Internal problem ID [8324]
Internal file name [OUTPUT/7257_Sunday_June_05_2022_05_40_07_PM_33549532/index.tex]
Book: Collection of Kovacic problems
Section: section 2. Solution found using all possible Kovacic cases
Problem number: 8.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_change_of_variable_on_y_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x^{2}-x \right ) y^{\prime \prime }-x y^{\prime }+y=0} \] Writing the ode as \begin {align*} \left (x^{2}-x \right ) y^{\prime \prime }-x y^{\prime }+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 &= x^{2}-x \\ B &= -x\tag {3} \\ C &= 1 \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end {align*}
Then (2) becomes \begin {align*} z''(x) = r z(x)\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 {-x +4}{4 \left (x -1\right )^{2} x}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= -x +4\\ t &= 4 \left (x -1\right )^{2} x \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {-x +4}{4 \left (x -1\right )^{2} x}\right ) z(x)\tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation \begin {align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
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) \\ &= 3 - 1 \\ &= 2 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (x -1\right )^{2} x\). There is a pole at \(x=1\) of order \(2\). There is a pole at \(x=0\) of order \(1\). 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 1. For the pole at \(x = 0\) of order 1 then \begin {align*} [\sqrt r]_c &= 0 \\ \alpha _c^+ &= 1 \\ \alpha _c^- &= 1 \end {align*}
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = -\frac {1}{x -1}+\frac {3}{4 \left (x -1\right )^{2}}+\frac {1}{x} \] For the pole at \(x=1\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -1\right )^{2}}\) 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 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) 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 {-x +4}{4 \left (x -1\right )^{2} x} \end {alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {1}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {-x +4}{4 \left (x -1\right )^{2} x} \]
| pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
| \(0\) | \(1\) | \(0\) | \(0\) | \(1\) |
| \(1\) | \(2\) | \(0\) | \(\frac {3}{2}\) | \(-{\frac {1}{2}}\) |
| Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
| \(2\) | \(0\) | \(\frac {1}{2}\) | \(\frac {1}{2}\) |
Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) 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 \(\alpha _\infty ^{-} = {\frac {1}{2}}\) then \begin {align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-}+\alpha _{c_2}^{-} \right ) \\ &= {\frac {1}{2}} - \left ( {\frac {1}{2}} \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)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}
The above gives \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right )+\left ( (-)[\sqrt r]_{c_2} + \frac { \alpha _{c_2}^{-} }{x- c_2}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {1}{x}-\frac {1}{2 \left (x -1\right )} + (-) \left ( 0 \right ) \\ &= \frac {1}{x}-\frac {1}{2 \left (x -1\right )}\\ &= \frac {x -2}{2 x \left (x -1\right )} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) 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(x) &= 1\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (\frac {1}{x}-\frac {1}{2 \left (x -1\right )}\right ) \left (0\right ) + \left ( \left (-\frac {1}{x^{2}}+\frac {1}{2 \left (x -1\right )^{2}}\right ) + \left (\frac {1}{x}-\frac {1}{2 \left (x -1\right )}\right )^2 - \left (\frac {-x +4}{4 \left (x -1\right )^{2} x}\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(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \left (\frac {1}{x}-\frac {1}{2 \left (x -1\right )}\right )d x}\\ &= \frac {x}{\sqrt {x -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} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-x}{x^{2}-x} \,dx} \\ &= z_1 e^{\frac {\ln \left (x -1\right )}{2}} \\ &= z_1 \left (\sqrt {x -1}\right ) \\ \end{align*} Which simplifies to \[ y_1 = x \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-x}{x^{2}-x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{\ln \left (x -1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\ln \left (x \right )+\frac {1}{x}\right ) \\ \end{align*} Therefore for kovacic case \(n=1\) the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x\right ) + c_{2} \left (x\left (\ln \left (x \right )+\frac {1}{x}\right )\right ) \\ \end{align*} Attempting to find a solution using case \(n=2\).
