1.615 problem 631

Internal problem ID [8753]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 631
Date solved : Monday, October 21, 2024 at 05:21:12 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

1.615.1 Solved as second order ode using Kovacic algorithm

Time used: 0.303 (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 {-3}{\left (t^{2}+1\right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

\begin{align*} z''(t) &= \left ( -\frac {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 615: 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 - 0 \\ &= 4 \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 \(t=i\) of order \(2\). There is a pole at \(t=-i\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(4\) 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 \(4\) 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

\[ r = \frac {3}{4 \left (t -i\right )^{2}}+\frac {3}{4 \left (t +i\right )^{2}}+\frac {3 i}{4 \left (t -i\right )}-\frac {3 i}{4 \left (t +i\right )} \]

For the pole at \(t=i\) let \(b\) be the coefficient of \(\frac {1}{ \left (t -i\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {3}{4}}\). Hence

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

For the pole at \(t=-i\) let \(b\) be the coefficient of \(\frac {1}{ \left (t +i\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {3}{4}}\). Hence

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

Since the order of \(r\) at \(\infty \) is \(4 > 2\) then

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

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

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

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

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(4\)	\(0\)	\(0\)	\(1\)

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 \(\alpha _\infty ^{-} = 1\) then

\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-}+\alpha _{c_2}^{+} \right ) \\ &= 1 - \left ( 1 \right ) \\ &= 0 \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*}

The above gives

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

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

Let

\begin{align*} p(t) &= 1\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

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

The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z_1(t) &= p e^{ \int \omega \,dt} \\ &= {\mathrm e}^{\int \left (-\frac {1}{2 \left (t -i\right )}+\frac {3}{2 \left (t +i\right )}\right )d t}\\ &= \frac {\left (t^{2}+1\right )^{{3}/{2}}}{\left (i t +1\right )^{2}} \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^{2}+1\right )}{2}} \\ &= z_1 \left (\sqrt {t^{2}+1}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = \frac {\left (t^{2}+1\right )^{2}}{\left (i t +1\right )^{2}} \]

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

1.615.2 Maple step by step solution

1.615.3 Maple trace

Methods for second order ODEs:
 

1.615.4 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 16

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

\[ y = c_2 \,t^{2}+c_1 t -c_2 \]

1.615.5 Mathematica DSolve solution

Solving time : 0.071 (sec)
Leaf size : 21

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

\[ y(t)\to c_2 t-c_1 (t-i)^2 \]