Internal problem ID [9723]
Internal file name [OUTPUT/8665_Monday_June_06_2022_04_44_22_AM_62972131/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1396.
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]]
\[ \boxed {y^{\prime \prime }+\frac {A y}{\left (a \,x^{2}+b x +c \right )^{2}}=0} \]
Writing the ode as \begin {align*} y^{\prime \prime } x^{2}+\frac {A y}{x^{2}} = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} y^{\prime \prime } x^{2}+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*} y^{\prime \prime } x^{2}+\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 &= \sqrt {A}\\ 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 {\sqrt {A}}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {A}}{x}}}-\frac {c_{2} \sqrt {x}\, \sqrt {2}\, \cos \left (\frac {\sqrt {A}}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {A}}{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 {\sqrt {A}}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {A}}{x}}}-\frac {c_{2} \sqrt {x}\, \sqrt {2}\, \cos \left (\frac {\sqrt {A}}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {A}}{x}}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \sqrt {x}\, \sqrt {2}\, \sin \left (\frac {\sqrt {A}}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {A}}{x}}}-\frac {c_{2} \sqrt {x}\, \sqrt {2}\, \cos \left (\frac {\sqrt {A}}{x}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {A}}{x}}} \] Verified OK.
Writing the ode as \begin {align*} y^{\prime \prime } \left (a \,x^{2}+b x +c \right )^{2}+A 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 \,x^{2}+b x +c \right )^{2} \\ B &= 0\tag {3} \\ C &= A \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 {-A}{\left (a \,x^{2}+b x +c \right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= -A\\ t &= \left (a \,x^{2}+b x +c \right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( -\frac {A}{\left (a \,x^{2}+b x +c \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 \(t=\left (a \,x^{2}+b x +c \right )^{2}\). There is a pole at \(x=-\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}\) of order \(2\). There is a pole at \(x=-\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}\) 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 \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = -\frac {A}{\left (-4 a c +b^{2}\right ) {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}}-\frac {A}{\left (-4 a c +b^{2}\right ) {\left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}}+\frac {2 a A}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}} \left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}-\frac {2 a A}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}} \left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}\right )} \] For the pole at \(x=-\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}\) let \(b\) be the coefficient of \(\frac {1}{ {\left (x +\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=\frac {A}{4 a c -b^{2}}\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2} \end {alignat*}
For the pole at \(x=-\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}\) let \(b\) be the coefficient of \(\frac {1}{ {\left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=\frac {A}{4 a c -b^{2}}\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{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 \[ r=-\frac {A}{\left (a \,x^{2}+b x +c \right )^{2}} \]
pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
\(-\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}\) | \(2\) | \(0\) | \(\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}\) | \(\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}\) |
\(-\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}\) | \(2\) | \(0\) | \(\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}\) | \(\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}\) |
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 {\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}}+\frac {\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}} + (-) \left ( 0 \right ) \\ &= \frac {\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}}+\frac {\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\\ &= \frac {2 x a -\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}+b}{2 a \,x^{2}+2 b x +2 c} \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 {\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}}+\frac {\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\right ) \left (0\right ) + \left ( \left (-\frac {\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{{\left (x +\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}}-\frac {\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{{\left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}}\right ) + \left (\frac {\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}}+\frac {\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\right )^2 - \left (-\frac {A}{\left (a \,x^{2}+b x +c \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 {\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b -\sqrt {-4 a c +b^{2}}}{2 a}}+\frac {\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}{x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\right )d x}\\ &= {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} \sqrt {\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}}\, {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} \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 \\ &= {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} \\ \end{align*} Which simplifies to \[ y_1 = {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Since \(B=0\) then the above becomes \begin{align*} y_2 &= y_1 \int \frac {1}{y_1^2} \,dx \\ &= {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}\int \frac {1}{{\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{1-\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{1+\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}} \,dx \\ &= {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}\left (\int {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{-1+\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{-1-\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}d x\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left ({\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}\right ) + c_{2} \left ({\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}\left (\int {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{-1+\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{-1-\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}d x\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}-c_{2} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} \sqrt {\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}}\, {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} a^{2} \left (\int -\frac {{\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{-\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}} {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}}{4 a \left (a \,x^{2}+b x +c \right )}d x \right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\frac {1}{2}-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}}-c_{2} {\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{\frac {1}{2}+\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} \sqrt {\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}}\, {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{-\frac {\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}{2}} a^{2} \left (\int -\frac {{\left (\frac {2 x a +\sqrt {-4 a c +b^{2}}+b}{a}\right )}^{-\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}} {\left (\frac {2 x a +b -\sqrt {-4 a c +b^{2}}}{a}\right )}^{\sqrt {\frac {4 a c -b^{2}+4 A}{4 a c -b^{2}}}}}{4 a \left (a \,x^{2}+b x +c \right )}d x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (a \,x^{2}+b x +c \right )^{2}+A 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 {A y}{\left (a \,x^{2}+b x +c \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 {A y}{\left (a \,x^{2}+b x +c \right )^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=0, P_{3}\left (x \right )=\frac {A}{\left (a \,x^{2}+b x +c \right )^{2}}\right ] \\ {} & \circ & \left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left (\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}}=0 \\ {} & \circ & {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left ({\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}}=0 \\ {} & \circ & x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (a \,x^{2}+b x +c \right )^{2}+A y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a^{2} u^{4}+2 a \,u^{3} \sqrt {-4 a c +b^{2}}-4 a c \,u^{2}+b^{2} u^{2}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+A y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-4 a c \,r^{2}+b^{2} r^{2}+4 a c r -b^{2} r +A \right ) u^{r}+\left (\left (-4 a c \,r^{2}+b^{2} r^{2}-4 a c r +b^{2} r +A \right ) a_{1}+2 a_{0} r \left (-1+r \right ) a \sqrt {-4 a c +b^{2}}\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (-4 a c \,k^{2}-8 a c k r -4 a c \,r^{2}+b^{2} k^{2}+2 b^{2} k r +b^{2} r^{2}+4 a c k +4 a c r -b^{2} k -b^{2} r +A \right )+2 a_{k -1} \left (k +r -1\right ) \left (k -2+r \right ) a \sqrt {-4 a c +b^{2}}+a_{k -2} \left (k -2+r \right ) \left (k -3+r \right ) a^{2}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -4 a c \,r^{2}+b^{2} r^{2}+4 a c r -b^{2} r +A =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}, \frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (-4 a c \,r^{2}+b^{2} r^{2}-4 a c r +b^{2} r +A \right ) a_{1}+2 a_{0} r \left (-1+r \right ) a \sqrt {-4 a c +b^{2}}=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {2 a_{0} r \left (-1+r \right ) a \sqrt {-4 a c +b^{2}}}{-4 a c \,r^{2}+b^{2} r^{2}-4 a c r +b^{2} r +A} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 a_{k -1} \left (k +r -1\right ) \left (k -2+r \right ) a \sqrt {-4 a c +b^{2}}+a_{k -2} \left (k -2+r \right ) \left (k -3+r \right ) a^{2}-4 c a_{k} \left (k +r \right ) \left (k +r -1\right ) a +a_{k} \left (b^{2} k^{2}+2 \left (r -\frac {1}{2}\right ) b^{2} k +b^{2} r^{2}-b^{2} r +A \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 2 a_{k +1} \left (k +1+r \right ) \left (k +r \right ) a \sqrt {-4 a c +b^{2}}+a_{k} \left (k +r \right ) \left (k +r -1\right ) a^{2}-4 c a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) a +a_{k +2} \left (b^{2} \left (k +2\right )^{2}+2 \left (r -\frac {1}{2}\right ) b^{2} \left (k +2\right )+b^{2} r^{2}-b^{2} r +A \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} a_{k +1}+4 \sqrt {-4 a c +b^{2}}\, k r a_{k +1}+2 \sqrt {-4 a c +b^{2}}\, r^{2} a_{k +1}+a \,k^{2} a_{k}+2 a k r a_{k}+a \,r^{2} a_{k}+2 \sqrt {-4 a c +b^{2}}\, k a_{k +1}+2 \sqrt {-4 a c +b^{2}}\, r a_{k +1}-a k a_{k}-a r a_{k}\right )}{-4 a c \,k^{2}-8 a c k r -4 a c \,r^{2}+b^{2} k^{2}+2 b^{2} k r +b^{2} r^{2}-12 a c k -12 a c r +3 b^{2} k +3 b^{2} r -8 a c +2 b^{2}+A} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )} \\ {} & {} & a_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} a_{k +1}-\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} a_{k}-\frac {a k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{4 a c -b^{2}}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k a_{k +1}-\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}-a k a_{k}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}+\frac {4 a c k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}-\frac {b^{2} k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k +\frac {6 a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k -\frac {3 b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}}, a_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} a_{k +1}-\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} a_{k}-\frac {a k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{4 a c -b^{2}}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k a_{k +1}-\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}-a k a_{k}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}+\frac {4 a c k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}-\frac {b^{2} k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k +\frac {6 a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k -\frac {3 b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A}, a_{1}=\frac {a_{0} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) \left (-1-\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}\right ) a \sqrt {-4 a c +b^{2}}}{\left (4 a c -b^{2}\right ) \left (-\frac {a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}+\frac {2 a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}+A \right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{k -\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}}, a_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} a_{k +1}-\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} a_{k}-\frac {a k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{4 a c -b^{2}}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k a_{k +1}-\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}-a k a_{k}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}+\frac {4 a c k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}-\frac {b^{2} k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k +\frac {6 a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k -\frac {3 b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A}, a_{1}=\frac {a_{0} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) \left (-1-\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}\right ) a \sqrt {-4 a c +b^{2}}}{\left (4 a c -b^{2}\right ) \left (-\frac {a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}+\frac {2 a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}+A \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )} \\ {} & {} & a_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} a_{k +1}+\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} a_{k}+\frac {a k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{4 a c -b^{2}}+\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k a_{k +1}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}-a k a_{k}-\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}-\frac {4 a c k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}+\frac {b^{2} k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k -\frac {6 a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k +\frac {3 b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}}, a_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} a_{k +1}+\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} a_{k}+\frac {a k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{4 a c -b^{2}}+\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k a_{k +1}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}-a k a_{k}-\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}-\frac {4 a c k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}+\frac {b^{2} k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k -\frac {6 a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k +\frac {3 b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A}, a_{1}=-\frac {a_{0} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) \left (-1+\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}\right ) a \sqrt {-4 a c +b^{2}}}{\left (4 a c -b^{2}\right ) \left (-\frac {a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-\frac {2 a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}+A \right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{k +\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}}, a_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} a_{k +1}+\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} a_{k}+\frac {a k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{4 a c -b^{2}}+\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} a_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k a_{k +1}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k +1}}{4 a c -b^{2}}-a k a_{k}-\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) a_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}-\frac {4 a c k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}+\frac {b^{2} k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k -\frac {6 a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k +\frac {3 b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A}, a_{1}=-\frac {a_{0} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) \left (-1+\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}\right ) a \sqrt {-4 a c +b^{2}}}{\left (4 a c -b^{2}\right ) \left (-\frac {a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-\frac {2 a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}+A \right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{k -\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{k +\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}}\right ), d_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} d_{k +1}-\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) d_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} d_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} d_{k}-\frac {a k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) d_{k}}{4 a c -b^{2}}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} d_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k d_{k +1}-\frac {\sqrt {-4 a c +b^{2}}\, \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) d_{k +1}}{4 a c -b^{2}}-a k d_{k}+\frac {a \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) d_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}+\frac {4 a c k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}-\frac {b^{2} k \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k +\frac {6 a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k -\frac {3 b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A}, d_{1}=\frac {d_{0} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) \left (-1-\frac {-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}\right ) a \sqrt {-4 a c +b^{2}}}{\left (4 a c -b^{2}\right ) \left (-\frac {a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}+\frac {2 a c \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {b^{2} \left (-4 a c +b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}+A \right )}, e_{k +2}=-\frac {a \left (2 \sqrt {-4 a c +b^{2}}\, k^{2} e_{k +1}+\frac {2 \sqrt {-4 a c +b^{2}}\, k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) e_{k +1}}{4 a c -b^{2}}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} e_{k +1}}{2 \left (4 a c -b^{2}\right )^{2}}+a \,k^{2} e_{k}+\frac {a k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) e_{k}}{4 a c -b^{2}}+\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2} e_{k}}{4 \left (4 a c -b^{2}\right )^{2}}+2 \sqrt {-4 a c +b^{2}}\, k e_{k +1}+\frac {\sqrt {-4 a c +b^{2}}\, \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) e_{k +1}}{4 a c -b^{2}}-a k e_{k}-\frac {a \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) e_{k}}{2 \left (4 a c -b^{2}\right )}\right )}{-4 a c \,k^{2}-\frac {4 a c k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}-\frac {a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+b^{2} k^{2}+\frac {b^{2} k \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-12 a c k -\frac {6 a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+3 b^{2} k +\frac {3 b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}-8 a c +2 b^{2}+A}, e_{1}=-\frac {e_{0} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right ) \left (-1+\frac {4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}}{2 \left (4 a c -b^{2}\right )}\right ) a \sqrt {-4 a c +b^{2}}}{\left (4 a c -b^{2}\right ) \left (-\frac {a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{\left (4 a c -b^{2}\right )^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )^{2}}{4 \left (4 a c -b^{2}\right )^{2}}-\frac {2 a c \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{4 a c -b^{2}}+\frac {b^{2} \left (4 a c -b^{2}+\sqrt {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}+16 A a c -4 A \,b^{2}}\right )}{2 \left (4 a c -b^{2}\right )}+A \right )}\right ] \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: 0.016 (sec). Leaf size: 178
dsolve(diff(diff(y(x),x),x) = -A/(a*x^2+b*x+c)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left ({\left (\frac {-b +i \sqrt {4 a c -b^{2}}-2 a x}{b +i \sqrt {4 a c -b^{2}}+2 a x}\right )}^{-\frac {a \sqrt {\frac {-4 a c +b^{2}-4 A}{a^{2}}}}{2 \sqrt {-4 a c +b^{2}}}} c_{2} +{\left (\frac {-b +i \sqrt {4 a c -b^{2}}-2 a x}{b +i \sqrt {4 a c -b^{2}}+2 a x}\right )}^{\frac {a \sqrt {\frac {-4 a c +b^{2}-4 A}{a^{2}}}}{2 \sqrt {-4 a c +b^{2}}}} c_{1} \right ) \sqrt {a \,x^{2}+b x +c} \]
✓ Solution by Mathematica
Time used: 1.298 (sec). Leaf size: 199
DSolve[y''[x] == -((A*y[x])/(c + b*x + a*x^2)^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {x (a x+b)+c} \exp \left (-\frac {\sqrt {4 a c-b^2} \sqrt {1-\frac {4 A}{b^2-4 a c}} \arctan \left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{\sqrt {b^2-4 a c}}\right ) \left (c_1 \exp \left (\frac {2 \sqrt {4 a c-b^2} \sqrt {1-\frac {4 A}{b^2-4 a c}} \arctan \left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{\sqrt {b^2-4 a c}}\right )+\frac {c_2}{\sqrt {b^2-4 a c} \sqrt {1-\frac {4 A}{b^2-4 a c}}}\right ) \]