Looking at poles of order 1. For the pole at \(x = 0\) of order 1 then \begin {align*} E_c &= \{4\} \end {align*}
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = -\frac {1}{x -1}+\frac {3}{4 \left (x -1\right )^{2}}+\frac {1}{x} \] For the pole at \(x=1\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {3}{4}}\). Hence \begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{-2, 2, 6\} \end {align*}
Since the order of \(r\) at \(\infty \) is 2 then let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) 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 {-x +4}{4 \left (x -1\right )^{2} x} \end {alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {1}{4}}\). Hence \begin {align*} E_\infty &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{2\} \end {align*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) for case 2 of Kovacic algorithm.
| pole \(c\) location | pole order | \(E_c\) |
| \(0\) | \(1\) | \(\{4\}\) |
| \(1\) | \(2\) | \(\{-2, 2, 6\}\) |
| Order of \(r\) at \(\infty \) | \(E_\infty \) |
| \(2\) | \(\{2\}\) |
Using the family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=4,\hspace {3pt} e_2=-2,\hspace {3pt} e_\infty =2 \] Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using \begin {align*} d &= \frac {1}{2} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {1}{2} \left ( 2 - \left (4+\left (-2\right )\right )\right )\\ &= 0 \end {align*}
We now form the following rational function \begin {align*} \theta &= \frac {1}{2} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {1}{2} \left (\frac {4}{\left (x-\left (0\right )\right )}+\frac {-2}{\left (x-\left (1\right )\right )}\right ) \\ &= \frac {2}{x}-\frac {1}{x -1} \end {align*}
Now we search for a monic polynomial \(p(x)\) of degree \(d=0\) such that \[ p'''+3 \theta p'' + \left (3 \theta ^2 + 3 \theta ' - 4 r\right )p' + \left (\theta '' + 3 \theta \theta ' + \theta ^3 - 4 r \theta - 2 r' \right ) p = 0 \tag {1A} \] Since \(d=0\), then letting \[ p = 1\tag {2A} \] Substituting \(p\) and \(\theta \) into Eq. (1A) gives \[ 0 = 0 \] And solving for \(p\) gives \[ p = 1 \] Now that \(p(x)\) is found let \begin {align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {2}{x}-\frac {1}{x -1} \end {align*}
Let \(\omega \) be the solution of \begin {align*} \omega ^2 - \phi \omega + \left ( \frac {1}{2} \phi ' + \frac {1}{2} \phi ^2 - r \right ) &= 0 \end {align*}
Substituting the values for \(\phi \) and \(r\) into the above equation gives \[ w^{2}-\left (\frac {2}{x}-\frac {1}{x -1}\right ) w +\frac {\left (x -2\right )^{2}}{4 x^{2} \left (x -1\right )^{2}} = 0 \] Solving for \(\omega \) gives \begin {align*} \omega &= \frac {x -2}{2 x \left (x -1\right )} \end {align*}
Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {x -2}{2 x \left (x -1\right )}d x}\\ &= \frac {x}{\sqrt {x -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} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-x}{x^{2}-x} \,dx} \\ &= z_1 e^{\frac {\ln \left (x -1\right )}{2}} \\ &= z_1 \left (\sqrt {x -1}\right ) \\ \end{align*} Which simplifies to \[ y_1 = x \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-x}{x^{2}-x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{\ln \left (x -1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\ln \left (x \right )+\frac {1}{x}\right ) \\ \end{align*} Therefore for kovacic case \(n=2\) the solution is \begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x\right ) + c_{2} \left (x\left (\ln \left (x \right )+\frac {1}{x}\right )\right ) \\ \end{align*} Attempting to find a solution using \(n=4\).
