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2.31.28 Problem 176

2.31.28.1 second order ode non constant coeff transformation on B
2.31.28.2 ✓Maple
2.31.28.3 ✓Mathematica
2.31.28.4 ✗Sympy

Internal problem ID [13837]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-5
Problem number : 176
Date solved : Thursday, January 01, 2026 at 02:54:48 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.31.28.1 second order ode non constant coeff transformation on B

1.993 (sec)

\begin{align*} \left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (k x +d \right ) y^{\prime }-k y&=0 \\ \end{align*}
Entering second order ode non constant coeff transformation on \(B\) solverGiven an ode of the form
\begin{align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end{align*}

This method reduces the order ode the ODE by one by applying the transformation

\begin{align*} y&= B v \end{align*}

This results in

\begin{align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end{align*}

And now the original ode becomes

\begin{align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end{align*}

If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve

\[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \]
By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is
\[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \]
The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).

This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that

\begin{align*} A &= a \,x^{2}+b x +c\\ B &= k x +d\\ C &= -k\\ F &= 0 \end{align*}

The above shows that for this ode

\begin{align*} AB''+BB'+CB &= \left (a \,x^{2}+b x +c\right ) \left (0\right ) + \left (k x +d\right ) \left (k\right ) + \left (-k\right ) \left (k x +d\right ) \\ &=0 \end{align*}

Hence the ode in \(v\) given in (1) now simplifies to

\begin{align*} \left (a \,x^{2}+b x +c \right ) \left (k x +d \right ) v'' +\left ( 2 k \left (a \,x^{2}+b x +c \right )+\left (k x +d \right )^{2}\right ) v' & =0 \end{align*}

Now by applying \(v'=u\) the above becomes

\begin{align*} \left (a \,x^{2}+b x +c \right ) \left (k x +d \right ) u^{\prime }\left (x \right )+2 u \left (x \right ) \left (\left (a +\frac {k}{2}\right ) k \,x^{2}+k \left (b +d \right ) x +c k +\frac {d^{2}}{2}\right ) = 0 \end{align*}

Which is now solved for \(u\). Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {\left (-2 a k -k^{2}\right ) x^{2}-2 k \left (b +d \right ) x -2 c k -d^{2}}{\left (a \,x^{2}+b x +c \right ) \left (k x +d \right )}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {\left (-2 a k -k^{2}\right ) x^{2}-2 k \left (b +d \right ) x -2 c k -d^{2}}{\left (a \,x^{2}+b x +c \right ) \left (k x +d \right )}d x}\\ &= \left (k x +d \right )^{2} \left (a \,x^{2}+b x +c \right )^{\frac {k}{2 a}} {\mathrm e}^{\frac {\left (2 a d -b k \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \left (k x +d \right )^{2} \left (a \,x^{2}+b x +c \right )^{\frac {k}{2 a}} {\mathrm e}^{\frac {\left (2 a d -b k \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \left (k x +d \right )^{2} \left (a \,x^{2}+b x +c \right )^{\frac {k}{2 a}} {\mathrm e}^{\frac {\left (2 a d -b k \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(\left (k x +d \right )^{2} \left (a \,x^{2}+b x +c \right )^{\frac {k}{2 a}} {\mathrm e}^{\frac {\left (2 a d -b k \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\) gives the final solution

\[ u \left (x \right ) = \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {\left (2 a d -b k \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} c_1}{\left (k x +d \right )^{2}} \]
Simplifying the above gives
\begin{align*} u \left (x \right ) &= \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \left (a d -\frac {b k}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} c_1}{\left (k x +d \right )^{2}} \\ \end{align*}
The ode for \(v\) now becomes
\[ v^{\prime }\left (x \right ) = \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \left (a d -\frac {b k}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} c_1}{\left (k x +d \right )^{2}} \]
Which is now solved for \(v\). Entering first order ode quadrature solverSince the ode has the form \(v^{\prime }\left (x \right )=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dv} &= \int {\frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \left (a d -\frac {b k}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} c_1}{\left (k x +d \right )^{2}}\, dx}\\ v \left (x \right ) &= \int \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \left (a d -\frac {b k}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} c_1}{\left (k x +d \right )^{2}}d x + c_2 \end{align*}
\begin{align*} v \left (x \right )&= \int \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \left (a d -\frac {b k}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} c_1}{\left (k x +d \right )^{2}}d x +c_2 \end{align*}

Replacing \(v \left (x \right )\) above by \(\frac {y}{k x +d}\), then the solution becomes

\begin{align*} y(x) &= B v\\ &= \left (\int \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \left (a d -\frac {b k}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} c_1}{\left (k x +d \right )^{2}}d x +c_2 \right ) \left (k x +d \right ) \end{align*}

Summary of solutions found

\begin{align*} y &= \left (\int \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \left (a d -\frac {b k}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} c_1}{\left (k x +d \right )^{2}}d x +c_2 \right ) \left (k x +d \right ) \\ \end{align*}
2.31.28.2 ✓ Maple. Time used: 0.027 (sec). Leaf size: 315
ode:=(a*x^2+b*x+c)*diff(diff(y(x),x),x)+(k*x+d)*diff(y(x),x)-k*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
\[ y = c_1 \left (k x +d \right )+c_2 {\left (2 \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, x \,a^{2}+\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, b a -4 a c +b^{2}\right )}^{\frac {a \left (a -\frac {k}{2}\right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+a d -\frac {b k}{2}}{\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a^{2}}} \operatorname {hypergeom}\left (\left [-\frac {k \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a -2 a d +b k}{2 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}}, \frac {a \left (a +\frac {k}{2}\right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+a d -\frac {b k}{2}}{\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a^{2}}\right ], \left [\frac {4 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}-k \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a +2 a d -b k}{2 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}}\right ], \frac {\left (-2 x \,a^{2}-b a \right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+4 a c -b^{2}}{8 a c -2 b^{2}}\right ) \]

