2.1.561 Problem 577

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9733]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 577
Date solved : Wednesday, March 05, 2025 at 07:58:10 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

4x2y+2x3y+(3x2+1)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.304 (sec)

Writing the ode as

(1)4x2y+2x3y+(3x2+1)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=4x2(3)B=2x3C=3x2+1

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=x48x2416x2

Comparing the above to (5) shows that

s=x48x24t=16x2

Therefore eq. (4) becomes

(7)z(x)=(x48x2416x2)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 2.561: 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)=24=2

The poles of r in eq. (7) and the order of each pole are determined by solving for the roots of t=16x2. 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. 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=x2161214x2

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 Or()=2 then

v=Or()2=22=1

[r] is the sum of terms involving xi for 0iv in the Laurent series for r at . Therefore

[r]=i=0vaixi(8)=i=01aixi

Let a be the coefficient of xv=x1 in the above sum. The Laurent series of r at is

(9)rx41x52x310x51052x7310x91965x1113060x13+

Comparing Eq. (9) with Eq. (8) shows that

a=14

From Eq. (9) the sum up to v=1 gives

[r]=i=01aixi(10)=x4

Now we need to find b, where b be the coefficient of xv1=x0=1 in r minus the coefficient of same term but in ([r])2 where [r] was found above in Eq (10). Hence

([r])2=x216

This shows that the coefficient of 1 in the above is 0. Now we need to find the coefficient of 1 in r. How this is done depends on if v=0 or not. Since v=1 which is not zero, then starting r=st, we do long division and write this in the form

r=Q+Rt

Where Q is the quotient and R is the remainder. Then the coefficient of 1 in r will be the coefficient this term in the quotient. Doing long division gives

r=st=x48x2416x2=Q+R16x2=(x21612)+(14x2)=x2161214x2

We see that the coefficient of the term x in the quotient is 12. Now b can be found.

b=(12)(0)=12

Hence

[r]=x4α+=12(bav)=12(12141)=32α=12(bav)=12(12141)=12

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

r=x48x2416x2

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

Order of r at [r] α+ α
2 x4 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+)=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+αc1+xc1)+()[r]=12x+()(x4)=12xx4=12xx4

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(12xx4)(0)+((12x214)+(12xx4)2(x48x2416x2))=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(12xx4)dx=xex28

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

y1=z1e12BAdx=z1e122x34x2dx=z1ex28=z1(ex28)

Which simplifies to

y1=ex24x

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e2x34x2dx(y1)2dx=y1ex24(y1)2dx=y1(Ei1(x24)2)

Therefore the solution is

y=c1y1+c2y2=c1(ex24x)+c2(ex24x(Ei1(x24)2))

Will add steps showing solving for IC soon.

Maple. Time used: 0.008 (sec). Leaf size: 25
ode:=4*x^2*diff(diff(y(x),x),x)+2*x^3*diff(y(x),x)+(3*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=xex24(c1+c2Ei1(x24))

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 solve4x2(d2dx2y(x))+2(ddxy(x))x3+(3x2+1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(3x2+1)y(x)4x2x(ddxy(x))2Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+x(ddxy(x))2+(3x2+1)y(x)4x2=0Check to see ifx0=0is a regular singular pointDefine functions[P2(x)=x2,P3(x)=3x2+14x2]xP2(x)is analytic atx=0(xP2(x))|x=0=0x2P3(x)is analytic atx=0(x2P3(x))|x=0=14x=0is a regular singular pointCheck to see ifx0=0is a regular singular pointx0=0Multiply by denominators4x2(d2dx2y(x))+2(ddxy(x))x3+(3x2+1)y(x)=0Assume series solution fory(x)y(x)=k=0akxk+rRewrite ODE with series expansionsConvertxmy(x)to series expansion form=0..2xmy(x)=k=0akxk+r+mShift index usingk>kmxmy(x)=k=makmxk+rConvertx3(ddxy(x))to series expansionx3(ddxy(x))=k=0ak(k+r)xk+r+2Shift index usingk>k2x3(ddxy(x))=k=2ak2(k2+r)xk+rConvertx2(d2dx2y(x))to series expansionx2(d2dx2y(x))=k=0ak(k+r)(k+r1)xk+rRewrite ODE with series expansionsa0(1+2r)2xr+a1(1+2r)2x1+r+(k=2(ak(2k+2r1)2+ak2(2k+2r1))xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(1+2r)2=0Values of r that satisfy the indicial equationr=12Each term must be 0a1(1+2r)2=0Solve for the dependent coefficient(s)a1=0Each term in the series must be 0, giving the recursion relationak(2k+2r1)2+ak2(2k+2r1)=0Shift index usingk>k+2ak+2(2k+2r+3)2+ak(2k+2r+3)=0Recursion relation that defines series solution to ODEak+2=ak2k+2r+3Recursion relation forr=12ak+2=ak2k+4Solution forr=12[y(x)=k=0akxk+12,ak+2=ak2k+4,a1=0]
Mathematica. Time used: 0.102 (sec). Leaf size: 44
ode=4*x^2*D[y[x],{x,2}]+2*x^3*D[y[x],x]+(1+3*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)12ex2412x(c2ExpIntegralEi(x24)+2ec1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**3*Derivative(y(x), x) + 4*x**2*Derivative(y(x), (x, 2)) + (3*x**2 + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False