Looking at poles of order 1. For the pole at \(x = 0\) of order 1 then \begin {align*} E_c &= \{12\} \end {align*}
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = -\frac {1}{x -1}+\frac {3}{4 \left (x -1\right )^{2}}+\frac {1}{x} \] For the pole at \(x=1\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b={\frac {3}{4}}\). Hence \begin {align*} E_c &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \end {align*}
Where \(n\) for case \(3\) is \(4,6\) or \(12\). For the current case \(n=4\). Hence the above becomes \begin {align*} E_c &= \{-6, 0, 6, 12, 18\} \end {align*}
Let \begin {align*} E_\infty &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \tag {B1} \end {align*}
Where \(b\) is the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series for \(r\) at \(\infty \) given by \[ r \approx -\frac {1}{4 x^{2}}+\frac {1}{2 x^{3}}+\frac {5}{4 x^{4}}+\frac {2}{x^{5}}+\frac {11}{4 x^{6}}+\frac {7}{2 x^{7}} + \cdots \] The above shows that \[ b = -{\frac {1}{4}} \] The value of \(n\) in eq. (B1) for case \(3\) is \(4,6\) or \(2\).For the current case \(n=4\). eq. (B1) simplifies to the following, after removing any duplicate and non integer entries in the set. \begin {align*} E_\infty &= \{6\} \end {align*}
The following table summarizes the results found so far for poles and for the order of \(r\) at \(\infty \) for case 3 of Kovacic algorithm using \(n=4\).
| pole \(c\) location | pole order | set \(\{E_c\}\) |
| \(0\) | \(1\) | \(\{12\}\) |
| \(1\) | \(2\) | \(\{-6, 0, 6, 12, 18\}\) |
| Order of \(r\) at \(\infty \) | set \(\{E_\infty \}\) |
| \(2\) | \(\{6\}\) |
Now that \(E_c\) sets for all poles are found and \(E_\infty \) set is found, the next step is to determine a non negative integer \(d\) using the following \begin {align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right ) \end {align*}
Where in the above \(e_c\) is a distinct element from each corresponding \(E_c\). This means all possible tuples \(\{e_{c_1},e_{c_2},\dots ,e_{c_n}\}\) are tried in the sum above, where \(e_{c_i}\) is one element of each \(E_c\) found earlier. Using the following family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=12,\hspace {3pt} e_2=-6,\hspace {3pt} e_\infty =6 \] Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using \begin {align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {4}{12} \left ( 6 - \left (12+\left (-6\right )\right )\right )\\ &= 0 \end {align*}
The following rational function is \begin {align*} \theta &= \frac {n}{12} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {4}{12} \left (\frac {12}{\left (x-\left (0\right )\right )}+\frac {-6}{\left (x-\left (1\right )\right )}\right ) \\ &= \frac {2 x -4}{x \left (x -1\right )} \end {align*}
And \begin {align*} S &= \prod _{c\in \Gamma } (x-c) \\ &= x \left (x -1\right ) \end {align*}
The polynomial \(p(x)\) is now determined. Since the degree of the polynomial is \(d=0\), then let \[ p(x) = 1 \] The following set of equations are set up in order to determine the coefficients \(a_i\) (if any) of the above polynomial \begin {align*} P_n &= - p(x) \\ &= - 1 \\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \qquad i=n,n-1,\dots , 0 \tag {1A} \end {align*}