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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> 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 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved -\ 
 returning with no uncomputed integrals 
   <- Kovacics algorithm successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (k x +d \right ) \left (\frac {d}{d x}y \left (x \right )\right )-k y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {k y \left (x \right )}{a \,x^{2}+b x +c}-\frac {\left (k x +d \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{a \,x^{2}+b x +c} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \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^{2}}{d x^{2}}y \left (x \right )+\frac {\left (k x +d \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{a \,x^{2}+b x +c}-\frac {k y \left (x \right )}{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 {k x +d}{a \,x^{2}+b x +c}, P_{3}\left (x \right )=-\frac {k}{a \,x^{2}+b x +c}\right ] \\ {} & \circ & \left (x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {\sqrt {-4 a c +b^{2}}-b}{2 a} \\ {} & {} & \left (\left (x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}}}}=0 \\ {} & \circ & {\left (x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {\sqrt {-4 a c +b^{2}}-b}{2 a} \\ {} & {} & \left ({\left (x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}}}}=0 \\ {} & \circ & x =\frac {\sqrt {-4 a c +b^{2}}-b}{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 {\sqrt {-4 a c +b^{2}}-b}{2 a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (k x +d \right ) \left (\frac {d}{d x}y \left (x \right )\right )-k y \left (x \right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a \,u^{2}+u \sqrt {-4 a c +b^{2}}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (k u +\frac {k \sqrt {-4 a c +b^{2}}}{2 a}-\frac {k b}{2 a}+d \right ) \left (\frac {d}{d u}y \left (u \right )\right )-k 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^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}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^{2}}{d u^{2}}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 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +k \sqrt {-4 a c +b^{2}}+2 a d -b k \right ) u^{r -1}}{2 a}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (\frac {a_{k +1} \left (k +1+r \right ) \left (2 \sqrt {-4 a c +b^{2}}\, a \left (k +1\right )+2 \sqrt {-4 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +k \sqrt {-4 a c +b^{2}}+2 a d -b k \right )}{2 a}+a_{k} \left (k +r -1\right ) \left (a k +a r +k \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 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +k \sqrt {-4 a c +b^{2}}+2 a d -b k \right )}{2 a}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {2 \left (k +1+r \right ) \left (\left (k +r \right ) a +\frac {k}{2}\right ) a_{k +1} \sqrt {-4 a c +b^{2}}+2 a_{k} \left (k +r \right ) \left (k +r -1\right ) a^{2}+\left (2 d \left (k +1+r \right ) a_{k +1}+2 k a_{k} \left (k +r -1\right )\right ) a -b k 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 +k k +k r -k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+4 \sqrt {-4 a c +b^{2}}\, k a r +2 \sqrt {-4 a c +b^{2}}\, a \,r^{2}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a r +\sqrt {-4 a c +b^{2}}\, k k +\sqrt {-4 a c +b^{2}}\, r k +2 a d k +2 a d r -b k k -b k r +k \sqrt {-4 a c +b^{2}}+2 a d -b k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}-a k +k k -k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 \sqrt {-4 a c +b^{2}}\, a k +\sqrt {-4 a c +b^{2}}\, k k +2 a d k -b k k +k \sqrt {-4 a c +b^{2}}+2 a d -b k} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {2 a a_{0} k}{k \sqrt {-4 a c +b^{2}}+2 a d -b k} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (1+\frac {2 a k u}{k \sqrt {-4 a c +b^{2}}+2 a d -b k}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a} \\ {} & {} & \left [y \left (x \right )=\frac {2 a_{0} a \left (k x +d \right )}{k \sqrt {-4 a c +b^{2}}+2 a d -b k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-a k -\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+k k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 a}+2 a d k +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k k -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-a k -\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+k k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 a}+2 a d k +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k k -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}\right )}^{k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-a k -\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+k k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 a}+2 a d k +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k k -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\frac {2 e_{0} a \left (k x +d \right )}{k \sqrt {-4 a c +b^{2}}+2 a d -b k}+\left (\moverset {\infty }{\munderset {m =0}{\sum }}f_{m} {\left (x -\frac {\sqrt {-4 a c +b^{2}}-b}{2 a}\right )}^{m +\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ), f_{m +1}=-\frac {2 a f_{m} \left (a \,m^{2}+\frac {m \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-a m -\frac {2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+m k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,m^{2}+2 m \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a m +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, m k +\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right ) k}{2 a}+2 a d m +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k m -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -k \sqrt {-4 a c +b^{2}}-2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ] \end {array} \]
2.31.28.3 ✓ Mathematica. Time used: 2.137 (sec). Leaf size: 107
ode=(a*x^2+b*x+c)*D[y[x],{x,2}]+(k*x+d)*D[y[x],x]-k*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to \frac {(d+k x) \left (c_2 \int _1^x\frac {\exp \left (\frac {(b k-2 a d) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) (c+K[1] (b+a K[1]))^{-\frac {k}{2 a}}}{(d+k K[1])^2}dK[1]+c_1\right )}{d} \end{align*}
2.31.28.4 ✗ Sympy
from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
d = symbols("d") 
k = symbols("k") 
y = Function("y") 
ode = Eq(-k*y(x) + (d + k*x)*Derivative(y(x), x) + (a*x**2 + b*x + c)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False

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