2.1.94 Problem 96

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9266]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 96
Date solved : Wednesday, March 05, 2025 at 07:46:37 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

2x2y+5xy+(1+x)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.278 (sec)

Writing the ode as

(1)2x2y+5xy+(1+x)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=2x2(3)B=5xC=1+x

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=38x16x2

Comparing the above to (5) shows that

s=38xt=16x2

Therefore eq. (4) becomes

(7)z(x)=(38x16x2)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 2.94: 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)=21=1

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 a pole of order 2 then necessary conditions for case two are met. Therefore

L=[2]

Attempting to find a solution using case n=2.

Looking at poles of order 2. The partial fractions decomposition of r is

r=316x212x

For the pole at x=0 let b be the coefficient of 1x2 in the partial fractions decomposition of r given above. Therefore b=316. Hence

Ec={2,2+21+4b,221+4b}={1,2,3}

Since the order of r at is 1<2 then

E={1}

The following table summarizes the findings so far for poles and for the order of r at for case 2 of Kovacic algorithm.

pole c location pole order Ec
0 2 {1,2,3}

Order of r at E
1 {1}

Using the family {e1,e2,,e} given by

e1=1,e=1

Gives a non negative integer d (the degree of the polynomial p(x)), which is generated using

d=12(ecΓec)=12(1(1))=0

We now form the following rational function

θ=12cΓecxc=12(1(x(0)))=12x

Now we search for a monic polynomial p(x) of degree d=0 such that

(1A)p+3θp+(3θ2+3θ4r)p+(θ+3θθ+θ34rθ2r)p=0

Since d=0, then letting

(2A)p=1

Substituting p and θ into Eq. (1A) gives

0=0

And solving for p gives

p=1

Now that p(x) is found let

ϕ=θ+pp=12x

Let ω be the solution of

ω2ϕω+(12ϕ+12ϕ2r)=0

Substituting the values for ϕ and r into the above equation gives

w2w2x+1+8x16x2=0

Solving for ω gives

ω=1+22x4x

Therefore the first solution to the ode z=rz is

z1(x)=eωdx=e1+22x4xdx=x1/4e2x

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

y1=z1e12BAdx=z1e125x2x2dx=z1e5ln(x)4=z1(1x5/4)

Which simplifies to

y1=e2xx

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

y2=y1eBAdxy12dx

Substituting gives

y2=y1e5x2x2dx(y1)2dx=y1e5ln(x)2(y1)2dx=y1(2x(1e22x)2x)

Therefore the solution is

y=c1y1+c2y2=c1(e2xx)+c2(e2xx(2x(1e22x)2x))

Will add steps showing solving for IC soon.

Maple. Time used: 0.008 (sec). Leaf size: 29
ode:=2*x^2*diff(diff(y(x),x),x)+5*x*diff(y(x),x)+y(x)*(x+1) = 0; 
dsolve(ode,y(x), singsol=all);
 
y=c1sin(x2)+c2cos(x2)x

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 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful`
 

Maple step by step

Let’s solve2x2(d2dx2y(x))+5x(ddxy(x))+(x+1)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(x+1)y(x)2x25(ddxy(x))2xGroup terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+5(ddxy(x))2x+(x+1)y(x)2x2=0Check to see ifx0=0is a regular singular pointDefine functions[P2(x)=52x,P3(x)=x+12x2]xP2(x)is analytic atx=0(xP2(x))|x=0=52x2P3(x)is analytic atx=0(x2P3(x))|x=0=12x=0is a regular singular pointCheck to see ifx0=0is a regular singular pointx0=0Multiply by denominators2x2(d2dx2y(x))+5x(ddxy(x))+(x+1)y(x)=0Assume series solution fory(x)y(x)=k=0akxk+rRewrite ODE with series expansionsConvertxmy(x)to series expansion form=0..1xmy(x)=k=0akxk+r+mShift index usingk>kmxmy(x)=k=makmxk+rConvertx(ddxy(x))to series expansionx(ddxy(x))=k=0ak(k+r)xk+rConvertx2(d2dx2y(x))to series expansionx2(d2dx2y(x))=k=0ak(k+r)(k+r1)xk+rRewrite ODE with series expansionsa0(1+r)(1+2r)xr+(k=1(ak(k+r+1)(2k+2r+1)+ak1)xk+r)=0a0cannot be 0 by assumption, giving the indicial equation(1+r)(1+2r)=0Values of r that satisfy the indicial equationr{1,12}Each term in the series must be 0, giving the recursion relation2(k+r+1)(k+r+12)ak+ak1=0Shift index usingk>k+12(k+2+r)(k+32+r)ak+1+ak=0Recursion relation that defines series solution to ODEak+1=ak(k+2+r)(2k+3+2r)Recursion relation forr=1ak+1=ak(k+1)(2k+1)Solution forr=1[y(x)=k=0akxk1,ak+1=ak(k+1)(2k+1)]Recursion relation forr=12ak+1=ak(k+32)(2k+2)Solution forr=12[y(x)=k=0akxk12,ak+1=ak(k+32)(2k+2)]Combine solutions and rename parameters[y(x)=(k=0akxk1)+(k=0bkxk12),ak+1=ak(k+1)(2k+1),bk+1=bk(k+32)(2k+2)]
Mathematica. Time used: 0.085 (sec). Leaf size: 60
ode=2*x^2*D[y[x],{x,2}]+5*x*D[y[x],x]+(1+x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)2c1ei2x+i2c2ei2x2x
Sympy. Time used: 0.206 (sec). Leaf size: 37
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**2*Derivative(y(x), (x, 2)) + 5*x*Derivative(y(x), x) + (x + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
y(x)=C1J12(2x)+C2Y12(2x)x34