2.31.27 Problem 175

Internal problem ID [13836]
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 : 175
Date solved : Thursday, January 01, 2026 at 02:52:03 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

2.31.27.1 second order linear exact ode

4.332 (sec)

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

Entering second order linear exact ode solverAn ode of the form

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

is exact if

\begin{align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end{align*}

For the given ode we have

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

Hence

\begin{align*} p''(x) &= 2 a\\ q'(x) &= d \end{align*}

Therefore (1) becomes

\begin{align*} 2 a- \left (d\right ) + \left (d -2 a\right )&=0 \end{align*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as

\begin{align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end{align*}

Integrating gives

\begin{align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end{align*}

Substituting the above values for \(p,q,r,s\) gives

\begin{align*} \left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (-2 a x +d x -b +k \right ) y&=c_1 \end{align*}

We now have a first order ode to solve which is

\begin{align*} \left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (-2 a x +d x -b +k \right ) y = c_1 \end{align*}

Entering first order ode linear solverIn canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {\left (-d +2 a \right ) x +b -k}{a \,x^{2}+b x +c}\\ p(x) &=\frac {c_1}{a \,x^{2}+b x +c} \end{align*}

The integrating factor \(\mu \) is

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

Integrating gives

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

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

\[ y = \left (a \,x^{2}+b x +c \right )^{-\frac {d -2 a}{2 a}} {\mathrm e}^{-\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (\int \frac {c_1 \left (a \,x^{2}+b x +c \right )^{\frac {d -2 a}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}}{a \,x^{2}+b x +c}d x +c_2 \right ) \]

Summary of solutions found

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

2.31.27.2 second order integrable as is

0.611 (sec)

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

Entering second order integrable as is solverIntegrating both sides of the ODE w.r.t \(x\) gives

\begin{align*} \int \left (\left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (d x +k \right ) y^{\prime }+\left (d -2 a \right ) y\right )d x &= 0 \\ \left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (-2 a x +d x -b +k \right ) y = c_1 \end{align*}

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

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

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {\left (-d +2 a \right ) x +b -k}{a \,x^{2}+b x +c}\\ p(x) &=\frac {c_1}{a \,x^{2}+b x +c} \end{align*}

The integrating factor \(\mu \) is

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

Integrating gives

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

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

\[ y = \left (a \,x^{2}+b x +c \right )^{-\frac {d -2 a}{2 a}} {\mathrm e}^{-\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (\int \frac {c_1 \left (a \,x^{2}+b x +c \right )^{\frac {d -2 a}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}}{a \,x^{2}+b x +c}d x +c_2 \right ) \]

Summary of solutions found

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

2.31.27.3 ✓ Maple. Time used: 0.043 (sec). Leaf size: 1400

ode:=(a*x^2+b*x+c)*diff(diff(y(x),x),x)+(d*x+k)*diff(y(x),x)+(d-2*a)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ \text {Expression too large to display} \]

