2.1.624 Problem 641

Internal problem ID [9796]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 641
Date solved : Sunday, March 30, 2025 at 02:46:49 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.283 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{t^2-2t-3}{16t^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =t^2-2t-3\\ t& =16t^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{t^2-2t-3}{16t^2}\right)z(t)\end{array}$$

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 2.624: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-2\\ & =0\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=16t^2$. There is a pole at $t=0$ of order $2$. 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. Since there is a pole of order $2$ then necessary conditions for case two are met. Therefore

$$\begin{array}{rl}L& =[1,2]\end{array}$$

Attempting to find a solution using case $n=1$.

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

$$r=\frac{1}{16}-\frac{1}{8t}-\frac{3}{16t^2}$$

For the pole at $t=0$ let $b$ be the coefficient of $\frac{1}{t^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{3}{16}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{4}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=0$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{0}{2}& & =0\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $t^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{t}^i\\ \text{(8)}& & =\sum _{i=0}^0{a}_i{t}^i\end{array}$$

Let $a$ be the coefficient of $t^v=t^0$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{1}{4}-\frac{1}{4t}-\frac{1}{2t^2}-\frac{1}{2t^3}-\frac{1}{t^4}-\frac{2}{t^5}-\frac{9}{2t^6}-\frac{21}{2t^7}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=\frac{1}{4}$$

From Eq. (9) the sum up to $v=0$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^0{a}_i{t}^i\\ \text{(10)}& & =\frac{1}{4}\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $t^{v-1}=t^{-1}=\frac{1}{t}$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{16}$$

This shows that the coefficient of $\frac{1}{t}$ in the above is $0$. Now we need to find the coefficient of $\frac{1}{t}$ in $r$. How this is done depends on if $v=0$ or not. Since $v=0$ then starting from $r=\frac{s}{t}$ and doing long division in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $\frac{1}{t}$ in $r$ will be the coefficient in $R$ of the term in $t$ of degree of $t$ minus one, divided by the leading coefficient in $t$. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{t^2-2t-3}{16t^2}\\ & =Q+\frac{R}{16t^2}\\ & =\left(\frac{1}{16}\right)+\left(\frac{-2t-3}{16t^2}\right)\\ & =\frac{1}{16}+\frac{-2t-3}{16t^2}\end{array}$$

Since the degree of $t$ is $2$, then we see that the coefficient of the term $t$ in the remainder $R$ is $-2$. Dividing this by leading coefficient in $t$ which is $16$ gives $-\frac{1}{8}$. Now $b$ can be found.

$$\begin{array}{rl}b& =(-\frac{1}{8})-\left(0\right)\\ & =-\frac{1}{8}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{1}{4}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-\frac{1}{8}}{\frac{1}{4}}-0)& & =-\frac{1}{4}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-\frac{1}{8}}{\frac{1}{4}}-0)& & =\frac{1}{4}\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ where $r$ is

$$r=\frac{t^2-2t-3}{16t^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$0$	$\frac{1}{4}$	$-\frac{1}{4}$	$\frac{1}{4}$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{-}=\frac{1}{4}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =\frac{1}{4}-\left(\frac{1}{4}\right)\\ & =0\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{t-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{t-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{1}{4t}+(-)\left(\frac{1}{4}\right)\\ & =\frac{1}{4t}-\frac{1}{4}\\ & =-\frac{t-1}{4t}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(t)$ of degree $d=0$ to solve the ode. The polynomial $p(t)$ needs to satisfy the equation

$$\begin{array}{}\text{(1A)}& p^{″}+2\omega p^{\prime }+({\omega }^{\prime }+{\omega }^2-r)p=0\end{array}$$

Let

$$\begin{array}{}\text{(2A)}& p(t)& =1\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(\frac{1}{4t}-\frac{1}{4})\left(0\right)+((-\frac{1}{4t^2})+{(\frac{1}{4t}-\frac{1}{4})}^2-\left(\frac{t^2-2t-3}{16t^2}\right))& =0\\ 0=0\end{array}$$

The equation is satisfied since both sides are zero. Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & ={\mathrm{e}}^{\int (\frac{1}{4t}-\frac{1}{4})dt}\\ & =t^{1/4}{\mathrm{e}}^{-\frac{t}{4}}\end{array}$$

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

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

Which simplifies to

$$y_1=\sqrt{t}{\mathrm{e}}^{-\frac{t}{2}}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{t^2-t}{2t^2}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{-\frac{t}{2}+\frac{\ln \left(t\right)}{2}}}{{\left(y_1\right)}^2}dt\\ & =y_1(-i\sqrt{\pi }\sqrt{2}\mathrm{erf}\left(\frac{i\sqrt{2}\sqrt{t}}{2}\right))\end{array}$$

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.053 (sec). Leaf size: 47

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

$$y=\frac{{\mathrm{e}}^{-\frac{t}{2}}(\mathrm{erf}\left(\frac{\sqrt{2}\sqrt{-t}}{2}\right)\sqrt{\pi }{2}^{3/4}{c}_{1}t+4c_2\sqrt{t}\sqrt{-t})}{4\sqrt{-t}}$$

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 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Whittaker 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
         <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
      <- Whittaker successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved -\ 
 returning with no uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 2t^2\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+(t^2-t)\left(\frac{d}{dt}y\left(t\right)\right)+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{y\left(t\right)}{2t^2}-\frac{(t-1)\left(\frac{d}{dt}y\left(t\right)\right)}{2t}\\ \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{(t-1)\left(\frac{d}{dt}y\left(t\right)\right)}{2t}+\frac{y\left(t\right)}{2t^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{t-1}{2t},P_3\left(t\right)=\frac{1}{2t^2}]\\ & \circ & t\cdot P_2\left(t\right)\text{is analytic at}t=0\\ & & (t\cdot P_2\left(t\right))|{}_{t=0}=-\frac{1}{2}\\ & \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}{2}\\ & \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}\\ & & 2t^2\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+t(t-1)\left(\frac{d}{dt}y\left(t\right)\right)+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 \left(\frac{d}{dt}y\left(t\right)\right)\text{to series expansion for}m=1..2\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)t^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+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+r)t^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_k(2k+2r-1)(k+r-1)+a_{k-1}(k+r-1))t^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (-1+2r)(-1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{1,\frac{1}{2}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 2(k+r-1)((k+r-\frac{1}{2})a_k+\frac{a_{k-1}}{2})=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & 2(k+r)((k+\frac{1}{2}+r)a_{k+1}+\frac{a_k}{2})=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{a_k}{2k+1+2r}\\ \bullet & & \text{Recursion relation for}r=1\\ & & a_{k+1}=-\frac{a_k}{2k+3}\\ \bullet & & \text{Solution for}r=1\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+1},a_{k+1}=-\frac{a_k}{2k+3}]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{2}\\ & & a_{k+1}=-\frac{a_k}{2k+2}\\ \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+1}=-\frac{a_k}{2k+2}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(t\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+1})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{t}^{k+\frac{1}{2}}),a_{k+1}=-\frac{a_k}{2k+3},b_{k+1}=-\frac{b_k}{2k+2}]\end{array}$$

✓ Mathematica. Time used: 0.037 (sec). Leaf size: 46

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

$$y(t)\to e^{-t/2}(c_2\sqrt{t}+\sqrt{2}{c}_1\sqrt{-t}\mathrm{\Gamma }(\frac{1}{2},-\frac{t}{2}))$$

✓ Sympy. Time used: 1.485 (sec). Leaf size: 474

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

$$\text{Solution too large to show}$$