Internal problem ID [9630]
Internal file name [OUTPUT/8572_Monday_June_06_2022_04_13_05_AM_7515344/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1303.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {\left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (d x +f \right ) y^{\prime }+g y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (d x +f \right ) y^{\prime }+g 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 {g y}{a \,x^{2}+b x +c}-\frac {\left (d x +f \right ) y^{\prime }}{a \,x^{2}+b x +c} \\ \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 {\left (d x +f \right ) y^{\prime }}{a \,x^{2}+b x +c}+\frac {g y}{a \,x^{2}+b x +c}=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 )=\frac {d x +f}{a \,x^{2}+b x +c}, P_{3}\left (x \right )=\frac {g}{a \,x^{2}+b x +c}\right ] \\ {} & \circ & \left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a} \\ {} & {} & \left (\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}}}}=0 \\ {} & \circ & {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a} \\ {} & {} & \left ({\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}}}}=0 \\ {} & \circ & x =\frac {-b +\sqrt {-4 c a +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 c a +b^{2}}}{2 a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (d x +f \right ) y^{\prime }+g y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a \,u^{2}+u \sqrt {-4 c a +b^{2}}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (d u -\frac {d b}{2 a}+\frac {d \sqrt {-4 c a +b^{2}}}{2 a}+f \right ) \left (\frac {d}{d u}y \left (u \right )\right )+g 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}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \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 =1..2 \\ {} & {} & 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} \\ {} & {} & \frac {a_{0} r \left (2 \sqrt {-4 c a +b^{2}}\, a r -2 \sqrt {-4 c a +b^{2}}\, a +d \sqrt {-4 c a +b^{2}}+2 a f -b d \right ) u^{-1+r}}{2 a}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (\frac {a_{k +1} \left (k +1+r \right ) \left (2 \sqrt {-4 c a +b^{2}}\, a \left (k +1\right )+2 \sqrt {-4 c a +b^{2}}\, a r -2 \sqrt {-4 c a +b^{2}}\, a +d \sqrt {-4 c a +b^{2}}+2 a f -b d \right )}{2 a}+a_{k} \left (a \,k^{2}+2 a k r +a \,r^{2}-a k -a r +d k +d r +g \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {r \left (2 \sqrt {-4 c a +b^{2}}\, a r -2 \sqrt {-4 c a +b^{2}}\, a +d \sqrt {-4 c a +b^{2}}+2 a f -b d \right )}{2 a}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}\, a}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {2 a_{k +1} \left (\left (k +r \right ) a +\frac {d}{2}\right ) \left (k +1+r \right ) \sqrt {-4 c a +b^{2}}+2 a_{k} \left (k +r \right ) \left (k +r -1\right ) a^{2}+\left (2 f \left (k +1+r \right ) a_{k +1}+2 a_{k} \left (d k +d r +g \right )\right ) a -b d a_{k +1} \left (k +1+r \right )}{2 a}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+2 a k r +a \,r^{2}-a k -a r +d k +d r +g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+4 \sqrt {-4 c a +b^{2}}\, a k r +2 \sqrt {-4 c a +b^{2}}\, a \,r^{2}+2 \sqrt {-4 c a +b^{2}}\, a k +2 \sqrt {-4 c a +b^{2}}\, a r +\sqrt {-4 c a +b^{2}}\, d k +\sqrt {-4 c a +b^{2}}\, d r +2 a f k +2 a f r -b d k -b d r +d \sqrt {-4 c a +b^{2}}+2 a f -b d} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}-a k +d k +g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 \sqrt {-4 c a +b^{2}}\, a k +\sqrt {-4 c a +b^{2}}\, d k +2 a f k -b d k +d \sqrt {-4 c a +b^{2}}+2 a f -b d} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}-a k +d k +g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 \sqrt {-4 c a +b^{2}}\, a k +\sqrt {-4 c a +b^{2}}\, d k +2 a f k -b d k +d \sqrt {-4 c a +b^{2}}+2 a f -b d}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{k}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}-a k +d k +g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 \sqrt {-4 c a +b^{2}}\, a k +\sqrt {-4 c a +b^{2}}\, d k +2 a f k -b d k +d \sqrt {-4 c a +b^{2}}+2 a f -b d}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}\, a} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{4 a \left (-4 c a +b^{2}\right )}-a k -\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}}+d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}+g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{2 \sqrt {-4 c a +b^{2}}\, a}+2 \sqrt {-4 c a +b^{2}}\, a k +2 \sqrt {-4 c a +b^{2}}\, a +\sqrt {-4 c a +b^{2}}\, d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 a}+2 a f k +\frac {f \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}-b d k -\frac {b d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}\, a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{4 a \left (-4 c a +b^{2}\right )}-a k -\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}}+d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}+g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{2 \sqrt {-4 c a +b^{2}}\, a}+2 \sqrt {-4 c a +b^{2}}\, a k +2 \sqrt {-4 c a +b^{2}}\, a +\sqrt {-4 c a +b^{2}}\, d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 a}+2 a f k +\frac {f \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}-b d k -\frac {b d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{k +\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{4 a \left (-4 c a +b^{2}\right )}-a k -\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}}+d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}+g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{2 \sqrt {-4 c a +b^{2}}\, a}+2 \sqrt {-4 c a +b^{2}}\, a k +2 \sqrt {-4 c a +b^{2}}\, a +\sqrt {-4 c a +b^{2}}\, d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 a}+2 a f k +\frac {f \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}-b d k -\frac {b d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}h_{k} {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{k +\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}\, a}}\right ), e_{k +1}=-\frac {2 a e_{k} \left (a \,k^{2}-a k +d k +g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 \sqrt {-4 c a +b^{2}}\, a k +\sqrt {-4 c a +b^{2}}\, d k +2 a f k -b d k +d \sqrt {-4 c a +b^{2}}+2 a f -b d}, h_{k +1}=-\frac {2 a h_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{4 a \left (-4 c a +b^{2}\right )}-a k -\frac {2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d}{2 \sqrt {-4 c a +b^{2}}}+d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}+g \right )}{2 \sqrt {-4 c a +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )+\frac {\left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )^{2}}{2 \sqrt {-4 c a +b^{2}}\, a}+2 \sqrt {-4 c a +b^{2}}\, a k +2 \sqrt {-4 c a +b^{2}}\, a +\sqrt {-4 c a +b^{2}}\, d k +\frac {d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 a}+2 a f k +\frac {f \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{\sqrt {-4 c a +b^{2}}}-b d k -\frac {b d \left (2 \sqrt {-4 c a +b^{2}}\, a -d \sqrt {-4 c a +b^{2}}-2 a f +b d \right )}{2 \sqrt {-4 c a +b^{2}}\, a}}\right ] \end {array} \]
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)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.188 (sec). Leaf size: 501
dsolve((a*x^2+b*x+c)*diff(diff(y(x),x),x)+(d*x+f)*diff(y(x),x)+g*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [-\frac {a -d +\sqrt {a^{2}+\left (-2 d -4 g \right ) a +d^{2}}}{2 a}, \frac {-a +d +\sqrt {a^{2}+\left (-2 d -4 g \right ) a +d^{2}}}{2 a}\right ], \left [\frac {d \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a -2 a f +b d}{2 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}}\right ], \frac {\left (-2 a^{2} x -a b \right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+4 a c -b^{2}}{8 a c -2 b^{2}}\right )+c_{2} {\left (2 \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, x \,a^{2}+\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, b a -4 a c +b^{2}\right )}^{\frac {a \left (a -\frac {d}{2}\right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+a f -\frac {b d}{2}}{\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a^{2}}} \operatorname {hypergeom}\left (\left [\frac {a \left (a -\sqrt {a^{2}+\left (-2 d -4 g \right ) a +d^{2}}\right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+2 a f -b d}{2 \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a^{2}}, \frac {a \left (a +\sqrt {a^{2}+\left (-2 d -4 g \right ) a +d^{2}}\right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+2 a f -b d}{2 \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a^{2}}\right ], \left [\frac {4 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}-d \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a +2 a f -b d}{2 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}}\right ], \frac {\left (-2 a^{2} x -a b \right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+4 a c -b^{2}}{8 a c -2 b^{2}}\right ) \]
✓ Solution by Mathematica
Time used: 4.122 (sec). Leaf size: 498
DSolve[g*y[x] + (f + d*x)*y'[x] + (c + b*x + a*x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \operatorname {Hypergeometric2F1}\left (-\frac {a-d+\sqrt {(a-d)^2-4 a g}}{2 a},\frac {-a+d+\sqrt {(a-d)^2-4 a g}}{2 a},\frac {\left (b+\sqrt {b^2-4 a c}\right ) d-2 a f}{2 a \sqrt {b^2-4 a c}},\frac {b+2 a x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )-c_2 2^{\frac {\frac {b d}{\sqrt {b^2-4 a c}}+d}{2 a}-\frac {f}{\sqrt {b^2-4 a c}}-1} \exp \left (-\frac {i \pi \left (d \left (\sqrt {b^2-4 a c}+b\right )-2 a f\right )}{2 a \sqrt {b^2-4 a c}}\right ) \left (\frac {\sqrt {b^2-4 a c}+2 a x+b}{\sqrt {b^2-4 a c}}\right )^{-\frac {\frac {b d}{\sqrt {b^2-4 a c}}+d}{2 a}+\frac {f}{\sqrt {b^2-4 a c}}+1} \operatorname {Hypergeometric2F1}\left (\frac {\frac {2 f a}{\sqrt {b^2-4 a c}}+a-\frac {b d}{\sqrt {b^2-4 a c}}-\sqrt {(a-d)^2-4 a g}}{2 a},\frac {\frac {2 f a}{\sqrt {b^2-4 a c}}+a-\frac {b d}{\sqrt {b^2-4 a c}}+\sqrt {(a-d)^2-4 a g}}{2 a},-\frac {\frac {b d}{\sqrt {b^2-4 a c}}+d+a \left (-\frac {2 f}{\sqrt {b^2-4 a c}}-4\right )}{2 a},\frac {b+2 a x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right ) \]