2.1.618 Problem 634

Internal problem ID [9790]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 634
Date solved : Sunday, March 30, 2025 at 02:46:42 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.186 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& t^2{y}^{\prime \prime }+t{y}^{\prime }+(t^2-\frac{1}{4})y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =t^2\\ \text{(3)}& B& =tC& =t^2-\frac{1}{4}\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(t)& =y{e}^{\int \frac{B}{2A}dt}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(t)=rz(t)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{-1}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-1\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =-z\left(t\right)\end{array}$$

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

$$\begin{array}{rl}y& =z\left(t\right)e^{-\int \frac{B}{2A}dt}\end{array}$$

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 $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
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	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.618: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0-0\\ & =0\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=-1$ 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

$$z_1(t)=\cos \left(t\right)$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{t}{t^2}dt}\\ & =z_1{e}^{-\frac{\ln \left(t\right)}{2}}\\ & =z_1\left(\frac{1}{\sqrt{t}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{\cos \left(t\right)}{\sqrt{t}}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dt}}{y_1^2}dt$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{t}{t^2}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{-\ln \left(t\right)}}{{\left(y_1\right)}^2}dt\\ & =y_1(\tan \left(t\right))\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{\cos \left(t\right)}{\sqrt{t}}\right)+c_2\left(\frac{\cos \left(t\right)}{\sqrt{t}}(\tan \left(t\right))\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.012 (sec). Leaf size: 17

ode:=t^2*diff(diff(y(t),t),t)+diff(y(t),t)*t+(t^2-1/4)*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 

$$y=\frac{c_1\sin \left(t\right)+c_2\cos \left(t\right)}{\sqrt{t}}$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & t^2\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+t\left(\frac{d}{dt}y\left(t\right)\right)+(t^2-\frac{1}{4})y\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)=-\frac{(4t^2-1)y\left(t\right)}{4t^2}-\frac{\frac{d}{dt}y\left(t\right)}{t}\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)+\frac{\frac{d}{dt}y\left(t\right)}{t}+\frac{(4t^2-1)y\left(t\right)}{4t^2}=0\\ ◻& & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(t\right)=\frac{1}{t},P_3\left(t\right)=\frac{4t^2-1}{4t^2}]\\ & \circ & t\cdot P_2\left(t\right)\text{is analytic at}t=0\\ & & (t\cdot P_2\left(t\right))|{}_{t=0}=1\\ & \circ & t^2\cdot P_3\left(t\right)\text{is analytic at}t=0\\ & & (t^2\cdot P_3\left(t\right))|{}_{t=0}=-\frac{1}{4}\\ & \circ & t=0\text{is a regular singular point}\\ & & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & & t_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & 4t^2\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+4t\left(\frac{d}{dt}y\left(t\right)\right)+(4t^2-1)y\left(t\right)=0\\ \bullet & & \text{Assume series solution for}y\left(t\right)\\ & & y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{t}^m\cdot y\left(t\right)\text{to series expansion for}m=0..2\\ & & t^m\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & t^m\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{t}^{k+r}\\ & \circ & \text{Convert}t\cdot \left(\frac{d}{dt}y\left(t\right)\right)\text{to series expansion}\\ & & t\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)t^{k+r}\\ & \circ & \text{Convert}{t}^2\cdot \left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)\text{to series expansion}\\ & & t^2\cdot \left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)t^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0(1+2r)(-1+2r)t^r+a_1(3+2r)(1+2r)t^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k(2k+2r+1)(2k+2r-1)+4a_{k-2})t^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (1+2r)(-1+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-\frac{1}{2},\frac{1}{2}\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(3+2r)(1+2r)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_k(4k^2+8kr+4r^2-1)+4a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+2}(4{(k+2)}^2+8(k+2)r+4r^2-1)+4a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{4a_k}{4k^2+8kr+4r^2+16k+16r+15}\\ \bullet & & \text{Recursion relation for}r=-\frac{1}{2}\\ & & a_{k+2}=-\frac{4a_k}{4k^2+12k+8}\\ \bullet & & \text{Solution for}r=-\frac{1}{2}\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k-\frac{1}{2}},a_{k+2}=-\frac{4a_k}{4k^2+12k+8},a_1=0]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{2}\\ & & a_{k+2}=-\frac{4a_k}{4k^2+20k+24}\\ \bullet & & \text{Solution for}r=\frac{1}{2}\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+\frac{1}{2}},a_{k+2}=-\frac{4a_k}{4k^2+20k+24},a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(t\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k-\frac{1}{2}})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{t}^{k+\frac{1}{2}}),a_{k+2}=-\frac{4a_k}{4k^2+12k+8},a_1=0,b_{k+2}=-\frac{4b_k}{4k^2+20k+24},b_1=0]\end{array}$$

✓ Mathematica. Time used: 0.034 (sec). Leaf size: 39

ode=t^2*D[y[t],{t,2}]+t*D[y[t],t]+(t^2-1/4)*y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to \frac{e^{-it}(2c_1-i{c}_2{e}^{2it})}{2\sqrt{t}}$$

✓ Sympy. Time used: 0.248 (sec). Leaf size: 15

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t**2*Derivative(y(t), (t, 2)) + t*Derivative(y(t), t) + (t**2 - 1/4)*y(t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=C_1{J}_{\frac{1}{2}}\left(t\right)+C_2{Y}_{\frac{1}{2}}\left(t\right)$$