[r]c=0αc+=12+1+4b=2αc=121+4b=1

For the pole at x=i let b be the coefficient of 1(xi)2 in the partial fractions decomposition of r given above. Therefore b=316. Hence

[r]c=0αc+=12+1+4b=34αc=121+4b=14

For the pole at x=i let b be the coefficient of 1(x+i)2 in the partial fractions decomposition of r given above. Therefore b=316. Hence

[r]c=0αc+=12+1+4b=34αc=121+4b=14

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=3x4+14x2+84(x3+x)2

Since the gcd(s,t)=1. This gives b=34. Hence

[r]=0α+=12+1+4b=32α=121+4b=12

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

r=3x4+14x2+84(x3+x)2

pole c location pole order [r]c αc+ αc
0 2 0 2 1
i 2 0 34 14
i 2 0 34 14

Order of r at [r] α+ α
2 0 32 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+αc2+αc3)=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]c2+αc2xc2)+(()[r]c3+αc3xc3)+()[r]=1x+14x4i+14x+4i+()(0)=1x+14x4i+14x+4i=1x+x2x2+2

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(1x+14x4i+14x+4i)(0)+((1x214(xi)214(x+i)2)+(1x+14x4i+14x+4i)2(3x4+14x2+84(x3+x)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(1x+14x4i+14x+4i)dx=(x2+1)1/4x

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

y1=z1e12BAdx=z1e127x2+4x3+xdx=z1e3ln(x2+1)42ln(x)=z1(1(x2+1)3/4x2)

Which simplifies to

y1=1x2+1x3

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e7x2+4x3+xdx(y1)2dx=y1e3ln(x2+1)24ln(x)(y1)2dx=y1(x5x2+1x33x2+1+4x7x2+1+8x93x2+1+xx2+12arcsinh(x)2+x2+1x334x5x2+138x7x2+13)

Therefore the solution is

y=c1y1+c2y2=c1(1x2+1x3)+c2(1x2+1x3(x5x2+1x33x2+1+4x7x2+1+8x93x2+1+xx2+12arcsinh(x)2+x2+1x334x5x2+138x7x2+13))

Will add steps showing solving for IC soon.

Maple. Time used: 0.037 (sec). Leaf size: 32
ode:=x*(x^2+1)*diff(diff(y(x),x),x)+(7*x^2+4)*diff(y(x),x)+8*x*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=x2+1c2x+arcsinh(x)c2+c1x2+1x3

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 solvex(x2+1)(d2dx2y(x))+(7x2+4)(ddxy(x))+8xy(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=8y(x)x2+1(7x2+4)(ddxy(x))x(x2+1)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+(7x2+4)(ddxy(x))x(x2+1)+8y(x)x2+1=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=7x2+4x(x2+1),P3(x)=8x2+1]xP2(x)is analytic atx=0(xP2(x))|x=0=4x2P3(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+1)(d2dx2y(x))+(7x2+4)(ddxy(x))+8xy(x)=0Assume series solution fory(x)y(x)=k=0akxk+rRewrite ODE with series expansionsConvertxy(x)to series expansionxy(x)=k=0akxk+r+1Shift index usingk>k1xy(x)=k=1ak1xk+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..3xm(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 expansionsa0r(3+r)x1+r+a1(1+r)(4+r)xr+(k=1(ak+1(k+r+1)(k+r+4)+ak1(k+r+3)(k+r+1))xk+r)=0a0cannot be 0 by assumption, giving the indicial equationr(3+r)=0Values of r that satisfy the indicial equationr{3,0}Each term must be 0a1(1+r)(4+r)=0Each term in the series must be 0, giving the recursion relation(k+r+1)(ak+1(k+r+4)+ak1(k+r+3))=0Shift index usingk>k+1(k+r+2)(ak+2(k+5+r)+ak(k+r+4))=0Recursion relation that defines series solution to ODEak+2=ak(k+r+4)k+5+rRecursion relation forr=3ak+2=ak(k+1)k+2Solution forr=3[y(x)=k=0akxk3,ak+2=ak(k+1)k+2,2a1=0]Recursion relation forr=0ak+2=ak(k+4)k+5Solution forr=0[y(x)=k=0akxk,ak+2=ak(k+4)k+5,4a1=0]Combine solutions and rename parameters[y(x)=(k=0akxk3)+(k=0bkxk),ak+2=ak(k+1)k+2,2a1=0,bk+2=bk(4+k)5+k,4b1=0]
Mathematica. Time used: 0.22 (sec). Leaf size: 108
ode=x*(1+x^2)*D[y[x],{x,2}]+(4+7*x^2)*D[y[x],x]+8*x*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)exp(1xK[1]2+22(K[1]3+K[1])dK[1]121x7K[2]2+4K[2]3+K[2]dK[2])(c21xexp(21K[3]K[1]2+22(K[1]3+K[1])dK[1])dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*(x**2 + 1)*Derivative(y(x), (x, 2)) + 8*x*y(x) + (7*x**2 + 4)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False