The coefficients \(a_i\) are solved for from \[ P_{-1} = 0 \tag {2A} \] By using method of undetermined coefficients. Carrying the above computation in eq. (1A) gives the following sequence of polynomials \(P_i\) (noting that \(n=4\) and \(r=\frac {-x +4}{4 \left (x -1\right )^{2} x}\)). \begin{align*} P_{4} &= - p &= -1 \\ P_{3} &= 2 x -4 \\ P_{2} &= -3 \left (x -2\right )^{2} \\ P_{1} &= 3 \left (x -2\right )^{3} \\ P_{0} &= -\frac {3 \left (x -2\right )^{4}}{2} \\ P_{-1} &= 0 \\ \end{align*} Because \(P_{-1} = 0\) then \(z=e^{\int \omega }\) is a solution. \(\omega \) is found by finding a solution to the equation generated by the following sum \begin {align*} \sum _{i=0}^{n} S^i \frac {P_i}{(n-i)!} \omega ^i &= 0 \\ \sum _{i=0}^{4} S^i \frac {P_i}{(4-i)!} \omega ^i &= 0 \end {align*}
Where the \(P_i\) are the polynomials found earlier. Computing the above sum gives \begin{equation} \tag{3A} {\frac {1}{16}}\left (-\left (2 \omega \,x^{2}-2 x \omega -x +2\right )^{4}\right ) =0 \end{equation} The solution \(\omega \) of eq. 3A is found as \[ \omega =\frac {1}{2 x \left (x -1\right )}\left (x -2\right ) \] This \(\omega \) is used to find a solution to \(z''=r z\). \[ z_1(x) = e^{ \int \omega \,dx}\tag {5A} \] Doing the integration gives in eq. (4A) gives \begin{align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {x -2}{2 x \left (x -1\right )}d x} \\ &= \frac {x}{\sqrt {x -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} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-x}{x^{2}-x} \,dx} \\ &= z_1 e^{\frac {\ln \left (x -1\right )}{2}} \\ &= z_1 \left (\sqrt {x -1}\right ) \\ \end{align*} Which simplifies to \[ y_1 = x \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-x}{x^{2}-x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{\ln \left (x -1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\ln \left (x \right )+\frac {1}{x}\right ) \\ \end{align*} Therefore for kovacic case \(n=4\) the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x\right ) + c_{2} \left (x\left (\ln \left (x \right )+\frac {1}{x}\right )\right ) \\ \end{align*} Attempting to find a solution using \(n=6\).
Looking at poles of order 1. For the pole at \(x = 0\) of order 1 then \begin {align*} E_c &= \{12\} \end {align*}
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = -\frac {1}{x -1}+\frac {3}{4 \left (x -1\right )^{2}}+\frac {1}{x} \] For the pole at \(x=1\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b={\frac {3}{4}}\). Hence \begin {align*} E_c &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \end {align*}
Where \(n\) for case \(3\) is \(4,6\) or \(12\). For the current case \(n=6\). Hence the above becomes \begin {align*} E_c &= \{-6, -2, 2, 6, 10, 14, 18\} \end {align*}
Let \begin {align*} E_\infty &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \tag {B1} \end {align*}
Where \(b\) is the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series for \(r\) at \(\infty \) given by \[ r \approx -\frac {1}{4 x^{2}}+\frac {1}{2 x^{3}}+\frac {5}{4 x^{4}}+\frac {2}{x^{5}}+\frac {11}{4 x^{6}}+\frac {7}{2 x^{7}} + \cdots \] The above shows that \[ b = -{\frac {1}{4}} \] The value of \(n\) in eq. (B1) for case \(3\) is \(4,6\) or \(2\).For the current case \(n=6\). eq. (B1) simplifies to the following, after removing any duplicate and non integer entries in the set. \begin {align*} E_\infty &= \{6\} \end {align*}
The following table summarizes the results found so far for poles and for the order of \(r\) at \(\infty \) for case 3 of Kovacic algorithm using \(n=6\).
