2.1.346 Problem 353

Internal problem ID [9518]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 353
Date solved : Wednesday, March 05, 2025 at 07:50:22 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}4(t^2-3t+2)y^{\prime \prime }-2y^{\prime }+y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.306 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (4t^2-12t+8)y^{\prime \prime }-2y^{\prime }+y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =4t^2-12t+8\\ \text{(3)}& B& =-2C& =1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-4t^2+20t-19}{16{(t^2-3t+2)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-4t^2+20t-19\\ t& =16{(t^2-3t+2)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{-4t^2+20t-19}{16{(t^2-3t+2)}^2}\right)z(t)\end{array}$$

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 2.346: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-2\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=16{(t^2-3t+2)}^2$. There is a pole at $t=2$ of order $2$. There is a pole at $t=1$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $2$ 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. Since pole order is not larger than $2$ and the order at $\mathrm{\infty }$ is $2$ then the necessary conditions for case three are met. Therefore

$$\begin{array}{rl}L& =[1,2,4,6,12]\end{array}$$

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

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

$$r=\frac{3}{8(t-1)}-\frac{3}{16{(t-1)}^2}-\frac{3}{8(t-2)}+\frac{5}{16{(t-2)}^2}$$

For the pole at $t=2$ let $b$ be the coefficient of $\frac{1}{{(t-2)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{5}{16}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{4}\end{array}$$

For the pole at $t=1$ let $b$ be the coefficient of $\frac{1}{{(t-1)}^2}$ 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 2 then $[\sqrt{r}{]}_{\mathrm{\infty }}=0$. Let $b$ be the coefficient of $\frac{1}{t^2}$ in the Laurent series expansion of $r$ at $\mathrm{\infty }$. which can be found by dividing the leading coefficient of $s$ by the leading coefficient of $t$ from

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-4t^2+20t-19}{16{(t^2-3t+2)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{1}{4}$. Hence

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

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

$$r=\frac{-4t^2+20t-19}{16{(t^2-3t+2)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$2$	$2$	$0$	$\frac{5}{4}$	$-\frac{1}{4}$
$1$	$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{1}{2}$	$\frac{1}{2}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{1}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})\\ & =\frac{1}{2}-\left(\frac{1}{2}\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{t-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{t-c_2})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{4(t-2)}+\frac{3}{4(t-1)}+(-)\left(0\right)\\ & =-\frac{1}{4(t-2)}+\frac{3}{4(t-1)}\\ & =\frac{2t-5}{4(t-1)(t-2)}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{4(t-2)}+\frac{3}{4(t-1)})\left(0\right)+((\frac{1}{4{(t-2)}^2}-\frac{3}{4{(t-1)}^2})+{(-\frac{1}{4(t-2)}+\frac{3}{4(t-1)})}^2-\left(\frac{-4t^2+20t-19}{16{(t^2-3t+2)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & ={\mathrm{e}}^{\int (-\frac{1}{4(t-2)}+\frac{3}{4(t-1)})dt}\\ & =\frac{{(t-1)}^{3/4}}{{(t-2)}^{1/4}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-2}{4t^2-12t+8}dt}\\ & =z_1{e}^{-\frac{\ln (t-1)}{4}+\frac{\ln (t-2)}{4}}\\ & =z_1\left(\frac{{(t-2)}^{1/4}}{{(t-1)}^{1/4}}\right)\end{array}$$

Which simplifies to

$$y_1=\sqrt{t-1}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-2}{4t^2-12t+8}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{-\frac{\ln (t-1)}{2}+\frac{\ln (t-2)}{2}}}{{\left(y_1\right)}^2}dt\\ & =y_1(-\frac{2\sqrt{t-2}}{\sqrt{t-1}}+\frac{\ln (-\frac{3}{2}+t+\sqrt{t^2-3t+2})\sqrt{(t-1)(t-2)}}{\sqrt{t-2}\sqrt{t-1}})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\sqrt{t-1}\right)+c_2\left(\sqrt{t-1}(-\frac{2\sqrt{t-2}}{\sqrt{t-1}}+\frac{\ln (-\frac{3}{2}+t+\sqrt{t^2-3t+2})\sqrt{(t-1)(t-2)}}{\sqrt{t-2}\sqrt{t-1}})\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.019 (sec). Leaf size: 56

ode:=4*(t^2-3*t+2)*diff(diff(y(t),t),t)-2*diff(y(t),t)+y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 

$$y=c_1\sqrt{t-1}+\frac{c_2(-\frac{\sqrt{t^2-3t+2}(-\ln \left(2\right)+\ln (-3+2t+2\sqrt{(t-1)(t-2)}))}{2}+t-2)}{\sqrt{t-2}}$$

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 
<- Kovacics algorithm successful`
 

Maple step by step

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

✓ Mathematica. Time used: 0.176 (sec). Leaf size: 112

ode=4*(t^2-3*t+2)*D[y[t],{t,2}]-2*D[y[t],t]+y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to \exp ({\int }_1^t\frac{2K[1]-5}{4(K[1{]}^2-3K[1]+2)}dK[1]-\frac{1}{2}{\int }_1^t-\frac{1}{2(K[2{]}^2-3K[2]+2)}dK[2])(c_2{\int }_1^t\exp (-2{\int }_1^{K[3]}\frac{2K[1]-5}{4(K[1{]}^2-3K[1]+2)}dK[1])dK[3]+c_1)$$

✓ Sympy. Time used: 0.471 (sec). Leaf size: 44

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq((4*t**2 - 12*t + 8)*Derivative(y(t), (t, 2)) + y(t) - 2*Derivative(y(t), t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=\sqrt{t-2}(C_1\sqrt{\frac{t-1}{t-2}}{}_1{F}_0\left(\begin{array}{c}0\end{array}\right|\frac{t-1}{t-2})+C_2{}_2{F}_1\left(\begin{array}{c}-\frac{1}{2},1\\ \frac{1}{2}\end{array}\right|\frac{t-1}{t-2}))$$