2.1.614 Problem 630

Internal problem ID [9786]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 630
Date solved : Wednesday, March 05, 2025 at 07:58:58 AM
CAS classification : [_Gegenbauer]

Solve

\begin{align*} \left (-t^{2}+1\right ) y^{\prime \prime }-2 t y^{\prime }+2 y&=0 \end{align*}

Solved as second order ode using Kovacic algorithm

Time used: 0.297 (sec)

Writing the ode as

\begin{align*} \left (-t^{2}+1\right ) y^{\prime \prime }-2 t y^{\prime }+2 y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= -t^{2}+1 \\ B &= -2 t\tag {3} \\ C &= 2 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(t) &= y e^{\int \frac {B}{2 A} \,dt} \end{align*}

Then (2) becomes

\begin{align*} z''(t) = r z(t)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

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

\begin{align*} r &= \frac {2 t^{2}-3}{\left (t^{2}-1\right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 2 t^{2}-3\\ t &= \left (t^{2}-1\right )^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= \left ( \frac {2 t^{2}-3}{\left (t^{2}-1\right )^{2}}\right ) z(t)\tag {7} \end{align*}

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

\begin{align*} y &= z \left (t \right ) e^{-\int \frac {B}{2 A} \,dt} \end{align*}

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 \(\infty \). The following table summarizes these cases.

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

Table 2.614: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 4 - 2 \\ &= 2 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=\left (t^{2}-1\right )^{2}\). There is a pole at \(t=1\) 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 \(\infty \) 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 \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore

\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}

Attempting to find a solution using case \(n=1\).

Looking at poles of order 2. The partial fractions decomposition of \(r\) is

\[ r = \frac {5}{4 \left (t -1\right )}-\frac {1}{4 \left (t -1\right )^{2}}-\frac {1}{4 \left (t +1\right )^{2}}-\frac {5}{4 \left (t +1\right )} \]

For the pole at \(t=1\) let \(b\) be the coefficient of \(\frac {1}{ \left (t -1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {1}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \end{alignat*}

For the pole at \(t=-1\) let \(b\) be the coefficient of \(\frac {1}{ \left (t +1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {1}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \end{alignat*}

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

\begin{alignat*}{2} r &= \frac {s}{t} &&= \frac {2 t^{2}-3}{\left (t^{2}-1\right )^{2}} \end{alignat*}

Since the \(\text {gcd}(s,t)=1\). This gives \(b=2\). Hence

\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= 2\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -1 \end{alignat*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is

\[ r=\frac {2 t^{2}-3}{\left (t^{2}-1\right )^{2}} \]

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

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(2\)	\(0\)	\(2\)	\(-1\)

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

\begin{align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end{align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{+} = 2\) then

\begin{align*} d &= \alpha _\infty ^{+} - \left ( \alpha _{c_1}^{-}+\alpha _{c_2}^{-} \right ) \\ &= 2 - \left ( 1 \right ) \\ &= 1 \end{align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using

\begin{align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{t-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

Substituting the above values in the above results in

\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{t- c_1}\right )+\left ( (-)[\sqrt r]_{c_2} + \frac { \alpha _{c_2}^{-} }{t- c_2}\right ) + (+) [\sqrt r]_\infty \\ &= \frac {1}{2 t -2}+\frac {1}{2 t +2} + \left ( 0 \right ) \\ &= \frac {1}{2 t -2}+\frac {1}{2 t +2}\\ &= \frac {t}{t^{2}-1} \end{align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(t)\) of degree \(d=1\) to solve the ode. The polynomial \(p(t)\) needs to satisfy the equation

\begin{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end{align*}

Let

\begin{align*} p(t) &= t +a_{0}\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

\begin{align*} \left (0\right ) + 2 \left (\frac {1}{2 t -2}+\frac {1}{2 t +2}\right ) \left (1\right ) + \left ( \left (-\frac {1}{2 \left (t -1\right )^{2}}-\frac {1}{2 \left (t +1\right )^{2}}\right ) + \left (\frac {1}{2 t -2}+\frac {1}{2 t +2}\right )^2 - \left (\frac {2 t^{2}-3}{\left (t^{2}-1\right )^{2}}\right ) \right ) &= 0\\ -\frac {2 a_{0}}{t^{2}-1} = 0 \end{align*}

