Internal problem ID [9625]
Internal file name [OUTPUT/8566_Monday_June_06_2022_03_53_09_AM_94473075/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1297.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {\left (a \,x^{2}+1\right ) y^{\prime \prime }+a x y^{\prime }+b y=0} \]
In normal form the ode \begin {align*} \left (a \,x^{2}+1\right ) y^{\prime \prime }+a x y^{\prime }+b 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 {a x}{a \,x^{2}+1}\\ q \left (x \right )&=\frac {b}{a \,x^{2}+1} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \frac {a x}{a \,x^{2}+1}d x \right )}d x\\ &= \int e^{-\frac {\ln \left (a \,x^{2}+1\right )}{2}} \,dx\\ &= \int \frac {1}{\sqrt {a \,x^{2}+1}}d x\\ &= \frac {\ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{\sqrt {a}}\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {b}{a \,x^{2}+1}}{\frac {1}{a \,x^{2}+1}}\\ &= b\tag {7} \end {align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+b y \left (\tau \right )&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=b\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+b \,{\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}+b = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=b\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (b\right )}\\ &= \pm \sqrt {-b} \end {align*}
Hence \begin{align*} \lambda _1 &= + \sqrt {-b} \\ \lambda _2 &= - \sqrt {-b} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= \sqrt {-b} \\ \lambda _2 &= -\sqrt {-b} \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} y \left (\tau \right ) &= c_{1} e^{\lambda _1 \tau } + c_{2} e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_{1} e^{\left (\sqrt {-b}\right )\tau } +c_{2} e^{\left (-\sqrt {-b}\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\sqrt {-b}\, \tau }+c_{2} {\mathrm e}^{-\sqrt {-b}\, \tau } \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{\frac {\sqrt {-b}}{\sqrt {a}}}+c_{2} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{-\frac {\sqrt {-b}}{\sqrt {a}}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{\frac {\sqrt {-b}}{\sqrt {a}}}+c_{2} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{-\frac {\sqrt {-b}}{\sqrt {a}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{\frac {\sqrt {-b}}{\sqrt {a}}}+c_{2} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{-\frac {\sqrt {-b}}{\sqrt {a}}} \] Verified OK.
In normal form the ode \begin {align*} \left (a \,x^{2}+1\right ) y^{\prime \prime }+a x y^{\prime }+b 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 {a x}{a \,x^{2}+1}\\ q \left (x \right )&=\frac {b}{a \,x^{2}+1} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) results \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {\frac {b}{a \,x^{2}+1}}}{c}\tag {6} \\ \tau '' &= -\frac {b a x}{c \sqrt {\frac {b}{a \,x^{2}+1}}\, \left (a \,x^{2}+1\right )^{2}} \end {align*}
Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {-\frac {b a x}{c \sqrt {\frac {b}{a \,x^{2}+1}}\, \left (a \,x^{2}+1\right )^{2}}+\frac {a x}{a \,x^{2}+1}\frac {\sqrt {\frac {b}{a \,x^{2}+1}}}{c}}{\left (\frac {\sqrt {\frac {b}{a \,x^{2}+1}}}{c}\right )^2} \\ &=0 \end {align*}
Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {\frac {b}{a \,x^{2}+1}}d x}{c}\\ &= \frac {\sqrt {\frac {b}{a \,x^{2}+1}}\, \sqrt {a \,x^{2}+1}\, \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{c \sqrt {a}} \end {align*}
Substituting the above into the solution obtained gives \[ y = c_{1} \cos \left (\frac {\sqrt {b}\, \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{\sqrt {a}}\right )+c_{2} \sin \left (\frac {\sqrt {b}\, \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{\sqrt {a}}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (\frac {\sqrt {b}\, \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{\sqrt {a}}\right )+c_{2} \sin \left (\frac {\sqrt {b}\, \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{\sqrt {a}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \cos \left (\frac {\sqrt {b}\, \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{\sqrt {a}}\right )+c_{2} \sin \left (\frac {\sqrt {b}\, \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )}{\sqrt {a}}\right ) \] Verified OK.
Writing the ode as \begin {align*} \left (a \,x^{2}+1\right ) y^{\prime \prime }+a x y^{\prime }+b 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 &= a \,x^{2}+1 \\ B &= a x\tag {3} \\ C &= b \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^{2} a^{2}-4 b a \,x^{2}+2 a -4 b}{4 \left (a \,x^{2}+1\right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= -x^{2} a^{2}-4 b a \,x^{2}+2 a -4 b\\ t &= 4 \left (a \,x^{2}+1\right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {-x^{2} a^{2}-4 b a \,x^{2}+2 a -4 b}{4 \left (a \,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 - 2 \\ &= 2 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (a \,x^{2}+1\right )^{2}\). There is a pole at \(x=\frac {\sqrt {-a}}{a}\) of order \(2\). There is a pole at \(x=-\frac {\sqrt {-a}}{a}\) of order \(2\). 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 \(n=1\).
Unable to find solution using case one
Attempting to find a solution using case \(n=2\).
Unable to find solution using case two.
