2.1.45 Problem 47

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9217]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 47
Date solved : Wednesday, March 05, 2025 at 07:37:57 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

(2+x)y+xy+3y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.346 (sec)

Writing the ode as

(1)(2+x)y+xy+3y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=2+x(3)B=xC=3

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=x212x204(2+x)2

Comparing the above to (5) shows that

s=x212x20t=4(2+x)2

Therefore eq. (4) becomes

(7)z(x)=(x212x204(2+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 2.45: 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=4(2+x)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+2(2+x)242+x

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

[r]c=0αc+=12+1+4b=2αc=121+4b=1

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)r124x6x272x3556x45440x555088x6586688x7+

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=x212x204x2+16x+16=Q+R4x2+16x+16=(14)+(16x244x2+16x+16)=14+16x244x2+16x+16

Since the degree of t is 2, then we see that the coefficient of the term x in the remainder R is 16. Dividing this by leading coefficient in t which is 4 gives 4. Now b can be found.

b=(4)(0)=4

Hence

[r]=12α+=12(bav)=12(4120)=4α=12(bav)=12(4120)=4

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

r=x212x204(2+x)2

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

Order of r at [r] α+ α
0 12 4 4

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=4 then

d=α(αc1+)=4(2)=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]

The above gives

ω=((+)[r]c1+αc1+xc1)+()[r]=22+x+()(12)=22+x12=x22(2+x)

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(22+x12)(2x+a1)+((2(2+x)2)+(22+x12)2(x212x204(2+x)2))=0(a1+6)x+2a0+2a1+42+x=0

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

{a0=4,a1=6}

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

p(x)=x26x+4

Therefore the first solution to the ode z=rz is

z1(x)=peωdx=(x26x+4)e(22+x12)dx=(x26x+4)ex2+2ln(2+x)=(x26x+4)(2+x)2ex2

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

y1=z1e12BAdx=z1e12x2+xdx=z1ex2+ln(2+x)=z1((2+x)ex2)

Which simplifies to

y1=(2+x)3ex(x26x+4)

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1ex2+xdx(y1)2dx=y1ex+2ln(2+x)(y1)2dx=y1(ex(x4x318x222x+8)240(x26x+4)(2+x)3e2Ei1(2x)240)

Therefore the solution is

y=c1y1+c2y2=c1((2+x)3ex(x26x+4))+c2((2+x)3ex(x26x+4)(ex(x4x318x222x+8)240(x26x+4)(2+x)3e2Ei1(2x)240))

Will add steps showing solving for IC soon.

Maple. Time used: 0.006 (sec). Leaf size: 72
ode:=(2+x)*diff(diff(y(x),x),x)+x*diff(y(x),x)+3*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=ex2c2(x26x+4)(x+2)3Ei1(x2)+c1ex(x26x+4)(x+2)3+c2(x4x318x222x+8)

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 solve(x+2)(d2dx2y(x))+x(ddxy(x))+3y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=3y(x)x+2(ddxy(x))xx+2Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+(ddxy(x))xx+2+3y(x)x+2=0Check to see ifx0=2is a regular singular pointDefine functions[P2(x)=xx+2,P3(x)=3x+2](x+2)P2(x)is analytic atx=2((x+2)P2(x))|x=2=2(x+2)2P3(x)is analytic atx=2((x+2)2P3(x))|x=2=0x=2is a regular singular pointCheck to see ifx0=2is a regular singular pointx0=2Multiply by denominators(x+2)(d2dx2y(x))+x(ddxy(x))+3y(x)=0Change variables usingx=u2so that the regular singular point is atu=0u(d2du2y(u))+(u2)(dduy(u))+3y(u)=0Assume series solution fory(u)y(u)=k=0akuk+rRewrite ODE with series expansionsConvertum(dduy(u))to series expansion form=0..1um(dduy(u))=k=0ak(k+r)uk+r1+mShift index usingk>k+1mum(dduy(u))=k=1+mak+1m(k+1m+r)uk+rConvertu(d2du2y(u))to series expansionu(d2du2y(u))=k=0ak(k+r)(k+r1)uk+r1Shift index usingk>k+1u(d2du2y(u))=k=1ak+1(k+1+r)(k+r)uk+rRewrite ODE with series expansionsa0r(3+r)u1+r+(k=0(ak+1(k+1+r)(k2+r)+ak(k+r+3))uk+r)=0a0cannot be 0 by assumption, giving the indicial equationr(3+r)=0Values of r that satisfy the indicial equationr{0,3}Each term in the series must be 0, giving the recursion relationak+1(k+1+r)(k2+r)+ak(k+r+3)=0Recursion relation that defines series solution to ODEak+1=ak(k+r+3)(k+1+r)(k2+r)Recursion relation forr=0ak+1=ak(k+3)(k+1)(k2)Series not valid forr=0, division by0in the recursion relation atk=2ak+1=ak(k+3)(k+1)(k2)Recursion relation forr=3ak+1=ak(k+6)(k+4)(k+1)Solution forr=3[y(u)=k=0akuk+3,ak+1=ak(k+6)(k+4)(k+1)]Revert the change of variablesu=x+2[y(x)=k=0ak(x+2)k+3,ak+1=ak(k+6)(k+4)(k+1)]
Mathematica. Time used: 0.688 (sec). Leaf size: 106
ode=(2+x)*D[y[x],{x,2}]+x*D[y[x],x]+3*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)(x26x+4)exp(1x(2K[1]+212)dK[1]121xK[2]K[2]+2dK[2])(c21xexp(21K[3](2K[1]+212)dK[1])(K[3]26K[3]+4)2dK[3]+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*Derivative(y(x), x) + (x + 2)*Derivative(y(x), (x, 2)) + 3*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False