Internal problem ID [9704]
Internal file name [OUTPUT/8646_Monday_June_06_2022_04_36_40_AM_36347535/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1377.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_bessel_ode"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {y^{\prime \prime }+\frac {b^{2} y}{\left (a^{2}+x^{2}\right )^{2}}=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+\frac {b^{2} y}{x^{2}} = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end {align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following \begin {align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end {align*}
With the standard solution \begin {align*} y&=x^{\alpha } \left (c_{1} \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_{2} \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end {align*}
Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives \begin {align*} \alpha &= {\frac {1}{2}}\\ \beta &= b\\ n &= {\frac {1}{2}}\\ \gamma &= -1 \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = \frac {c_{1} \sqrt {x}\, \sqrt {2}\, \sin \left (\frac {b}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {b}{x}}}-\frac {c_{2} \sqrt {x}\, \sqrt {2}\, \cos \left (\frac {b}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {b}{x}}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \sqrt {x}\, \sqrt {2}\, \sin \left (\frac {b}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {b}{x}}}-\frac {c_{2} \sqrt {x}\, \sqrt {2}\, \cos \left (\frac {b}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {b}{x}}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \sqrt {x}\, \sqrt {2}\, \sin \left (\frac {b}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {b}{x}}}-\frac {c_{2} \sqrt {x}\, \sqrt {2}\, \cos \left (\frac {b}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {b}{x}}} \] Verified OK.
Writing the ode as \begin {align*} y^{\prime \prime } \left (a^{2}+x^{2}\right )^{2}+b^{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 &= \left (a^{2}+x^{2}\right )^{2} \\ B &= 0\tag {3} \\ C &= b^{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 {-b^{2}}{\left (a^{2}+x^{2}\right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= -b^{2}\\ t &= \left (a^{2}+x^{2}\right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( -\frac {b^{2}}{\left (a^{2}+x^{2}\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 \(t=\left (a^{2}+x^{2}\right )^{2}\). There is a pole at \(x=i a\) of order \(2\). There is a pole at \(x=-i a\) 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 \(n=1\).
Unable to find solution using case one
Attempting to find a solution using case \(n=2\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {b^{2}}{4 a^{2} \left (x -\sqrt {-a^{2}}\right )^{2}}+\frac {b^{2}}{4 a^{2} \left (x +\sqrt {-a^{2}}\right )^{2}}+\frac {b^{2}}{4 \left (-a^{2}\right )^{\frac {3}{2}} \left (x -\sqrt {-a^{2}}\right )}-\frac {b^{2}}{4 \left (-a^{2}\right )^{\frac {3}{2}} \left (x +\sqrt {-a^{2}}\right )} \] For the pole at \(x=i a\) let \(b\) be the coefficient of \(\frac {1}{ \left (-i a +x \right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=0\). Hence \begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{0, 2, 4\} \end {align*}
For the pole at \(x=-i a\) let \(b\) be the coefficient of \(\frac {1}{ \left (i a +x \right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=0\). Hence \begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{0, 2, 4\} \end {align*}
Now since the order of \(r\) at \(\infty \) is \(4 > 2\) then \begin {align*} E_\infty = \{0,2,4\} \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 | \(E_c\) |
| \(i a\) | \(2\) | \(\{0, 2, 4\}\) |
| \(-i a\) | \(2\) | \(\{0, 2, 4\}\) |
| Order of \(r\) at \(\infty \) | \(E_\infty \) |
| \(4\) | \(\{0, 2, 4\}\) |
Using the family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=2,\hspace {3pt} e_2=2,\hspace {3pt} e_\infty =4 \] 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 ( 4 - \left (2+\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 {2}{\left (x-\left (i a\right )\right )}+\frac {2}{\left (x-\left (-i a\right )\right )}\right ) \\ &= \frac {1}{-i a +x}+\frac {1}{i a +x} \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 {1}{-i a +x}+\frac {1}{i a +x} \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 \[ w^{2}-\left (\frac {1}{-i a +x}+\frac {1}{i a +x}\right ) w +\frac {a^{2}+b^{2}+x^{2}}{\left (a^{2}+x^{2}\right )^{2}} = 0 \] Solving for \(\omega \) gives \begin {align*} \omega &= \frac {x +\sqrt {-a^{2}-b^{2}}}{a^{2}+x^{2}} \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 +\sqrt {-a^{2}-b^{2}}}{a^{2}+x^{2}}d x}\\ &= \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}} \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 \\ &= \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}} \\ \end{align*} Which simplifies to \[ y_1 = \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}} \] 