2.1.613 Problem 629

Solved as second order ode using Kovacic algorithm
Maple
Mathematica
Sympy

Internal problem ID [9785]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 629
Date solved : Wednesday, March 05, 2025 at 07:58:57 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

y4ty+(4t22)y=0

Solved as second order ode using Kovacic algorithm

Time used: 0.054 (sec)

Writing the ode as

(1)y4ty+(4t22)y=0(2)Ay+By+Cy=0

Comparing (1) and (2) shows that

A=1(3)B=4tC=4t22

Applying the Liouville transformation on the dependent variable gives

z(t)=yeB2Adt

Then (2) becomes

(4)z(t)=rz(t)

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(t)=0

Equation (7) is now solved. After finding z(t) then y is found using the inverse transformation

y=z(t)eB2Adt

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 2.613: 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 t, then there is no need run Kovacic algorithm to obtain a solution for transformed ode z=rz as one solution is

z1(t)=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=z1e12BAdt=z1e124t1dt=z1et2=z1(et2)

Which simplifies to

y1=et2

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

y2=y1eBAdty12dt

Substituting gives

y2=y1e4t1dt(y1)2dt=y1e2t2(y1)2dt=y1(t)

Therefore the solution is

y=c1y1+c2y2=c1(et2)+c2(et2(t))

Will add steps showing solving for IC soon.

Maple. Time used: 0.002 (sec). Leaf size: 14
ode:=diff(diff(y(t),t),t)-4*t*diff(y(t),t)+(4*t^2-2)*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 
y=et2(c2t+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 solved2dt2y(t)4t(ddty(t))+(4t22)y(t)=0Highest derivative means the order of the ODE is2d2dt2y(t)Assume series solution fory(t)y(t)=k=0aktkRewrite ODE with series expansionsConverttmy(t)to series expansion form=0..2tmy(t)=k=max(0,m)aktk+mShift index usingk>kmtmy(t)=k=max(0,m)+makmtkConvertt(ddty(t))to series expansiont(ddty(t))=k=0akktkConvertd2dt2y(t)to series expansiond2dt2y(t)=k=2akk(k1)tk2Shift index usingk>k+2d2dt2y(t)=k=0ak+2(k+2)(k+1)tkRewrite ODE with series expansions2a22a0+(6a36a1)t+(k=2(ak+2(k+2)(k+1)2ak(2k+1)+4ak2)tk)=0The coefficients of each power oftmust be 0[2a22a0=0,6a36a1=0]Solve for the dependent coefficient(s){a2=a0,a3=a1}Each term in the series must be 0, giving the recursion relation(k2+3k+2)ak+24akk2ak+4ak2=0Shift index usingk>k+2((k+2)2+3k+8)ak+44ak+2(k+2)2ak+2+4ak=0Recursion relation that defines the series solution to the ODE[y(t)=k=0aktk,ak+4=2(2kak+22ak+5ak+2)k2+7k+12,a2=a0,a3=a1]
Mathematica. Time used: 0.02 (sec). Leaf size: 18
ode=D[y[t],{t,2}]-4*t*D[y[t],t]+(4*t^2-2)*y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 
y(t)et2(c2t+c1)
Sympy. Time used: 0.921 (sec). Leaf size: 37
from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-4*t*Derivative(y(t), t) + (4*t**2 - 2)*y(t) + Derivative(y(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 
y(t)=7t5r(3)10+C2(t42+t2+1)+C1t(1t45)+O(t6)