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)] 
   One independent solution has integrals. Trying a hypergeometric solution fre\ 
e of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
   -> Trying to convert hypergeometric functions to elementary form... 
   <- elementary form is not straightforward to achieve - returning hypergeomet\ 
ric solution free of uncomputed integrals 
<- linear_1 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 (d x +k \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (d -2 a \right ) 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 {\left (-d +2 a \right ) y \left (x \right )}{a \,x^{2}+b x +c}-\frac {\left (d x +k \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 (d x +k \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{a \,x^{2}+b x +c}-\frac {\left (-d +2 a \right ) 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 {d x +k}{a \,x^{2}+b x +c}, P_{3}\left (x \right )=-\frac {-d +2 a}{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^{2}}{d x^{2}}y \left (x \right )\right )+\left (d x +k \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (d -2 a \right ) y \left (x \right )=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^{2}}{d u^{2}}y \left (u \right )\right )+\left (d u -\frac {b d}{2 a}+\frac {d \sqrt {-4 c a +b^{2}}}{2 a}+k \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (d -2 a \right ) 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 a \sqrt {-4 c a +b^{2}}\, r -2 a \sqrt {-4 c a +b^{2}}+\sqrt {-4 c a +b^{2}}\, d +2 a k -b d \right ) u^{-1+r}}{2 a}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (\frac {a_{k +1} \left (k +r +1\right ) \left (2 a \sqrt {-4 c a +b^{2}}\, \left (k +1\right )+2 a \sqrt {-4 c a +b^{2}}\, r -2 a \sqrt {-4 c a +b^{2}}+\sqrt {-4 c a +b^{2}}\, d +2 a k -b d \right )}{2 a}+a_{k} \left (k +r +1\right ) \left (a k +a r -2 a +d \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 a \sqrt {-4 c a +b^{2}}\, r -2 a \sqrt {-4 c a +b^{2}}+\sqrt {-4 c a +b^{2}}\, d +2 a k -b d \right )}{2 a}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 a \sqrt {-4 c a +b^{2}}}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {\left (\left (\left (k +r \right ) a +\frac {d}{2}\right ) a_{k +1} \sqrt {-4 c a +b^{2}}+a_{k} \left (k +r -2\right ) a^{2}+\left (d a_{k}+k a_{k +1}\right ) a -\frac {b d a_{k +1}}{2}\right ) \left (k +r +1\right )}{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 +a r -2 a +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, k +2 a \sqrt {-4 c a +b^{2}}\, r +\sqrt {-4 c a +b^{2}}\, d +2 a k -b d} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a k -2 a +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, k +\sqrt {-4 c a +b^{2}}\, d +2 a k -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 +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, k +\sqrt {-4 c a +b^{2}}\, d +2 a k -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 \left (x \right )=\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 +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, k +\sqrt {-4 c a +b^{2}}\, d +2 a k -b d}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 a \sqrt {-4 c a +b^{2}}} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a k +\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 c a +b^{2}}}-2 a +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, k +2 a \sqrt {-4 c a +b^{2}}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 a \sqrt {-4 c a +b^{2}}} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 a \sqrt {-4 c a +b^{2}}}}, a_{k +1}=-\frac {2 a a_{k} \left (a k +\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 c a +b^{2}}}-2 a +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, k +2 a \sqrt {-4 c a +b^{2}}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{k +\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 a \sqrt {-4 c a +b^{2}}}}, a_{k +1}=-\frac {2 a a_{k} \left (a k +\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 c a +b^{2}}}-2 a +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, k +2 a \sqrt {-4 c a +b^{2}}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {m =0}{\sum }}e_{m} {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{m}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}f_{m} {\left (x -\frac {-b +\sqrt {-4 c a +b^{2}}}{2 a}\right )}^{m +\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 a \sqrt {-4 c a +b^{2}}}}\right ), e_{m +1}=-\frac {2 a e_{m} \left (a m -2 a +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, m +\sqrt {-4 c a +b^{2}}\, d +2 a k -b d}, f_{m +1}=-\frac {2 a f_{m} \left (a m +\frac {2 a \sqrt {-4 c a +b^{2}}-\sqrt {-4 c a +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 c a +b^{2}}}-2 a +d \right )}{2 a \sqrt {-4 c a +b^{2}}\, m +2 a \sqrt {-4 c a +b^{2}}}\right ] \end {array} \]

2.31.27.4 ✓ Mathematica. Time used: 6.507 (sec). Leaf size: 164

ode=(a*x^2+b*x+c)*D[y[x],{x,2}]+(d*x+k)*D[y[x],x]+(d-2*a)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to (x (a x+b)+c)^{1-\frac {d}{2 a}} \exp \left (\frac {(b d-2 a k) \arctan \left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) \left (c_2 \int _1^x\exp \left (\frac {(d-4 a) \log (c+K[1] (b+a K[1]))-\frac {2 (b d-2 a k) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}}{2 a}\right )dK[1]+c_1\right ) \end{align*}

2.31.27.5 ✗ 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((-2*a + d)*y(x) + (d*x + k)*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