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 \\ &= \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}\int \frac {1}{\left (a^{2}+x^{2}\right ) {\mathrm e}^{\frac {2 \sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}} \,dx \\ &= \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}\left (-\frac {{\mathrm e}^{-\frac {2 \sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}}{2 \sqrt {-a^{2}-b^{2}}}\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}\right ) + c_{2} \left (\sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}\left (-\frac {{\mathrm e}^{-\frac {2 \sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}}{2 \sqrt {-a^{2}-b^{2}}}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}-\frac {c_{2} \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}}{2 \sqrt {-a^{2}-b^{2}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}-\frac {c_{2} \sqrt {a^{2}+x^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}-b^{2}}\, \arctan \left (\frac {x}{a}\right )}{a}}}{2 \sqrt {-a^{2}-b^{2}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (a^{2}+x^{2}\right )^{2}+b^{2} y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {b^{2} y}{\left (a^{2}+x^{2}\right )^{2}} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\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}{d x}y^{\prime }+\frac {b^{2} y}{\left (a^{2}+x^{2}\right )^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (a^{2}+x^{2}\right )^{2}+b^{2} y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+\left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & \left (\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right ) \left (a^{2}+x^{2}\right )^{2}+b^{2} y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )-\frac {d}{d t}y \left (t \right )\right ) \left (a^{2}+x^{2}\right )^{2}}{x^{2}}+b^{2} y \left (t \right )=0 \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}y \left (t \right )=-\frac {b^{2} x^{2} y \left (t \right )}{\left (a^{2}+x^{2}\right )^{2}}+\frac {d}{d t}y \left (t \right ) \\ \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}{d t}\frac {d}{d t}y \left (t \right )+\frac {b^{2} x^{2} y \left (t \right )}{\left (a^{2}+x^{2}\right )^{2}}-\frac {d}{d t}y \left (t \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+\frac {b^{2} x^{2}}{\left (a^{2}+x^{2}\right )^{2}}-r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {a^{4} r^{2}+2 a^{2} r^{2} x^{2}+r^{2} x^{4}-a^{4} r -2 a^{2} r \,x^{2}-r \,x^{4}+b^{2} x^{2}}{\left (a^{2}+x^{2}\right )^{2}}=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\frac {\frac {a^{2}}{2}+\frac {x^{2}}{2}+\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}}{a^{2}+x^{2}}, \frac {\frac {a^{2}}{2}+\frac {x^{2}}{2}-\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}}{a^{2}+x^{2}}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{\frac {\left (\frac {a^{2}}{2}+\frac {x^{2}}{2}+\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}\right ) t}{a^{2}+x^{2}}} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\frac {\left (\frac {a^{2}}{2}+\frac {x^{2}}{2}-\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}\right ) t}{a^{2}+x^{2}}} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} {\mathrm e}^{\frac {\left (\frac {a^{2}}{2}+\frac {x^{2}}{2}+\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}\right ) t}{a^{2}+x^{2}}}+c_{2} {\mathrm e}^{\frac {\left (\frac {a^{2}}{2}+\frac {x^{2}}{2}-\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}\right ) t}{a^{2}+x^{2}}} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=c_{1} {\mathrm e}^{\frac {\left (\frac {a^{2}}{2}+\frac {x^{2}}{2}+\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}\right ) \ln \left (x \right )}{a^{2}+x^{2}}}+c_{2} {\mathrm e}^{\frac {\left (\frac {a^{2}}{2}+\frac {x^{2}}{2}-\frac {\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2}\right ) \ln \left (x \right )}{a^{2}+x^{2}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=c_{1} x^{\frac {a^{2}+x^{2}+\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2 a^{2}+2 x^{2}}}+c_{2} x^{\frac {a^{2}+x^{2}-\sqrt {a^{4}+2 a^{2} x^{2}-4 b^{2} x^{2}+x^{4}}}{2 a^{2}+2 x^{2}}} \end {array} \]
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 Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 83
dsolve(diff(diff(y(x),x),x) = -b^2/(a^2+x^2)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (\left (\frac {i x -a}{i x +a}\right )^{\frac {\sqrt {a^{2}+b^{2}}}{2 a}} c_{1} +\left (\frac {i x -a}{i x +a}\right )^{-\frac {\sqrt {a^{2}+b^{2}}}{2 a}} c_{2} \right ) \sqrt {a^{2}+x^{2}} \]
✓ Solution by Mathematica
Time used: 0.841 (sec). Leaf size: 97
DSolve[y''[x] == -((b^2*y[x])/(a^2 + x^2)^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \sqrt {a^2+x^2} e^{-i \sqrt {\frac {b^2}{a^2}+1} \arctan \left (\frac {a}{x}\right )} \left (\frac {i c_2 e^{2 i \sqrt {\frac {b^2}{a^2}+1} \arctan \left (\frac {a}{x}\right )}}{a \sqrt {\frac {b^2}{a^2}+1}}+2 c_1\right ) \]