Solving for the coefficients \(a_i\) in the above using method of undetermined coefficients gives

\[ \{a_{0} = 0\} \]

Substituting these coefficients in \(p(t)\) in eq. (2A) results in

\begin{align*} p(t) &= t \end{align*}

Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z_1(t) &= p e^{ \int \omega \,dt}\\ & = \left (t\right ) {\mathrm e}^{\int \left (\frac {1}{2 t -2}+\frac {1}{2 t +2}\right )d t}\\ & = \left (t\right ) \sqrt {\left (t -1\right ) \left (t +1\right )}\\ & = t \sqrt {t^{2}-1} \end{align*}

The first solution to the original ode in \(y\) is found from

\begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-2 t}{-t^{2}+1} \,dt} \\ &= z_1 e^{-\frac {\ln \left (t -1\right )}{2}-\frac {\ln \left (t +1\right )}{2}} \\ &= z_1 \left (\frac {1}{\sqrt {t -1}\, \sqrt {t +1}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = \frac {t \sqrt {t^{2}-1}}{\sqrt {t -1}\, \sqrt {t +1}} \]

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{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-2 t}{-t^{2}+1} \,dt}}{\left (y_1\right )^2} \,dt \\ &= y_1 \int \frac { e^{-\ln \left (t -1\right )-\ln \left (t +1\right )}}{\left (y_1\right )^2} \,dt \\ &= y_1 \left (\frac {\ln \left (t -1\right )}{2}-\frac {\ln \left (t +1\right )}{2}+\frac {1}{t}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (\frac {t \sqrt {t^{2}-1}}{\sqrt {t -1}\, \sqrt {t +1}}\right ) + c_2 \left (\frac {t \sqrt {t^{2}-1}}{\sqrt {t -1}\, \sqrt {t +1}}\left (\frac {\ln \left (t -1\right )}{2}-\frac {\ln \left (t +1\right )}{2}+\frac {1}{t}\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.006 (sec). Leaf size: 25

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

\[ y = -\frac {c_{2} \ln \left (t +1\right ) t}{2}+\frac {c_{2} \ln \left (t -1\right ) t}{2}+c_{1} t +c_{2} \]

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} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-t^{2}+1\right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )-2 t \left (\frac {d}{d t}y \left (t \right )\right )+2 y \left (t \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=\frac {2 y \left (t \right )}{t^{2}-1}-\frac {2 \left (\frac {d}{d t}y \left (t \right )\right ) t}{t^{2}-1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right ) t}{t^{2}-1}-\frac {2 y \left (t \right )}{t^{2}-1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=\frac {2 t}{t^{2}-1}, P_{3}\left (t \right )=-\frac {2}{t^{2}-1}\right ] \\ {} & \circ & \left (t +1\right )\cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =-1 \\ {} & {} & \left (\left (t +1\right )\cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}-1}}}=1 \\ {} & \circ & \left (t +1\right )^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =-1 \\ {} & {} & \left (\left (t +1\right )^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}-1}}}=0 \\ {} & \circ & t =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (t^{2}-1\right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+2 t \left (\frac {d}{d t}y \left (t \right )\right )-2 y \left (t \right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} t =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-2 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (2 u -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )-2 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r^{2} u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k +r +2\right ) \left (k +r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 a_{k +1} \left (k +1\right )^{2}+a_{k} \left (k +2\right ) \left (k -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right ) \left (k -1\right )}{2 \left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right ) \left (k -1\right )}{2 \left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (-u +1\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =t +1 \\ {} & {} & \left [y \left (t \right )=-a_{0} t \right ] \end {array} \]

✓ Mathematica. Time used: 0.02 (sec). Leaf size: 33

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

\[ y(t)\to c_1 t-\frac {1}{2} c_2 (t \log (1-t)-t \log (t+1)+2) \]

✓ Sympy. Time used: 0.869 (sec). Leaf size: 20

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

\[ y{\left (t \right )} = C_{2} \left (- \frac {t^{4}}{3} - t^{2} + 1\right ) + C_{1} t + O\left (t^{6}\right ) \]