2.1.613 Problem 629

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

$$\begin{array}{rl}{y}^{\prime \prime }-4t{y}^{\prime }+(4t^2-2)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.054 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{0}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =0\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =0\end{array}$$

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 2.613: Necessary conditions for each Kovacic case

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

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

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 $infinity$ then the necessary conditions for case one are met. Therefore

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

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

$$z_1(t)=1$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-4t}{1}dt}\\ & =z_1{e}^{t^2}\\ & =z_1\left({\mathrm{e}}^{t^2}\right)\end{array}$$

Which simplifies to

$$y_1={\mathrm{e}}^{t^2}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-4t}{1}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{2t^2}}{{\left(y_1\right)}^2}dt\\ & =y_1\left(t\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left({\mathrm{e}}^{t^2}\right)+c_2\left({\mathrm{e}}^{t^2}\left(t\right)\right)\end{array}$$

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);
 

$$y={\mathrm{e}}^{t^2}(c_{2}t+c_1)$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)-4t\left(\frac{d}{dt}y\left(t\right)\right)+(4t^2-2)y\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{t}^2}y\left(t\right)\\ \bullet & & \text{Assume series solution for}y\left(t\right)\\ & & y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k\\ ◻& & \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=max(0,-m)}}{a}_k{t}^{k+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & t^m\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)+m}}{a}_{k-m}{t}^k\\ & \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{t}^k\\ & \circ & \text{Convert}\frac{d^2}{d{t}^2}y\left(t\right)\text{to series expansion}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=2}}{a}_{k}k(k-1)t^{k-2}\\ & \circ & \text{Shift index using}k->k+2\\ & & \frac{d^2}{d{t}^2}y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k+2}(k+2)(k+1)t^k\\ & & \text{Rewrite ODE with series expansions}\\ & & 2a_2-2a_0+(6a_3-6a_1)t+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_{k+2}(k+2)(k+1)-2a_k(2k+1)+4a_{k-2})t^k)=0\\ \bullet & & \text{The coefficients of each power of}t\text{must be 0}\\ & & [2a_2-2a_0=0,6a_3-6a_1=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_2=a_0,a_3=a_1\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k^2+3k+2)a_{k+2}-4a_{k}k-2a_k+4a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & ({(k+2)}^2+3k+8)a_{k+4}-4a_{k+2}(k+2)-2a_{k+2}+4a_k=0\\ \bullet & & \text{Recursion relation that defines the series solution to the ODE}\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k,a_{k+4}=\frac{2(2k{a}_{k+2}-2a_k+5a_{k+2})}{k^2+7k+12},a_2=a_0,a_3=a_1]\end{array}$$

✓ 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]
 

$$y(t)\to e^{t^2}(c_{2}t+c_1)$$

✓ 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)
 

$$y\left(t\right)=\frac{7t^{5}r\left(3\right)}{10}+C_2(\frac{t^4}{2}+t^2+1)+C_{1}t(1-\frac{t^4}{5})+O\left(t^6\right)$$