| pole \(c\) location | pole order | set \(\{E_c\}\) |
| \(0\) | \(1\) | \(\{12\}\) |
| \(1\) | \(2\) | \(\{-6, -2, 2, 6, 10, 14, 18\}\) |
| Order of \(r\) at \(\infty \) | set \(\{E_\infty \}\) |
| \(2\) | \(\{6\}\) |
Now that \(E_c\) sets for all poles are found and \(E_\infty \) set is found, the next step is to determine a non negative integer \(d\) using the following \begin {align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right ) \end {align*}
Where in the above \(e_c\) is a distinct element from each corresponding \(E_c\). This means all possible tuples \(\{e_{c_1},e_{c_2},\dots ,e_{c_n}\}\) are tried in the sum above, where \(e_{c_i}\) is one element of each \(E_c\) found earlier. Using the following family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=12,\hspace {3pt} e_2=-6,\hspace {3pt} e_\infty =6 \] Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using \begin {align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {6}{12} \left ( 6 - \left (12+\left (-6\right )\right )\right )\\ &= 0 \end {align*}
The following rational function is \begin {align*} \theta &= \frac {n}{12} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {6}{12} \left (\frac {12}{\left (x-\left (0\right )\right )}+\frac {-6}{\left (x-\left (1\right )\right )}\right ) \\ &= \frac {3 x -6}{x \left (x -1\right )} \end {align*}
And \begin {align*} S &= \prod _{c\in \Gamma } (x-c) \\ &= x \left (x -1\right ) \end {align*}
The polynomial \(p(x)\) is now determined. Since the degree of the polynomial is \(d=0\), then let \[ p(x) = 1 \] The following set of equations are set up in order to determine the coefficients \(a_i\) (if any) of the above polynomial \begin {align*} P_n &= - p(x) \\ &= - 1 \\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \qquad i=n,n-1,\dots , 0 \tag {1A} \end {align*}
The coefficients \(a_i\) are solved for from \[ P_{-1} = 0 \tag {2A} \] By using method of undetermined coefficients. Carrying the above computation in eq. (1A) gives the following sequence of polynomials \(P_i\) (noting that \(n=6\) and \(r=\frac {-x +4}{4 \left (x -1\right )^{2} x}\)). \begin{align*} P_{6} &= - p &= -1 \\ P_{5} &= 3 x -6 \\ P_{4} &= -\frac {15 \left (x -2\right )^{2}}{2} \\ P_{3} &= 15 \left (x -2\right )^{3} \\ P_{2} &= -\frac {45 \left (x -2\right )^{4}}{2} \\ P_{1} &= \frac {45 \left (x -2\right )^{5}}{2} \\ P_{0} &= -\frac {45 \left (x -2\right )^{6}}{4} \\ P_{-1} &= 0 \\ \end{align*} Because \(P_{-1} = 0\) then \(z=e^{\int \omega }\) is a solution. \(\omega \) is found by finding a solution to the equation generated by the following sum \begin {align*} \sum _{i=0}^{n} S^i \frac {P_i}{(n-i)!} \omega ^i &= 0 \\ \sum _{i=0}^{6} S^i \frac {P_i}{(6-i)!} \omega ^i &= 0 \end {align*}
Where the \(P_i\) are the polynomials found earlier. Computing the above sum gives \begin{equation} \tag{3A} {\frac {1}{64}}\left (-\left (2 \omega \,x^{2}-2 x \omega -x +2\right )^{6}\right ) =0 \end{equation} The solution \(\omega \) of eq. 3A is found as \[ \omega =\frac {1}{2 x \left (x -1\right )}\left (x -2\right ) \] This \(\omega \) is used to find a solution to \(z''=r z\). \[ z_1(x) = e^{ \int \omega \,dx}\tag {5A} \] Doing the integration gives in eq. (4A) gives \begin{align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {x -2}{2 x \left (x -1\right )}d x} \\ &= \frac {x}{\sqrt {x -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} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-x}{x^{2}-x} \,dx} \\ &= z_1 e^{\frac {\ln \left (x -1\right )}{2}} \\ &= z_1 \left (\sqrt {x -1}\right ) \\ \end{align*} Which simplifies to \[ y_1 = x \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-x}{x^{2}-x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{\ln \left (x -1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\ln \left (x \right )+\frac {1}{x}\right ) \\ \end{align*} Therefore for kovacic case \(n=6\) the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x\right ) + c_{2} \left (x\left (\ln \left (x \right )+\frac {1}{x}\right )\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x\right ) + c_{2} \left (x\left (\ln \left (x \right )+\frac {1}{x}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +c_{2} \left (\ln \left (x \right ) x +1\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +c_{2} \left (\ln \left (x \right ) x +1\right ) \] Verified OK.
Maple trace Kovacic algorithm successful
`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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 16
dsolve((x^2-x)*diff(y(x), x$2)-x*diff(y(x), x)+y(x) = 0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} x +c_{2} \left (x \ln \left (x \right )+1\right ) \]
✓ Solution by Mathematica
Time used: 0.029 (sec). Leaf size: 20
DSolve[(x^2-x)*y''[x]-x*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x-c_2 (x \log (x)+1) \]