Internal problem ID [12189]
Internal file name [OUTPUT/10842_Thursday_September_21_2023_05_47_55_AM_26792915/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 36.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }-6 y=1} \]
Writing the ode as \begin {align*} \left (x^{2}-1\right ) y^{\prime \prime }-6 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}-1 \\ B &= 0\tag {3} \\ C &= -6 \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 {6}{x^{2}-1}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 6\\ t &= x^{2}-1 \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {6}{x^{2}-1}\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) \\ &= 2 - 0 \\ &= 2 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=x^{2}-1\). There is a pole at \(x=1\) of order \(1\). There is a pole at \(x=-1\) 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 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, 4, 6, 12] \end {align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 1. For the pole at \(x = 1\) of order 1 then \begin {align*} [\sqrt r]_c &= 0 \\ \alpha _c^+ &= 1 \\ \alpha _c^- &= 1 \end {align*}
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 {6}{x^{2}-1} \end {alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=6\). Hence \begin {alignat*} {2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= 3\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -2 \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {6}{x^{2}-1} \]
| pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
| \(1\) | \(1\) | \(0\) | \(0\) | \(1\) |
| Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
| \(2\) | \(0\) | \(3\) | \(-2\) |
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 \(\alpha _\infty ^{+} = 3\) then \begin {align*} d &= \alpha _\infty ^{+} - \left ( \alpha _{c_1}^{-} \right ) \\ &= 3 - \left ( 1 \right ) \\ &= 2 \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*}
Substituting the above values in the above results in \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (+) [\sqrt r]_\infty \\ &= \frac {1}{x -1} + \left ( 0 \right ) \\ &= \frac {1}{x -1}\\ &= \frac {1}{x -1} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=2\) 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) &= x^{2}+a_{1} x +a_{0}\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (2\right ) + 2 \left (\frac {1}{x -1}\right ) \left (2 x +a_{1}\right ) + \left ( \left (-\frac {1}{\left (x -1\right )^{2}}\right ) + \left (\frac {1}{x -1}\right )^2 - \left (\frac {6}{x^{2}-1}\right ) \right ) &= 0\\ \frac {\left (-4 a_{1}+4\right ) x -6 a_{0}+2 a_{1}-2}{x^{2}-1} = 0 \end {align*}
Solving for the coefficients \(a_i\) in the above using method of undetermined coefficients gives \[ \{a_{0} = 0, a_{1} = 1\} \] Substituting these coefficients in \(p(x)\) in eq. (2A) results in \begin {align*} p(x) &= x^{2}+x \end {align*}
Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= p e^{ \int \omega \,dx}\\ & = \left (x^{2}+x\right ) {\mathrm e}^{\int \frac {1}{x -1}d x}\\ & = \left (x^{2}+x\right ) x -1\\ & = x^{3}-x \end {align*}
The first solution to the original ode in \(y\) is found from \[ y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \]
Since \(B=0\) then the above reduces to \begin{align*} y_1 &= z_1 \\ &= x^{3}-x \\ \end{align*} Which simplifies to \[ y_1 = x^{3}-x \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Since \(B=0\) then the above becomes \begin{align*} y_2 &= y_1 \int \frac {1}{y_1^2} \,dx \\ &= x^{3}-x\int \frac {1}{\left (x^{3}-x \right )^{2}} \,dx \\ &= x^{3}-x\left (-\frac {1}{x}-\frac {1}{4 x -4}-\frac {3 \ln \left (x -1\right )}{4}-\frac {1}{4 x +4}+\frac {3 \ln \left (x +1\right )}{4}\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x^{3}-x\right ) + c_{2} \left (x^{3}-x\left (-\frac {1}{x}-\frac {1}{4 x -4}-\frac {3 \ln \left (x -1\right )}{4}-\frac {1}{4 x +4}+\frac {3 \ln \left (x +1\right )}{4}\right )\right ) \\ \end{align*} This is second order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the nonhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ \left (x^{2}-1\right ) y^{\prime \prime }-6 y = 0 \] The homogeneous solution is found using the Kovacic algorithm which results in \[ y_h = c_{1} \left (x^{3}-x \right )+c_{2} \left (\frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1\right ) \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= x^{3}-x \\ y_2 &= \frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1 \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} x^{3}-x & \frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1 \\ \frac {d}{dx}\left (x^{3}-x\right ) & \frac {d}{dx}\left (\frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x^{3}-x & \frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1 \\ 3 x^{2}-1 & \frac {\left (-9 x^{2}+3\right ) \ln \left (x -1\right )}{4}+\frac {-3 x^{3}+3 x}{4 x -4}+\frac {\left (9 x^{2}-3\right ) \ln \left (x +1\right )}{4}+\frac {3 x^{3}-3 x}{4 x +4}-3 x \end {vmatrix} \] Therefore \[ W = \left (x^{3}-x\right )\left (\frac {\left (-9 x^{2}+3\right ) \ln \left (x -1\right )}{4}+\frac {-3 x^{3}+3 x}{4 x -4}+\frac {\left (9 x^{2}-3\right ) \ln \left (x +1\right )}{4}+\frac {3 x^{3}-3 x}{4 x +4}-3 x\right ) - \left (\frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1\right )\left (3 x^{2}-1\right ) \] Which simplifies to \[ W = 1 \] Which simplifies to \[ W = 1 \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1}{x^{2}-1}\,dx \] Which simplifies to \[ u_1 = - \int \frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )+\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )-6 x^{2}+4}{4 x^{2}-4}d x \] Hence \[ u_1 = \frac {3 x}{4}+\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}+\frac {3 \left (x -1\right )^{2} \ln \left (x -1\right )}{8}+\frac {3 \ln \left (x -1\right ) \left (x -1\right )}{4}-\frac {3 \left (x +1\right )^{2} \ln \left (x +1\right )}{8}+\frac {3 \ln \left (x +1\right ) \left (x +1\right )}{4} \] And Eq. (3) becomes \[ u_2 = \int \frac {x^{3}-x}{x^{2}-1}\,dx \] Which simplifies to \[ u_2 = \int x d x \] Hence \[ u_2 = \frac {x^{2}}{2} \] Which simplifies to \begin{align*} u_1 &= \frac {3 \ln \left (x -1\right ) x^{2}}{8}-\frac {3 \ln \left (x +1\right ) x^{2}}{8}-\frac {\ln \left (x -1\right )}{8}+\frac {\ln \left (x +1\right )}{8}+\frac {3 x}{4} \\ u_2 &= \frac {x^{2}}{2} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \left (\frac {3 \ln \left (x -1\right ) x^{2}}{8}-\frac {3 \ln \left (x +1\right ) x^{2}}{8}-\frac {\ln \left (x -1\right )}{8}+\frac {\ln \left (x +1\right )}{8}+\frac {3 x}{4}\right ) \left (x^{3}-x \right )+\frac {x^{2} \left (\frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1\right )}{2} \] Which simplifies to \[ y_p(x) = -\frac {\left (\left (x^{2}-1\right ) \ln \left (x -1\right )-\ln \left (x +1\right ) x^{2}+2 x +\ln \left (x +1\right )\right ) x}{8} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} \left (x^{3}-x \right )+c_{2} \left (\frac {\left (-3 x^{3}+3 x \right ) \ln \left (x -1\right )}{4}+\frac {\left (3 x^{3}-3 x \right ) \ln \left (x +1\right )}{4}-\frac {3 x^{2}}{2}+1\right )\right ) + \left (-\frac {\left (\left (x^{2}-1\right ) \ln \left (x -1\right )-\ln \left (x +1\right ) x^{2}+2 x +\ln \left (x +1\right )\right ) x}{8}\right ) \\ \end{align*} Which simplifies to \[ y = \frac {3 c_{2} \left (-x^{3}+x \right ) \ln \left (x -1\right )}{4}+\frac {3 \left (x^{3}-x \right ) c_{2} \ln \left (x +1\right )}{4}+c_{1} x^{3}-\frac {3 c_{2} x^{2}}{2}-c_{1} x +c_{2} -\frac {\left (\left (x^{2}-1\right ) \ln \left (x -1\right )-\ln \left (x +1\right ) x^{2}+2 x +\ln \left (x +1\right )\right ) x}{8} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {3 c_{2} \left (-x^{3}+x \right ) \ln \left (x -1\right )}{4}+\frac {3 \left (x^{3}-x \right ) c_{2} \ln \left (x +1\right )}{4}+c_{1} x^{3}-\frac {3 c_{2} x^{2}}{2}-c_{1} x +c_{2} -\frac {\left (\left (x^{2}-1\right ) \ln \left (x -1\right )-\ln \left (x +1\right ) x^{2}+2 x +\ln \left (x +1\right )\right ) x}{8} \\ \end{align*}
Verification of solutions
\[ y = \frac {3 c_{2} \left (-x^{3}+x \right ) \ln \left (x -1\right )}{4}+\frac {3 \left (x^{3}-x \right ) c_{2} \ln \left (x +1\right )}{4}+c_{1} x^{3}-\frac {3 c_{2} x^{2}}{2}-c_{1} x +c_{2} -\frac {\left (\left (x^{2}-1\right ) \ln \left (x -1\right )-\ln \left (x +1\right ) x^{2}+2 x +\ln \left (x +1\right )\right ) x}{8} \] Verified OK.
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] -> Try solving first the homogeneous part of the ODE 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 <- solving first the homogeneous part of the ODE successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 52
dsolve((x^2-1)*diff(y(x),x$2)-6*y(x)=1,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {1}{6}+\frac {3 \left (x^{3}-x \right ) c_{1} \ln \left (-1+x \right )}{4}+\frac {3 c_{1} \left (-x^{3}+x \right ) \ln \left (1+x \right )}{4}+c_{2} x^{3}+\frac {3 c_{1} x^{2}}{2}-c_{2} x -c_{1} \]
✓ Solution by Mathematica
Time used: 0.086 (sec). Leaf size: 67
DSolve[(x^2-1)*y''[x]-6*y[x]==1,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{12} \left (-9 c_2 x \left (x^2-1\right ) \log (1-x)+9 c_2 x \left (x^2-1\right ) \log (x+1)+2 \left (6 c_1 x^3-9 c_2 x^2-6 c_1 x-1+6 c_2\right )\right ) \]