2.1.40 Problem 42

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9212]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 42
Date solved : Wednesday, March 05, 2025 at 07:37:53 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

x2y2x(x+2)y+(x2+4x+6)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.068 (sec)

Writing the ode as

(1)x2y+(2x24x)y+(x2+4x+6)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=x2(3)B=2x24xC=x2+4x+6

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=01

Comparing the above to (5) shows that

s=0t=1

Therefore eq. (4) becomes

(7)z(x)=0

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 2.40: 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)=0=

There are no poles in r. Therefore the set of poles Γ is empty. Since there is no odd order pole larger than 2 and the order at is infinity then the necessary conditions for case one are met. Therefore

L=[1]

Since r=0 is not a function of x, then there is no need run Kovacic algorithm to obtain a solution for transformed ode z=rz as one solution is

z1(x)=1

Using the above, the solution for the original ode can now be found. The first solution to the original ode in y is found from

y1=z1e12BAdx=z1e122x24xx2dx=z1ex+2ln(x)=z1(x2ex)

Which simplifies to

y1=x2ex

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e2x24xx2dx(y1)2dx=y1e2x+4ln(x)(y1)2dx=y1(x)

Therefore the solution is

y=c1y1+c2y2=c1(x2ex)+c2(x2ex(x))

Will add steps showing solving for IC soon.

Maple. Time used: 0.003 (sec). Leaf size: 15
ode:=x^2*diff(diff(y(x),x),x)-2*x*(2+x)*diff(y(x),x)+(x^2+4*x+6)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=exx2(c2x+c1)

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) 
<- Kovacics algorithm successful`
 

Maple step by step

Let’s solvex2(d2dx2y(x))2x(x+2)(ddxy(x))+(x2+4x+6)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(x2+4x+6)y(x)x2+2(x+2)(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)2(x+2)(ddxy(x))x+(x2+4x+6)y(x)x2=0Check to see ifx0=0is a regular singular pointDefine functions[P2(x)=2(x+2)x,P3(x)=x2+4x+6x2]xP2(x)is analytic atx=0(xP2(x))|x=0=4x2P3(x)is analytic atx=0(x2P3(x))|x=0=6x=0is a regular singular pointCheck to see ifx0=0is a regular singular pointx0=0Multiply by denominatorsx2(d2dx2y(x))2x(x+2)(ddxy(x))+(x2+4x+6)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+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(2+r)(3+r)xr+(a1(1+r)(2+r)2a0(2+r))x1+r+(k=2(ak(k+r2)(k+r3)2ak1(k+r3)+ak2)xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(2+r)(3+r)=0Values of r that satisfy the indicial equationr{2,3}Each term must be 0a1(1+r)(2+r)2a0(2+r)=0Solve for the dependent coefficient(s)a1=2a01+rEach term in the series must be 0, giving the recursion relationak(k+r2)(k+r3)2ak1k2ak1r+ak2+6ak1=0Shift index usingk>k+2ak+2(k+r)(k+r1)2ak+1(k+2)2ak+1r+ak+6ak+1=0Recursion relation that defines series solution to ODEak+2=2kak+1+2ak+1rak2ak+1(k+r)(k+r1)Recursion relation forr=2ak+2=2kak+1ak+2ak+1(k+2)(k+1)Solution forr=2[y(x)=k=0akxk+2,ak+2=2kak+1ak+2ak+1(k+2)(k+1),a1=2a0]Recursion relation forr=3ak+2=2kak+1ak+4ak+1(k+3)(k+2)Solution forr=3[y(x)=k=0akxk+3,ak+2=2kak+1ak+4ak+1(k+3)(k+2),a1=a0]Combine solutions and rename parameters[y(x)=(k=0akxk+2)+(k=0bkxk+3),ak+2=2kak+1ak+2ak+1(k+2)(k+1),a1=2a0,bk+2=2kbk+1bk+4bk+1(k+3)(k+2),b1=b0]
Mathematica. Time used: 0.049 (sec). Leaf size: 21
ode=x^2*D[y[x],{x,2}]-2*x*(x+2)*D[y[x],x]+(x^2+4*x+6)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)ex+2x2(c2x+c1)
Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*Derivative(y(x), (x, 2)) - 2*x*(x + 2)*Derivative(y(x), x) + (x**2 + 4*x + 6)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
False