Attempting to find a solution using \(n=4\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = -\frac {3}{16 \left (x -\sqrt {-\frac {1}{a}}\right )^{2}}-\frac {3}{16 \left (x +\sqrt {-\frac {1}{a}}\right )^{2}}+\frac {-a +8 b}{16 a^{2} \left (-\frac {1}{a}\right )^{\frac {3}{2}} \left (x -\sqrt {-\frac {1}{a}}\right )}-\frac {-a +8 b}{16 a^{2} \left (-\frac {1}{a}\right )^{\frac {3}{2}} \left (x +\sqrt {-\frac {1}{a}}\right )} \] For the pole at \(x=\frac {\sqrt {-a}}{a}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -\frac {\sqrt {-a}}{a}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b=0\). 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 &= \{0, 3, 6, 9, 12\} \end {align*}
For the pole at \(x=-\frac {\sqrt {-a}}{a}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x +\frac {\sqrt {-a}}{a}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b=0\). 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 &= \{0, 3, 6, 9, 12\} \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 \[ r \approx \frac {-a^{2}-4 b a}{4 a^{2} x^{2}}+\frac {-\frac {-a^{2}-4 b a}{2 a^{3}}+\frac {2 a -4 b}{4 a^{2}}}{x^{4}}+\frac {\frac {-\frac {3}{4} a^{2}-3 b a}{a^{4}}-\frac {2 a -4 b}{2 a^{3}}}{x^{6}}+\frac {-\frac {-a^{2}-4 b a}{a^{5}}+\frac {\frac {3 a}{2}-3 b}{a^{4}}}{x^{8}}+\frac {\frac {-\frac {5}{4} a^{2}-5 b a}{a^{6}}-\frac {2 a -4 b}{a^{5}}}{x^{10}}+\frac {-\frac {3 \left (-a^{2}-4 b a \right )}{2 a^{7}}+\frac {\frac {5 a}{2}-5 b}{a^{6}}}{x^{12}} + \cdots \] 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 \(\{E_c\}\) |
| \(\frac {\sqrt {-a}}{a}\) | \(2\) | \(\{0, 3, 6, 9, 12\}\) |
| \(-\frac {\sqrt {-a}}{a}\) | \(2\) | \(\{0, 3, 6, 9, 12\}\) |
| 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 \[ e_1=3,\hspace {3pt} e_2=3,\hspace {3pt} e_\infty =6 \] 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 (3+\left (3\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 {3}{\left (x-\left (\frac {\sqrt {-a}}{a}\right )\right )}+\frac {3}{\left (x-\left (-\frac {\sqrt {-a}}{a}\right )\right )}\right ) \\ &= \frac {2 a x}{a \,x^{2}+1} \end {align*}
And \begin {align*} S &= \prod _{c\in \Gamma } (x-c) \\ &= \left (x -\frac {\sqrt {-a}}{a}\right ) \left (x +\frac {\sqrt {-a}}{a}\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 \(r=\frac {-x^{2} a^{2}-4 b a \,x^{2}+2 a -4 b}{4 \left (a \,x^{2}+1\right )^{2}}\)). \begin{align*} P_{4} &= - p &= -1 \\ P_{3} &= 2 x \\ P_{2} &= \frac {-3 x^{2} a^{2}-4 b a \,x^{2}-4 b}{a^{2}} \\ P_{1} &= \frac {3 \left (a \left (a +4 b \right ) x^{2}+4 b \right ) x}{a^{2}} \\ P_{0} &= -\frac {3 \left (x^{2} a^{2}+4 b a \,x^{2}+4 b \right )^{2}}{2 a^{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}^{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 a^{4}}\left (-\left (4 x^{4} a^{2} \omega ^{2}-4 x^{3} a^{2} \omega +8 a \,\omega ^{2} x^{2}+x^{2} a^{2}+4 b a \,x^{2}-4 a \omega x +4 \omega ^{2}+4 b \right )^{2}\right ) =0 \end{equation} The solution \(\omega \) of eq. 3A is found as \[ \omega =\frac {1}{2 a \,x^{2}+2}\left (a x -2 \sqrt {-\left (a \,x^{2}+1\right ) b}\right ) \] 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 {a x -2 \sqrt {-\left (a \,x^{2}+1\right ) b}}{2 a \,x^{2}+2}d x} \\ &= \left (a \,x^{2}+1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {b \arctan \left (\frac {\sqrt {b a}\, x}{\sqrt {-\left (a \,x^{2}+1\right ) b}}\right )}{\sqrt {b a}}} \\ \end{align*} Which simplifies to \[ z_1(x) = \left (a \,x^{2}+1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}} \] 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 {a x}{a \,x^{2}+1} \,dx} \\ &= z_1 e^{-\frac {\ln \left (a \,x^{2}+1\right )}{4}} \\ &= z_1 \left (\frac {1}{\left (a \,x^{2}+1\right )^{\frac {1}{4}}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = {\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {a x}{a \,x^{2}+1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\frac {\ln \left (a \,x^{2}+1\right )}{2}}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {{\mathrm e}^{-\frac {2 b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}}{\sqrt {a \,x^{2}+1}}d x\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left ({\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}\right ) + c_{2} \left ({\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}\left (\int \frac {{\mathrm e}^{-\frac {2 b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}}{\sqrt {a \,x^{2}+1}}d x\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}+c_{2} {\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}} \left (\int \frac {{\mathrm e}^{-\frac {2 b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}}{\sqrt {a \,x^{2}+1}}d x \right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}+c_{2} {\mathrm e}^{\frac {b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}} \left (\int \frac {{\mathrm e}^{-\frac {2 b \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+1}}\right )}{\sqrt {b a}}}}{\sqrt {a \,x^{2}+1}}d x \right ) \] Verified OK.
Maple trace
`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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 59
dsolve((a*x^2+1)*diff(diff(y(x),x),x)+a*x*diff(y(x),x)+b*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{\frac {i \sqrt {b}}{\sqrt {a}}}+c_{2} \left (x \sqrt {a}+\sqrt {a \,x^{2}+1}\right )^{-\frac {i \sqrt {b}}{\sqrt {a}}} \]
✓ Solution by Mathematica
Time used: 0.188 (sec). Leaf size: 84
DSolve[b*y[x] + a*x*y'[x] + (1 + a*x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \cos \left (\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+1}-1}\right )}{\sqrt {a}}\right )+c_2 \sin \left (\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+1}-1}\right )}{\sqrt {a}}\right ) \]