Internal problem ID [9556]
Internal file name [OUTPUT/8496_Monday_June_06_2022_03_14_09_AM_33568322/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1227.
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]]
\[ \boxed {\left (x^{2}+1\right ) y^{\prime \prime }-2 y^{\prime } x +2 y=0} \]
In normal form the ode \begin {align*} \left (x^{2}+1\right ) y^{\prime \prime }-2 y^{\prime } x +2 y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=-\frac {2 x}{x^{2}+1}\\ q \left (x \right )&=\frac {2}{x^{2}+1} \end {align*}
Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\). \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (x \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{x^{2}}-\frac {2 n}{x^{2}+1}+\frac {2}{x^{2}+1}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-\frac {2 x}{x^{2}+1}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {2 v^{\prime }\left (x \right )}{x \left (x^{2}+1\right )}&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (x \right )+\frac {2 u \left (x \right )}{x \left (x^{2}+1\right )} = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {2 u}{x \left (x^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x \left (x^{2}+1\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2}{x \left (x^{2}+1\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2}{x \left (x^{2}+1\right )} \,d x}\\ \ln \left (u \right )&=\ln \left (x^{2}+1\right )-2 \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{\ln \left (x^{2}+1\right )-2 \ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{\ln \left (x^{2}+1\right )-2 \ln \left (x \right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = c_{1} \left (\frac {1}{x^{2}}+1\right ) \] Now that \(u \left (x \right )\) is known, then \begin {align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_{2}\\ &= c_{1} \left (x -\frac {1}{x}\right )+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (c_{1} \left (x -\frac {1}{x}\right )+c_{2} \right ) x\\ &= c_{1} x^{2}+c_{2} x -c_{1}\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{1} \left (x -\frac {1}{x}\right )+c_{2} \right ) x \\ \end{align*}
Verification of solutions
\[ y = \left (c_{1} \left (x -\frac {1}{x}\right )+c_{2} \right ) x \] Verified OK.
Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= x^{2}+1\\ B &= -2 x\\ C &= 2\\ F &= 0 \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (x^{2}+1\right ) \left (0\right ) + \left (-2 x\right ) \left (-2\right ) + \left (2\right ) \left (-2 x\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} -2 x \left (x^{2}+1\right ) v'' +\left ( -4\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} \left (-2 x^{3}-2 x \right ) u^{\prime }\left (x \right )-4 u \left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {2 u}{x \left (x^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x \left (x^{2}+1\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2}{x \left (x^{2}+1\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2}{x \left (x^{2}+1\right )} \,d x}\\ \ln \left (u \right )&=\ln \left (x^{2}+1\right )-2 \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{\ln \left (x^{2}+1\right )-2 \ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{\ln \left (x^{2}+1\right )-2 \ln \left (x \right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = c_{1} \left (\frac {1}{x^{2}}+1\right ) \] The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} \left (\frac {1}{x^{2}}+1\right ) \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{1} \left (x^{2}+1\right )}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= c_{1} \left (x -\frac {1}{x}\right )+c_{2} \end {align*}
Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (-2 x\right ) \left (c_{1} \left (x -\frac {1}{x}\right )+c_{2}\right ) \\ &= -2 c_{1} x^{2}-2 c_{2} x +2 c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -2 c_{1} x^{2}-2 c_{2} x +2 c_{1} \\ \end{align*}
Verification of solutions
\[ y = -2 c_{1} x^{2}-2 c_{2} x +2 c_{1} \] Verified OK.
Writing the ode as \begin {align*} \left (x^{2}+1\right ) y^{\prime \prime }-2 y^{\prime } x +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 &= x^{2}+1 \\ B &= -2 x\tag {3} \\ C &= 2 \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 {-3}{\left (x^{2}+1\right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= -3\\ t &= \left (x^{2}+1\right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( -\frac {3}{\left (x^{2}+1\right )^{2}}\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) \\ &= 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 \(t=\left (x^{2}+1\right )^{2}\). There is a pole at \(x=i\) of order \(2\). There is a pole at \(x=-i\) 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 \[ r = \frac {3}{4 \left (x -i\right )^{2}}+\frac {3}{4 \left (x +i\right )^{2}}+\frac {3 i}{4 \left (x -i\right )}-\frac {3 i}{4 \left (x +i\right )} \] For the pole at \(x=i\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -i\right )^{2}}\) 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 \(x=-i\) let \(b\) be the coefficient of \(\frac {1}{ \left (x +i\right )^{2}}\) 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 \[ r=-\frac {3}{\left (x^{2}+1\right )^{2}} \]
| 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)}}{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}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )} + (-) \left ( 0 \right ) \\ &= -\frac {1}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\\ &= \frac {x -2 i}{x^{2}+1} \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}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\right ) \left (0\right ) + \left ( \left (\frac {1}{2 \left (x -i\right )^{2}}-\frac {3}{2 \left (x +i\right )^{2}}\right ) + \left (-\frac {1}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\right )^2 - \left (-\frac {3}{\left (x^{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(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \left (-\frac {1}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\right )d x}\\ &= \frac {\left (x^{2}+1\right )^{\frac {3}{2}}}{\left (i x +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} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-2 x}{x^{2}+1} \,dx} \\ &= z_1 e^{\frac {\ln \left (x^{2}+1\right )}{2}} \\ &= z_1 \left (\sqrt {x^{2}+1}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {\left (x^{2}+1\right )^{2}}{\left (i x +1\right )^{2}} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-2 x}{x^{2}+1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{\ln \left (x^{2}+1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (-\frac {x}{\left (x +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 (x^{2}+1\right )^{2}}{\left (i x +1\right )^{2}}\right ) + c_{2} \left (\frac {\left (x^{2}+1\right )^{2}}{\left (i x +1\right )^{2}}\left (-\frac {x}{\left (x +i\right )^{2}}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (x^{2}+1\right )^{2}}{\left (i x +1\right )^{2}}+\frac {c_{2} \left (x^{2}+1\right )^{2} x}{\left (-x +i\right )^{2} \left (x +i\right )^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (x^{2}+1\right )^{2}}{\left (i x +1\right )^{2}}+\frac {c_{2} \left (x^{2}+1\right )^{2} x}{\left (-x +i\right )^{2} \left (x +i\right )^{2}} \] 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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 16
dsolve((x^2+1)*diff(diff(y(x),x),x)-2*x*diff(y(x),x)+2*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{2} x^{2}+c_{1} x -c_{2} \]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 21
DSolve[2*y[x] - 2*x*y'[x] + (1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 x-c_1 (x-i)^2 \]